Automatic proof generation in an axiomatic system for $$\mathsf{CPL}$$ by means of the method of Socratic proofs

Automatic proof generation in an axiomatic system for $$\mathsf{CPL}$$ by means of the method of... Abstract The aim of this article is to describe an algorithm for the automatic generation of proofs in an axiomatic system for Classical Propositional Logic. The idea of the algorithm was taken from a book by Helena Rasiowa and Roman Sikorski [31], where the authors suggest using the method of diagrams to automatically obtain proofs in Classical Propositional Logic. However, in this article the method of diagrams developed by Rasiowa and Sikorski was replaced by a right-sided erotetic calculus developed by Andrzej Wiśniewski. Erotetic calculi have been used before in designing similar algorithms. The proofs are presented together with the estimations of their lengths, widths and sizes — measures introduced for the purposes of the article. The estimations are then used to derive the conclusion that the axiomatic system simulates polynomially the erotetic calculus. 1 Introduction Despite the intensive development of the discipline known as proof-theory, axiomatic systems, also called ‘Frege systems’ or ‘Hilbert systems’,1 are still one of the most universal proof systems, capable of capturing a very wide range of logics for which no other proof systems are known (see [19, p.26]). Although it is acknowledged that in the case of Classical Propositional Logic ($$\mathsf{CPL}$$ for short) purely automatic proof generation is obtainable, e.g. by translating natural deduction proofs into axiomatic system proofs (see e.g. [33, p.34], more references are given below), axiomatic systems are often regarded as difficult to implement or resulting in inefficient algorithms (see [3, p.18]). Probably for this reason there is little interest in the subject, contrary to proof-search by — the still unquestionable leader in this field — resolution, or by tableau and related methods.2 On the other hand, an important motivation to study axiomatic systems comes from theory of computational complexity. During the last few decades, researchers focusing on the computational complexity of proof systems for $$\mathsf{CPL}$$ have shown, int.al., that in the case of $$\mathsf{CPL}$$ axiomatic systems, d.a.g.-like (from directed acyclic graph) natural deduction systems, and sequent calculi with cut belong to the same complexity class ([9], [7], [8], [29]). Moreover, it is known that these systems for $$\mathsf{CPL}$$ are essentially more efficient than sequent systems without cut, resolution or tableau methods3 for $$\mathsf{CPL}$$ (except for [9], [7], [8], [29] see also [2], [34], [6]). After the first presentation of an axiomatic system by Frege in 1879, the discussed results make axiomatic systems for $$\mathsf{CPL}$$ — again — a highly attractive research subject. Translating natural deduction proofs into axiomatic system proofs is actually the content of the proof of Deduction Theorem. The reader will find many constructions of this kind in logic textbooks such as [33] (where the construction is described for the case of intuitionistic logic with the classical case left as an exercise), [13] or the classic textbook [20]. The book [33, p. 34] also mentions a possible translation from natural deduction proofs to axiomatic proofs via$$\lambda$$-terms of type theory. Obviously, a proof-search algorithm for natural deduction must be added to these results to achieve a proof-search algorithm for axiomatic systems. This is not a problem (see e.g. [19]), but algorithms based on a version of the method of diagrams may seem more straightforward. In their joint paper [30], and later in the monograph [31] published in 1963, the authors — Helena Rasiowa and Roman Sikorski — present the method of diagrams of formulas, which nowadays is variously called the Rasiowa-Sikorski method, Rasiowa-Sikorski diagrams, or simply the R-S system. The R-S system for $$\mathsf{CPL}$$ may be used, as the authors claim [31, p. 269], to generate a proof in an axiomatic system for $$\mathsf{CPL}$$ in a fully automatic way. It is worth noticing that the R-S systems have been developed for many logics and found various important applications, among others, in the area of computer science — see e.g. [28], [15], [21], [22], [14]. The researchers working with R-S systems have introduced the term ‘dual tableaux’ for it, to emphasize the duality of R-S systems with respect to analytic tableau systems. The former may be interpreted as deriving the Conjunctive Normal Form of a formula $$A$$, and is a ‘validity checker’, whereas the latter attempts to build a proof of $$A$$ by deriving the Disjunctive Normal Form of $$\lnot A$$ and is thus an ‘unsatisfiability checker’. The idea of exploring the relation between R-S systems and axiomatic systems has been undertaken by Ewa Orłowska and Joanna Golińska-Pilarek in [28] on the level of the First-Order Logic ($$\mathsf{FOL}$$ for short). In the book, the authors present an effective procedure ([28, pp. 22–24]) transforming an RS-proof (i.e., one conducted in the framework of the R-S system) of a formula of the language of $$\mathsf{FOL}$$ into a proof of the formula in an axiomatic system. The focus of the approach in [28] is on the first-order level, therefore the propositional level is, in a way, neglected, as it is assumed that every formula of the language of $$\mathsf{FOL}$$ which falls under a tautological schema is an axiom of the axiomatic system used in their work. Moreover, the authors introduce into their axiomatic system derived rules which are ‘induced by’ the rules of the R-S system, i.e., which map the applications of the R-S rules. On the propositional level, the effective procedure then becomes a quite straightforward imitation of the applications of R-S rules inside the axiomatic system. Thus we have not found in [28] the answer to the question of how to generate a proof in an axiomatic system for $$\mathsf{CPL}$$ with a finite number of axioms and with the rule of substitution — the book was not intended to answer the question; this article is. The right-sided erotetic calculus $$\mathbb{E}^{**}$$ is a proof system grounded in Inferential Erotetic Logic, developed by Andrzej Wiśniewski. $$\mathbb{E}^{**}$$ is the propositional part of the calculus $$\mathbf{E^{RPQ}}$$ for $$\mathsf{FOL}$$ described in [38].4 As we will show in this article, calculus $$\mathbb{E}^{**}$$ may be interpreted as an erotetic version of the R-S system. In fact, the results obtained in this article with the use of the erotetic calculus can be easily reformulated for the case of the R-S system. We emphasize this fact again at the end of this article. The reason to choose the method of Socratic proofs is that it has been used successfully in designing similar algorithms as the one presented in this article, hence the results presented here constitute a part of a wider research project. In [26] the method of Socratic proofs in its both-sided version has been used to simulate the derivation process in sequent calculi both for $$\mathsf{CPL}$$ and $$\mathsf{FOL}$$. In [25] (unpublished) this result is extended to propositional modal logics. Moreover, in [25] the left-sided version of the method of Socratic proofs is used to simulate the derivation process in an analytic tableau system for $$\mathsf{CPL}$$ and $$\mathsf{FOL}$$. There is also another variant of the method of Socratic proofs, based on the so-called reversed sequents, which has been used in [5] to simulate the method of resolution for $$\mathsf{CPL}$$ and some non-classical propositional logics.5 The summarized results show that due to the use of various sequents — or in fact, due to the potential of the structure of a sequent and the numerous interplays between its sides — it is possible to obtain many results concerning the relations between different proof methods, and we think it is important to obtain the results in a uniform manner, i.e., with the use of one proof method. Taking this all into account, as well as the fact that we have at our disposal some algorithms of proof-search by means of the method of Socratic proofs together with an implementation in Prolog,6 we have decided to use $$\mathbb{E}^{**}$$ instead of the R-S system. The formal result of this article is a procedure which transforms a Socratic proof of a sequent $$\vdash A$$ in erotetic calculus $$\mathbb{E}^{**}$$ into a proof of a formula $$A$$ in the axiomatic system $$\mathbb{A}$$ for $$\mathsf{CPL}$$. We introduce three measures to estimate the complexity of the presented proofs: length (the number of lines of a proof), width (the maximum number of occurrences of symbols in a line of a proof), and size (the product of the previous two). Apart from the procedure itself and the detailed calculations which may probably be useful in the implementation of the procedure, the calculations presented in this article yield that the axiomatic system $$\mathbb{A}$$polynomially simulates the erotetic calculus $$\mathbb{E}^{**}$$. Taking into account what we know about the relative complexity of various proof systems this result does not seem surprising. Nevertheless, there are no such detailed results concerning the relation between the axiomatic systems and the erotetic calculi. To the best of our knowledge, there are also no results concerning the relation between the axiomatic systems and the R-S systems, except for that presented in [28]. The article is structured as follows. In Section 2, we describe an axiomatic system for $$\mathsf{CPL}$$, namely $$\mathbb{A}$$, and we recall the original, neat presentation of the R-S system for $$\mathsf{CPL}$$. In Section 3, we describe the method of Socratic proofs, and in Section 4 — the necessary proofs obtained in the system $$\mathbb{A}$$ together with the estimations of their sizes. Section 5 contains a sketch of the whole procedure and discusses some complexity issues. We end up with some conclusions and perspectives for further research. 2 Classical propositional logic: axiomatic system and the Rasiowa-Sikorski system 2.1 Language and semantics Let $$\mathcal{L}$$ be the language of $$\mathsf{CPL}$$ with countably infinitely many propositional variables: $$p_{1}, p_{2}, p_{3}, \ldots$$, connectives: negation ($$\neg $$), implication ($$\to$$), conjunction ($$\wedge$$), disjunction ($$\vee$$), equivalence ($$\leftrightarrow$$) and brackets: $$(,)$$. By FORM$$_{\mathcal{L}}$$ we mean the set of all formulas of $$\mathcal{L}$$, defined in a usual way.7 Till the end of Section 2, instead of ‘formula of $$\mathcal{L}$$’ we will write simply ‘formula’. For simplicity, the letters $$p, q, r, s$$ will be used as propositional variables, and $$p_i$$ will be used as a metavariable. As metavariables for formulas we will use $$A, B, C, D, F$$, possibly with subscripts. Symbols $$X, X_{1}, X_{2}$$ will stand for sets of formulas. We also adopt the usual conventions concerning saving on parentheses. $$\mathcal{L}$$ is equipped with the standard semantics based on Boolean valuations. 2.2 Axiomatic system $$\mathbb{A}$$ for $$\mathsf{CPL}$$ We use the system presented in [1, pp. 42–43],8 for which we shall use the symbol $$\mathbb{A}$$.    Axioms for implication     Axioms for negation  1.  $$p \to (q \to p)$$  3.  $$(p \to q) \to (\neg q \to \neg p)$$  2.  $$( p \to (q \to r)) \to ((p \to q) \to (p \to r))$$  4.  $$\neg \neg p \to p$$        5.  $$p \to \neg \neg p$$     Axioms for conjunction     Axioms for disjunction  6.  $$p \wedge q \to p$$  9.  $$p \to p \vee q $$  7.  $$p \wedge q \to q$$  10.  $$q \to p \vee q$$  8.  $$(p \to q) \to ((p \to r) \to (p \to q \wedge r))$$  11.  $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$     Axioms for equivalence        12.  $$(p\leftrightarrow q) \to (p \to q)$$        13.  $$(p \leftrightarrow q) \to (q \to p)$$        14.  $$(p\to q)\to ((q \to p)\to (p\leftrightarrow q))$$           Axioms for implication     Axioms for negation  1.  $$p \to (q \to p)$$  3.  $$(p \to q) \to (\neg q \to \neg p)$$  2.  $$( p \to (q \to r)) \to ((p \to q) \to (p \to r))$$  4.  $$\neg \neg p \to p$$        5.  $$p \to \neg \neg p$$     Axioms for conjunction     Axioms for disjunction  6.  $$p \wedge q \to p$$  9.  $$p \to p \vee q $$  7.  $$p \wedge q \to q$$  10.  $$q \to p \vee q$$  8.  $$(p \to q) \to ((p \to r) \to (p \to q \wedge r))$$  11.  $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$     Axioms for equivalence        12.  $$(p\leftrightarrow q) \to (p \to q)$$        13.  $$(p \leftrightarrow q) \to (q \to p)$$        14.  $$(p\to q)\to ((q \to p)\to (p\leftrightarrow q))$$        The set of the 14 axioms presented above will be referred to by $$Apc$$ (this is the English-language version of the symbol $$Arz$$ used in [1]), whereas the set of all valid formulas of $$\mathsf{CPL}$$ will be represented as $$Tpc$$ (again, instead of $$Trz$$ used in [1]). Theorems of $$\mathsf{CPL}$$ are derived in $$\mathbb{A}$$ from the above axioms by the rule of modus ponens (MP) and / or the rule of substitution (Sub):   \[\begin{matrix} \frac{\begin{matrix} \begin{matrix} A\to B \\ A \\ \end{matrix} \\ \end{matrix}}{B} & {} & {} & \frac{B}{S(A,{{p}_{i}},B)}, \\ \end{matrix}\] where the operation of substitution of formula$$A$$for variable$$p_{i}$$in formula$$B$$ is defined in a usual way. Proof of formula$$A$$from set$$X$$of formulas in$$\mathbb{A}$$ is also defined in a standard way: as a finite sequence of formulas of $$\mathcal{L}$$ ending with $$A$$ and such that each formula of the sequence is an element of $$X$$ or results from some previous term(s) of the sequence by MP or Sub. If there exists a proof of $$A$$ from $$X$$ in $$\mathbb{A}$$, then we write $$A\in Cn(X)$$ ($$Cn$$ stands for the consequence operator). Obviously, the following holds: Theorem 1 (Soundness and completeness of $$\mathbb{A}$$ with respect to Boolean semantics)   $$Cn(Apc) = Tpc.$$ 2.3 The Rasiowa and Sikorski system Below we describe the elegant and simple R-S system for $$\mathsf{CPL}$$. This presentation is strictly based on [31, Chapter VII], we adjust the notation, however, to that used in this article. In this subsection, $$\Gamma$$ will be used as a metavariable for (possibly empty) sequences of formulas of the form:   \begin{equation}\label{1} A_{1},...,A_{m}. \end{equation} (1) If $$\Gamma$$ is of the form (1), $$\Gamma^\sharp$$ is a sequence of formulas: $$B_{1},...,B_{n}$$ and $$A$$, $$B$$ are formulas, then the following expressions:   \begin{eqnarray*} &\Gamma \ ' \ \Gamma^\sharp&\nonumber\\[3pt] &\Gamma \ ' \ A \ ' \ \Gamma^\sharp&\nonumber\\[3pt] &\Gamma \ ' \ A \ ' \ B \ ' \ \Gamma^\sharp& \end{eqnarray*} will denote, respectively, the following sequences:   \begin{eqnarray*} &A_{1},...,A_{m}, B_{1},...,B_{n}&\\[3pt] &A_{1},...,A_{m}, A, B_{1},...,B_{n}&\\[3pt] &A_{1},...,A_{m}, A, B, B_{1},...,B_{n}.& \end{eqnarray*} The symbol ‘$$'$$’ is therefore used for the concatenation of finite sequences of formulas, where the cases of single formulas $$A$$, $$B$$ are treated as cases of single-term sequences $$\langle A \rangle$$, $$\langle B \rangle$$. We say that a formula is indecomposable if it is either a propositional variable or the negation of a propositional variable. A sequence (1) is said to be indecomposable, if its terms are indecomposable formulas; it is said to be fundamental if it contains a formula and its negation. By a rule schema we shall understand a pair $$\langle \Gamma,\Gamma^{0} \rangle$$ or a triple $$\langle \Gamma,\Gamma^{0}, \Gamma^{1}\rangle$$ of finite, non-empty sequences of formulas. These are presented as follows:   \begin{equation}\label{2} \frac{\Gamma}{\Gamma^{0}} \end{equation} (2)  \begin{equation}\label{3} \frac{\Gamma}{\Gamma^{0}~;~\Gamma^{1}}. \end{equation} (3) The authors of the method call $$\Gamma$$ the conclusion of the rule schema, and $$\Gamma^{0},\Gamma^{1}$$ — its premises. In the case of schema (3), $$\Gamma^{0},\Gamma^{1}$$ are called the left and the right premise, respectively. Let us observe that, as in sequent calculi, this naming preserves the direction of proving, though not the direction of proof-search; the process of proving starts with the premises (axioms, or fundamental sequences) to reach the final conclusion (a thesis), whereas the process of proof-search goes in the opposite direction. The R-S system for $$\mathsf{CPL}$$ has no axioms and is composed of the following rule schemas (symbol $$\Gamma^*$$ always denotes an indecomposable sequence, which may be empty):   \begin{align*} &(D) ~~~~~~~~& \frac{\Gamma^*,(A \lor B), \Gamma^{**}}{\Gamma^*, A, B, \Gamma^{**}} & & & (\lnot D) ~~~~~~~~& \frac{\Gamma^*,\lnot(A \lor B), \Gamma^{**}}{\Gamma^*,\lnot A,\Gamma^{**}~;~\Gamma^*,\lnot B, \Gamma^{**}}\\ \\ &(\lnot C) & \frac{\Gamma^*,\lnot(A \land B), \Gamma^{**}}{\Gamma^*,\lnot A,\lnot B, \Gamma^{**}} & & & (C) & \frac{\Gamma^*,(A \land B), \Gamma^{**}}{\Gamma^*,A,\Gamma^{**}~;~\Gamma^*,B, \Gamma^{**}}\\ \\ &(I) & \frac{\Gamma^*,(A \rightarrow B), \Gamma^{**}}{\Gamma^*,\lnot A, B, \Gamma^{**}} & & & (\lnot I) & \frac{\Gamma^*,\lnot(A \rightarrow B), \Gamma^{**}}{\Gamma^*,A,\Gamma^{**}~;~\Gamma^*,\lnot B, \Gamma^{**}}\\ \\ &(\lnot N) & \frac{\Gamma^*,\lnot \lnot A, \Gamma^{**}}{\Gamma^*, A, \Gamma^{**}} \end{align*} As we can see, the rules of the forms $$(D), (\lnot C), (I), (\lnot N)$$ are of type (2), whereas the rules of the forms $$(\lnot D), (C), (\lnot I)$$ are of type (3). Observe that if a sequence $$\Gamma$$ of formulas is not indecomposable, then $$\Gamma$$ is the conclusion of one of these schemas. The assumption that $$\Gamma^*$$ denotes exclusively indecomposable sequences warrants that $$\Gamma$$ is the conclusion of exactly one of the rule schemas. As we shall see, the notion of diagram of a formula will be introduced as a certain partial mapping; the mappings will be defined on finite binary sequences. Therefore, the letters i, j will denote finite, possibly empty, sequences:   \begin{equation}\label{4} i_{1},...,i_{n} \end{equation} (4) of integers $$0$$, $$1$$. We will write j$$\le$$i, if j is an initial (proper or improper) segment of i. If i is of the form (4), then i, 0 and i, 1 will denote the sequences:   $$i_{1},...,i_{n}, 0$$ and   $$i_{1},...,i_{n}, 1$$ respectively. The empty sequence will be denoted by $$\mathbf{O}$$. By definition, $$\mathbf{O} \le$$i for each i. Now we are in a position to introduce the definition of diagram. The following is almost an exact quotation from [31, p. 266]. Definition 1 By the diagram of a formula$$A$$ we shall mean any mapping which, with certain sequences $${\boldsymbol{i}}$$, associates non-empty finite sequences $$\Gamma_{\boldsymbol{i}}$$ of formulas, and which is defined by induction as follows: (1) $$\Gamma_{\mathbf{O}}$$ is the sequence formed only of the formula $$A$$. (2) If $$\Gamma_{\boldsymbol{i}}$$ is defined but is either fundamental or indecomposable, then $$\Gamma_{\boldsymbol{i},0}$$ and $$\Gamma_{\boldsymbol{i},1}$$ are not defined. (3) If $$\Gamma_{\boldsymbol{i}}$$ is defined and is neither fundamental nor indecomposable, then $$\Gamma_{\boldsymbol{i}}$$ is the conclusion of exactly one of the schemas of rules. If $$\Gamma_{\boldsymbol{i}}$$ is the conclusion of schema $$(D), (\lnot C), (I)$$ or $$(\lnot N)$$, then $$\Gamma_{\boldsymbol{i},0}$$ is the only premise of this schema and $$\Gamma_{\boldsymbol{i},1}$$ is not defined. If $$\Gamma_{\boldsymbol{i}}$$ is the conclusion of schema $$(\lnot D), (C)$$ or $$(\lnot I)$$, then $$\Gamma_{\boldsymbol{i},0}$$ and $$\Gamma_{\boldsymbol{i},1}$$ are the left and the right premises of this schema. (4) If $$\Gamma_{\boldsymbol{i}}$$ is not defined, then $$\Gamma_{\boldsymbol{i},0}$$ and $$\Gamma_{\boldsymbol{i},1}$$ are not defined. As we have observed above, diagrams are partial mappings defined on the set of all finite binary sequences. Now we may also observe that for each formula $$A$$, the diagram of $$A$$ is uniquely determined by the formula. Moreover, the diagram of any formula is always finite, i.e. $$\Gamma_{{\boldsymbol{i}}}$$ is defined only for a finite number of binary sequences. We will say that $$\Gamma_{{\boldsymbol{i}}}$$ is an end sequence of the diagram of $$A$$ if $$\Gamma_{{\boldsymbol{i}}}$$ is fundamental or indecomposable, i.e. if $$\Gamma_{{\boldsymbol{i}},0}$$ and $$\Gamma_{{\boldsymbol{i}},1}$$ are not defined. Finally: Theorem 2 (Completeness of the R-S system) A formula $$A$$ is valid iff all end sequences in the diagram of $$A$$ are fundamental. The proof of this theorem presented in [31] is based on the fact that the diagram of a formula is uniquely determined by the rules of the R-S system. This means that for an arbitrary formula its diagram may be obtained ‘mechanically’; then one may determine validity of the formula by inspection of the end sequences of its diagram. Here are two examples of diagrams. Example 1 The diagram for formula $$((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r):$$ The formula is valid, and the end sequences of its diagram are fundamental. Example 2 The diagram for formula $$((p \rightarrow q) \land (r \rightarrow q)) \rightarrow \lnot (p \land \lnot r):$$ The formula is not valid, as witnessed by its diagram. 2.4 A task crazy enough to undertake In their summary of the method of diagrams, Rasiowa and Sikorski state that the diagram of a valid formula $$A$$ may be used to yield a construction of a proof of $$A$$ in the axiomatic system presented in their book. Their argumentation runs as follows. By Theorem 2, we know that each end sequence of the diagram of $$A$$ is fundamental. Moreover, if $$\Gamma$$ is a fundamental sequence, then the disjunction of all terms of $$\Gamma$$ is a valid formula. ‘The construction is based on the fact that for every fundamental formula $$A$$ we can effectively give a formal proof of $$A$$ from (…) [axioms] by means of modus ponens.’ [31, p. 269].9 Next, if sequence $$\Gamma$$ results from sequence $$\Gamma_{0}$$ by a rule falling under the schema (2), or results from $$\Gamma_{0}$$ and $$\Gamma_{1}$$ by a rule falling under the schema (3), then the proof of the formula denoting the disjunction of terms of $$\Gamma_{0}$$ (respectively, $$\Gamma_0$$ and $$\Gamma_{1}$$) may be transformed into the proof of disjunction of the terms of $$\Gamma$$. However, the authors state ‘We shall not describe these formal proofs in detail because the description is rather long.’ The task seemed crazy enough to undertake. 3 The method of Socratic proofs The method of Socratic proofs has been created as an explication of the — Socratic indeed — idea of solving questions by pure questioning. In the most basic version, the questions concern the validity of formulas of language of $$\mathsf{CPL}$$. Through progressive simplification of the logical structure of consecutive questions we are led to a question concerning some basic properties of the relation of entailment, like its reflexivity or ex falso quodlibet. The answer to such a question may be considered ‘obvious’. Since questions come into play, the method must be grounded in the logic of questions which analyses inferential relations between questions. This is the basic task of Inferential Erotetic Logic. 3.1 Inferential Erotetic Logic Inferential Erotetic10 Logic (hereafter $$\mathsf{IEL}$$) focuses on the so-called erotetic inferences — inferences in which questions play the role of conclusions and/or premises — and proposes criteria of validity of such inferences. $$\mathsf{IEL}$$ was invented by Andrzej Wiśniewski in the late 1980s (see [35]) and then further developed in various directions (see [36], [37], [40], [39], [27] — to mention the most important publications). It originated as an alternative with respect to the widespread (in that time) understanding of the logic of questions as focusing on analyses of the answerhood relation, and also in opposition to the interrogative model of inquiry developed by Jaakko Hintikka (see [18]). The method of Socratic proofs illustrates the importance of questions in proof theory. It has already been described for Classical Logic and various non-classical logics (see [37], [41], [40], [23], [24], [5], [4]). Below we present the method of Socratic proofs for $$\mathsf{CPL}$$ in the right-sided version (see [38], where $$\mathbf{E^{RPQ}}$$ for $$\mathsf{FOL}$$ has been introduced; as we have pointed out in the introduction, calculus $$\mathbb{E}^{**}$$ is the propositional part of calculus $$\mathbf{E^{RPQ}}$$ for $$\mathsf{FOL}$$). The choice of this version has been motivated by the aims of this article, i.e., by the strict correspondence between the right-sided erotetic calculus $$\mathbb{E^{**}}$$ and the R-S system. We describe formal language $$\mathcal{L_{\vdash}^{?}}$$ equipped with questions, and then we present erotetic calculus $$\mathbb{E^{**}}$$. Since this article concentrates on proof-theoretical issues we do not focus on the erotetic interpretation of the calculus. The Reader will find it in the cited works by Andrzej Wiśniewski. 3.2 Language $$\mathcal{L}_{\vdash}^{?}$$ Right-sided erotetic calculus $$\mathbb{E^{**}}$$ is worded in language $$\mathcal{L}_{\vdash}^{?}$$ and its declarative part is based on right-sided sequents. Formulas of language $$\mathcal{L_{\vdash}^{?}}$$ are built of the formulas of $$\mathcal{L}$$ and the following additional symbols: $$\vdash, ?, \&, \underline{ng}$$ and the comma. Finite, possibly empty sequences of formulas of language $$\mathcal{L}$$ will be represented by capital letters $$S, T, U, W$$. Each formula of language $$\mathcal{L}_{\vdash}^{?}$$ is either a declarative formula, or a question, and each declarative formula of language $$\mathcal{L}_{\vdash}^{?}$$ is either atomic or complex. Atomic declarative formulas of language$$\mathcal{L_{\vdash}^{?}}$$ are right-sided sequents, i.e., expressions of the form:   \begin{equation}\label{sequent} \vdash S, \end{equation} (5) where $$S$$ is a finite and non-empty sequence of formulas of language $$\mathcal{L}$$. If $$S = \langle A_1, \ldots, A_n\rangle$$, then we will represent (5) as   \begin{equation}\label{sequentA} \vdash A_{1},...,A_{n} \end{equation} (6) omitting the angle brackets. In the sequel, by terms of sequent (6) we shall mean formulas $$A_1, \ldots, A_n$$, i.e. the terms of sequence $$S$$ in $$\vdash S$$. We will say that sequent (5) is valid iff there is no valuation $$V$$ such that each term of $$S$$ is false under $$V$$. Thus sequent (6) is valid iff formula $$A_{1} \lor (A_{2}\lor \ldots(A_{n-1} \lor A_{n})\ldots)$$ is valid. As we can see, semantically the right-sided sequents may be interpreted as disjunctions of formulas. Questions of language$$\mathcal{L_{\vdash}^{?}}$$ are expressions of the form $$?(\Phi)$$, where $$\Phi$$ is a finite and non-empty sequence of atomic declarative formulas of language $$\mathcal{L_{\vdash}^{?}}$$ (sequents). Again, if $$\Phi = \langle \vdash~S_1, \ldots, \vdash~S_n \rangle$$, then we represent question $$?(\Phi)$$ as:   \begin{equation}\label{question} ?(\vdash S_1, \ldots, \vdash S_n) \end{equation} (7) and we say that question $$?(\Phi)$$ is based on sequents$$\vdash S_1, \ldots, \vdash S_n$$. We shall use $$Q, Q_1, Q^*$$ as metavariables for questions. Complex declarative formulas of language$$\mathcal{L_{\vdash}^{?}}$$ are built from atomic declarative formulas by connectives ‘$$\&$$’ and/or ‘$$\underline{ng}$$’. The complex formulas are used mainly to express answers to questions of $$\mathcal{L_{\vdash}^{?}}$$. All questions of $$\mathcal{L_{\vdash}^{?}}$$ are polar, and thus an affirmative answer to (7) states, roughly speaking, that all terms of (7) are valid sequents. The negative answer, in turn, states that it is not the case. As we have noted above, in this article we will not consider the strictly ‘erotetic’ issues of the method of Socratic proofs, as all this may be found in the works by Andrzej Wiśniewski. We will use the semicolon ‘;’ for the concatenation of sequences of sequents. We adopt conventions analogous to that pertaining to the concatenation of sequences of formulas of $$\mathcal{L}$$ (see page 5). 3.3 Erotetic calculus $$\mathbb{E^{**}}$$ Calculus $$\mathbb{E^{**}}$$ is a set of erotetic rules, i.e., rules transforming questions. There are no axioms. Table 1 presents the rules’ schemas (in the sequel we use the terms ‘schema of a rule’ and ‘rule’ interchangeably). Table 1 the rules of $$\mathbb{E}^{**}$$.   $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A\to B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A\to B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \land B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \land B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \lor B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \lor B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A\to B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A\to B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \land B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \land B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \lor B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \lor B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:\neg\neg A\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\Psi)} {\textbf{R}_{\neg\neg}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:\neg\neg A\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\Psi)} {\textbf{R}_{\neg\neg}}$$   The question in the numerator of an erotetic rule is called the question-premise, and the one in the denominator — the question-conclusion. Thus, contrary to the rules of the R-S system, the direction of proving as defined by the erotetic rules coincides with the direction of proof-search. The sequent which is distinguished in the question-premise of a rule will be called its sequent-premise. The sequent(s) distinguished in the question-conclusion of a rule will be called its sequent(s)-conclusion(s). Finite sequences of questions transformed by the above rules are called Socratic transformations. Formally: Definition 2 (Socratic transformation) Let $$Q$$ be a question of language $$\mathcal{L_{\vdash}^{?}}$$. A finite sequence of questions $$\langle Q_{1}, Q_{2},...,Q_{n} \rangle$$ is a Socratic transformation of question$$Q$$ via the rules of $$\mathbb{E^{**}}$$ iff $$Q = Q_{1}$$ and each question of the sequence, except for the first one, results from the previous question by an application of one of the rules of $$\mathbb{E^{**}}$$. As we have written above, the method of Socratic proofs is an explication of the idea of answering questions by pure questioning. In accordance with this idea we will regard a Socratic transformation as successful, if its last question is, in a way, a trivial one. The notion of Socratic proof embodies this idea in the following way. Definition 3 (Socratic proof) Let $$\ \vdash S$$ be a sequent of language $$\mathcal{L_{\vdash}^{?}}$$. A Socratic proof of sequent$$\vdash S$$in$$\mathbb{E^{**}}$$ is a Socratic transformation of question $$?(\vdash S)$$ such that each constituent of its last question is of one of the following forms: $$\vdash T\: ' \: B\: ' \: U \: ' \: \neg B\: ' \: W,$$ $$\vdash T\: ' \: \neg B\: ' \: U \: ' \: B\: ' \: W.$$ Sequents of one of the two forms indicated in the definition are obviously valid sequents (as far as the law of excluded middle is an obvious one); in this sense the answer to the last question of a Socratic proof is obviously affirmative. Further, we shall say that sequents of one of the two forms are base sequents. The following theorem is true: Theorem 3 There is a Socratic proof of sequent $$\vdash A$$ of language $$\mathcal{L_{\vdash}^{?}}$$ in $$\mathbb{E^{**}}$$ iff formula $$A$$ of language $$\mathcal{L}$$ is valid. The Reader will find the relevant proofs in [38]. Let us now present two examples of Socratic transformations. The first one is a Socratic proof of the corresponding sequent, the second one is not. Example 3 Socratic transformation of question $$?~(~\vdash~ ((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r) ~):$$  $$ \frac {?~(~\vdash~ ((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r)~)} {?~(~\vdash~ \lnot((p \rightarrow q) \land (q \rightarrow r)), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ \lnot(p \rightarrow q), \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot (p \land \lnot r) ~;~ \vdash~ p, \lnot r, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, q, \lnot (p \land \lnot r) ~;~ \vdash ~ \lnot q, \lnot r, \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, q, \lnot (p \land \lnot r) ~;~ \vdash ~ \lnot q, \lnot r, \lnot p, \lnot\lnot r) ~)} . $$ The Socratic transformation is a Socratic proof of sequent $$\vdash~ ((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r)$$. Let us emphasize that the last question is based on sequents containing exactly the same sequences of formulas which occur in the leaves of the R-S diagram for formula $$((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r)$$ (compare Example 1). This example illustrates a close connection between calculus $$\mathbb{E^{**}}$$ and the method of Rasiowa and Sikorski, which justifies us in saying that calculus $$\mathbb{E^{**}}$$ is an erotetic version of the R-S system. The connection, in turn, derives from the fact that the two methods are based on deriving Conjunctive Normal Form of a formula. Example 4 Socratic transformation of question $$?~(~\vdash~ ((p \rightarrow q) \land (r \rightarrow q)) \rightarrow \lnot (p \land \lnot r) ~):$$   $$ \frac{?~(~\vdash~ \lnot((p \rightarrow q) \land (r \rightarrow q)), \lnot (p \land \lnot r) ~)} {?~(~\vdash~ ((p \rightarrow q) \land (r \rightarrow q)) \rightarrow \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ \lnot(p \rightarrow q), \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~) }\\ \frac {} {?~(~\vdash~ p, r, \lnot (p \land \lnot r) ~;~ \vdash~ p, \lnot q, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot (p \land \lnot r) ~;~ \vdash ~ \lnot q, \lnot q, \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, \lnot \lnot r ~;~ \vdash ~ \lnot q, \lnot q, \lnot(p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, r ~;~ \vdash ~ \lnot q, \lnot q, \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, r ~;~ \vdash ~ \lnot q, \lnot q, \lnot p, \lnot\lnot r ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, r ~;~ \vdash ~ \lnot q, \lnot q, \lnot p, r ~)} .$$ Again, the transformation should be compared with Example 2. 4 From Socratic proofs in $$\mathbb{E^{**}}$$ to proofs in $$\mathbb{A}$$ 4.1 Base sequents This subsection is devoted to proof of the following theorem: Theorem 4 If $$\vdash A_{1},...,A_{n}$$ is a base sequent, then formula: $$(*)$$$$A_{1}\vee (A_{2}\vee ...\vee (A_{n-1} \vee A_{n})...)$$ is a thesis of $$\mathbb{A}$$. Further, we will say that formula $$(*)$$ corresponds to sequent $$\vdash A_{1},...,A_{n}$$. So the above theorem states that a formula which corresponds to a base sequent is provable in $$\mathbb{A}$$. Before we proceed to the proof of Theorem 4, however, we will prove several laws of $$\mathsf{CPL}$$ for further use. In the first account of this work, we wanted to obtain ‘clear’ proofs in $$\mathbb{A}$$ without making use of the derived rules except for the rule of simultaneous substitution — and that is how the work is done in [16]. In this article, however, shortcuts are necessary, so for some of the laws of $$\mathsf{CPL}$$ proved below we will introduce a derived rule based on the law; but then we will also estimate the cost of applying such a rule in a proof. The application of the standard rule of substitution will be indicated by: [no. of line Sub: $$p_i/B_i$$], whereas the application of the rule of simultaneous substitution will be indicated by: [no. of line SSub: $$p_{i_1}/B_1, \ldots, p_{i_n}/B_n$$]. For each proof $$\mathbf{p}$$ in $$\mathbb{A}$$ we calculate (or estimate the upper bound for) the length of proof$$\mathbf{p}$$, symbolically $$l_\mathbf{p}$$, understood as the number of lines of $$\mathbf{p}$$. In the case of the application of a derived rule we want to calculate (or at least estimate) the number of lines that the proof would have, were the applications of the derived rules replaced with applications of the primary rules. As we know, when the application of the rule of simultaneous substitution is replaced with the applications of the standard substitution rule, for some variables an intermediate variable must be introduced to derive the result correctly. So the cost of the application of the simultaneous substitution rule is, in the worst case, twice as large as the cost of the subsequent applications of the standard substitution rule. More specifically, suppose that the rule of simultaneous substitution of formula $$B_1$$ for $$p_{i_1}$$, …, $$B_n$$ for $$p_{i_n}$$ is used to formula $$A$$ (where $$n \geq 2$$). Let $$p_{k_1}, \ldots, p_{k_n}$$ stand for $$n$$ pairwise distinct variables which do not occur in $$A$$ nor in any of $$B_1, \ldots, B_n$$. First, we apply the rule of substitution $$n$$ times:   $$C = S(p_{k_n}, p_{i_n}, S(\ \ldots \ S(p_{k_1}, p_{i_1}, A) \ \ldots ))$$ and then another $$n$$ times starting with formula $$C$$:   $$D = S(B_n, {p_{k_n}}, S(\ \ldots \ S(B_1, p_{k_1}, C) \ \ldots \ ))$$ which gives us $$2\cdot n$$ applications of the rule of substitution — i.e. $$2\cdot n$$ lines in the proof — to get from $$A$$ to $$D$$. Another useful measure will be that of width of proof in$$\mathbb{A}$$. Let $$|A|$$ stand for the number of occurrences of signs in formula $$A$$ without parentheses. $$|A|$$ is called the length of$$A$$. Then the width of proof$$\mathbf{p}$$in$$\mathbb{A}$$, symbolically $$w_\mathbf{p}$$, is understood as the maximum:   \begin{align*} w_\mathbf{p} = \max \{|A|: A~ \text{is a term of}~ \mathbf{p} \}. \end{align*} Given the two measures we may give an upper bound for the total number of signs occurring in a proof. By size of proof$$\mathbf{p}$$in$$\mathbb{A}$$, symbolically $$s_\mathbf{p}$$, we shall mean the product $$l_\mathbf{p}\cdot w_\mathbf{p}$$. This kind of measure corresponds well to time-complexity (see [29, p. 551]). In the proofs presented below we shall indicate, as the first piece of information, the number of lines ‘abbreviated’ whenever a derived rule is used. So the second line of the proof below starts with ‘(4)’, as this application of simultaneous substitution abbreviates 4 applications of Sub. In bold — as in the fifth line — we give the length of formula which is the longest formula in the proof (if there are more formulas of the same length, as it happens in this proof, we indicate only the first one). This number is the width of the proof. On the right side, in square brackets, we leave the usual didascalia. First, the proof of the hypothetical syllogism in $$\mathbb{A}$$: The law of hypothetical syllogism (1) $$p \to (q \to p)$$$$\hspace{0,2 cm}$$[Ax $$1.$$] (2) (4) $$((p\to (q\to r))\to ((p\to q) \to (p\to r)))\to ((q\to r) \to ((p\to (q\to r))\to ((p\to q) \to$$ $$\quad (p\to r))))$$$$\hspace{0,2 cm}$$ [$$1$$ SSub: $$p/(p\to (q\to r))\to ((p\to q) \to (p\to r)),q/ q\to r$$] (3) $$( p \to (q \to r)) \to ((p \to q) \to( p \to r))$$$$\hspace{0,2 cm}$$[Ax $$2.$$] (4) $$(q\to r) \to ((p\to (q\to r))\to ((p\to q) \to (p\to r)))$$$$\hspace{0,2 cm}$$ [$$2,3$$ MP] (5) (6) 39$$( (q\to r) \to ((p\to (q\to r)) \to ((p\to q) \to (p\to r)))) \to (((q\to r) \to (p\to (q\to r))) \to$$ $$\quad((q\to r) \to ((p\to q) \to (p\to r))))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q\to r, q/p\to (q\to r), r/(p\to q) \to (p\to r)$$] (6) $$((q\to r) \to (p\to (q\to r))) \to ((q\to r) \to ((p\to q) \to (p\to r)))$$$$\hspace{0,2 cm}$$[$$5,4$$ MP] (7) (4) $$(q\to r)\to (p \to (q\to r))$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/ q\to r, q/p$$] (8) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[$$6,7$$ MP] (9) (6) $$(\neg \neg p \to p) \to ((\neg q\to \neg \neg p) \to (\neg q \to p))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$p/\neg q, q/\neg\neg p, r/p$$] (10) $$\neg \neg p \to p$$$$\hspace{0,2 cm}$$[Ax $$4.$$] (11) $$(\neg q\to \neg \neg p) \to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$9,10$$ MP] (12) (6) $$((q\to r) \to ((p \to q) \to (p\to r))) \to (((q\to r) \to (p \to q)) \to ((q\to r) \to (p\to r)))$$ $$\quad$$ [$$3$$ SSub: $$p/q\to r, q/ p \to q, r/ p\to r$$] (13) $$((q\to r) \to (p \to q)) \to ((q\to r) \to (p\to r ))$$$$\hspace{0,2 cm}$$[$$12,8$$ MP] (14) (6) $$(((q\to r) \to (p\to q))\to ((q\to r) \to (p \to r))) \to (((p\to q)\to ((q\to r) \to (p\to q)))\to$$ $$\quad((p\to q)\to ((q\to r) \to (p\to r))))$$$$\hspace{0,2 cm}$$ [$$8$$ SSub: $$q/(q\to r) \to (p\to q) , r/(q\to r) \to (p \to r),$$ $$\quad p/p\to q$$] (15) $$((p\to q)\to ((q\to r) \to (p\to q))) \to ((p\to q)\to ((q\to r) \to (p\to r)))$$$$\hspace{0,2 cm}$$[$$14,13$$ MP] (16) (4) $$(p\to q) \to ((q \to r) \to (p\to q))$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/ p\to q, q/ q \to r$$] (17) $$(p\to q)\to ((q\to r) \to (p\to r))$$$$\hspace{0,2 cm}$$[$$15,16$$ MP] We will use $$l_{hyp~syl}$$ for the length of this proof and $$w_{hyp~syl}$$ for its width. As we can see, $$l_{hyp~syl}=46$$ and $$w_{hyp~syl}=39$$. Moreover, in the $$8^{th}$$ line there is a thesis which will be used later, so let us note that the length of its proof equals $$19$$ and the width of its proof equals $$39$$. The use of the derived rule based on the hypothetical syllogism will be indicated by [$$a,~b$$ HS], where line no. $$a$$ contains the result of substitution to formula $$(p \to q)$$, line no. $$b$$ – to $$(q \to r)$$, and the result is the suitable substitution to $$(p\to r)$$. To sum up, the use of HS yields 6 applications of Sub and two applications of MP (i.e. 8 lines). We must remember that the first use of HS in a proof also yields the above 46 lines and a formula of length 39. Two laws of associativity of disjunction The proof of $$(p\lor q) \lor r \to p\lor(q\lor r):$$ (1) $$p \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$9.$$] (2) (4) $$q \rightarrow q \lor r$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/q, q/r$$] (3) $$q \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$10.$$] (4) $$q \lor r \rightarrow p \lor (q \lor r)$$$$\hspace{0,2 cm}$$[$$3$$ Sub: $$q/q\lor r$$] (5) (46+8) 39$$q \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[$$2, 4$$ HS] (6) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax $$11.$$] (7) $$(p \to p\lor (q\lor r)) \to ((q \to p\lor (q\lor r)) \to (p \vee q \to p\lor (q\lor r)))$$$$\hspace{0,2 cm}$$[$$6$$ Sub: $$r/p\lor (q\lor r)$$] (8) $$p \rightarrow p \lor (q\lor r)$$$$\hspace{0,2 cm}$$[$$1$$ Sub: $$q/q\lor r$$] (9) $$(q \to p\lor (q\lor r)) \to (p \vee q \to p\lor (q\lor r))$$$$\hspace{0,2 cm}$$[$$7, 8$$ MP] (10) $$p \vee q \to p\lor (q\lor r)$$$$\hspace{0,2 cm}$$[$$9, 5$$ MP] (11) (4) $$r \rightarrow q \lor r$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q, q/r$$] (12) (8) $$r\to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[$$11, 4$$ HS] (13) (6) $$(p\lor q \to p\lor(q\lor r)) \to ((r \to p\lor(q\lor r)) \to ((p\lor q) \lor r \to p\lor(q\lor r)))$$ $$\quad$$ [$$6$$ SSub: $$p/p\lor q, q/r , r/p\lor(q\lor r)$$] (14) $$(r \to p\lor(q\lor r)) \to ((p\lor q) \lor r \to p\lor(q\lor r))$$$$\hspace{0,2 cm}$$[$$13, 10$$ MP] (15) $$(p\lor q) \lor r \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[$$14, 12$$ MP] Again, we use the self-explanatory notation and write $$l_{as~dis~1}=86$$, $$w_{as~dis~1} = 39$$. The proof of $$p\lor (q \lor r) \to (p\lor q)\lor r$$: (1) $$p \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$9.$$] (2) (4) $$p\lor q \rightarrow (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/p\lor q, q/r$$] (3) (46+8) 39$$p\to (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$1, 2$$ HS] (4) $$q \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$10.$$] (5) (8) $$q\to (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$4,2$$ HS] (6) (4) $$r \rightarrow (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$4$$ SSub: $$p/p\lor q, q/r $$] (7) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax $$11.$$] (8) (6) $$(q \to (p\lor q) \lor r) \to ((r \to (p\lor q) \lor r) \to (q \lor r \to (p\lor q) \lor r))$$ $$\quad$$ [$$7$$ SSub: $$p/q, q/r, r/(p\lor q) \lor r$$] (9) $$(r \to (p\lor q) \lor r) \to (q \lor r \to (p\lor q) \lor r)$$$$\hspace{0,2 cm}$$[$$8, 5$$ MP] (10) $$q \lor r \to (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$9, 6$$ MP] (11) (4) $$(p \to (p\lor q) \lor r) \to ((q\lor r \to (p\lor q) \lor r) \to (p \vee (q\lor r) \to (p\lor q) \lor r))$$ $$\quad$$ [$$7$$ SSub: $$q/q\lor r, r/(p\lor q) \lor r$$] (12) $$(q\lor r \to (p\lor q) \lor r) \to (p \vee (q\lor r) \to (p\lor q) \lor r)$$$$\hspace{0,2 cm}$$[$$11, 3$$ MP] (13) $$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[$$12, 10$$ MP] We have: $$l_{as~dis~2}=87$$ and $$w_{as~dis~2}=39$$. Further, we will also need the proofs of formulas: $$p\vee \neg p$$, $$\neg p\vee p$$, $$p\vee (q \vee \neg p)$$, $$\neg p\vee (q \vee p)$$. Below we present a proof of the first formula (the law of excluded middle). The proof is long, but it contains some important intermediate steps, like reaching the laws of commutation and importation. Further we will use the derived rules based on these laws. The laws of commutation and importation, the Duns Scotus law, the law of identity, the law of excluded middle (1) $$p \to (q \to p)$$$$\hspace{0,2 cm}$$[Ax $$1.$$] (2) $$(p \to (q \to r)) \to ((p \to q) \to( p \to r))$$$$\hspace{0,2 cm}$$[Ax $$2.$$] (3) (46) 39$$(p\to q)\to ((q\to r) \to (p\to r))$$$$\hspace{0,2 cm}$$[hypothetical syllogism] (4) (6) $$(q\to (p\to q))\to (((p\to q)\to (p\to r)) \to (q\to (p\to r)))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q, q/p\to q, r/p\to r$$] (5) (4) $$q \to (p \to q)$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/q, q/p$$] (6) $$((p\to q)\to (p\to r)) \to (q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$4,5$$ MP] (7) (8) $$(p\to (q\to r))\to (q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$2,6$$ HS] [the law of commutation] The seven lines above constitute a proof of the law of commutation with $$l_{com}=67$$ and $$w_{com}=39$$. Further we will use a derived rule based on the law. The application of the derived rule will be indicated by $$[a$$ Com$$]$$, where line no. $$a$$ contains the result of substitution to $$p\to(q\to r)$$, and the result of the application of Com is the suitable substitution to $$q\to(p\to r)$$. The application of Com yields $$67$$ lines of the proof, one application of the rule of simultaneous substitution (6 lines) and one application of MP, i.e., $$74$$ lines. We continue the proof of the law of excluded middle. (8) (19) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[thesis, see the proof of the hypothetical syllogism, line 8] (9) (6) $$(\neg \neg p \to p) \to ((\neg q\to \neg \neg p) \to (\neg q \to p))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$p/\neg q, q/\neg \neg p, r/p $$] (10) $$\neg \neg p \to p$$$$\hspace{0,2 cm}$$[Ax $$4.$$] (11) $$(\neg q\to \neg \neg p) \to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$9,10$$ MP] (12) $$(p \to q)\to (\neg q \to \neg p)$$$$\hspace{0,2 cm}$$ [Ax $$3.$$] (13) $$(\neg p \to q)\to (\neg q \to \neg \neg p)$$$$\hspace{0,2 cm}$$ [$$12$$ Sub: $$p/ \neg p$$] (14) (8) $$(\neg p \to q)\to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$13,11$$ HS] (15) $$(p \to (p \vee q))\to (\neg (p \vee q) \to \neg p)$$$$\hspace{0,2 cm}$$ [$$12$$ Sub: $$q/ p \vee q$$] (16) $$p \to p \vee q$$$$\hspace{0,2 cm}$$[Ax $$9.$$] (17) $$\neg (p \vee q) \to \neg p$$$$\hspace{0,2 cm}$$[$$15,16$$ MP] (18) (4) $$(q \to p\vee q)\to (\neg (p\vee q) \to \neg q)$$$$\hspace{0,2 cm}$$ [$$12$$ SSub: $$p/q, q/p \lor q$$] (19) $$q\to p \vee q $$$$\hspace{0,2 cm}$$[Ax $$10.$$] (20) $$\neg (p\vee q) \to \neg q$$$$\hspace{0,2 cm}$$[$$18,19$$ MP] (21) $$(p \to q) \to ((p \to r) \to (p \to q \wedge r))$$$$\hspace{0,2 cm}$$ [Ax $$8.$$] (22) (6) $$(\neg (p\vee q) \to \neg p) \to ((\neg (p\vee q) \to \neg q) \to (\neg (p\vee q) \to \neg p \wedge \neg q))$$ $$\quad$$ [$$21$$ SSub: $$p/\neg (p\vee q), q/\neg p, r/\neg q$$] (23) $$(\neg (p\vee q) \to \neg q) \to (\neg (p\vee q) \to \neg p \wedge \neg q)$$$$\hspace{0,2 cm}$$[$$22,17$$ MP] (24) $$\neg (p\vee q) \to \neg p \wedge \neg q$$$$\hspace{0,2 cm}$$[$$23,20$$ MP] (25) (4) $$(\neg (p\vee q) \to \neg p \wedge \neg q)\to (\neg (\neg p \wedge \neg q) \to p\vee q)$$$$\hspace{0,2 cm}$$[$$14$$ SSub: $$p/p\vee q, q/\neg p \wedge \neg q$$] (26) $$\neg (\neg p \wedge \neg q) \to p\vee q$$$$\hspace{0,2 cm}$$[$$25,24$$ MP] (27) $$(p\wedge q\to q)\to ((q\to r) \to (p\wedge q\to r))$$$$\hspace{0,2 cm}$$[$$3$$ Sub: $$p/p\wedge q$$] (28) $$p \wedge q \to q$$$$\hspace{0,2 cm}$$[Ax $$7.$$] (29) $$(q\to r) \to (p\wedge q\to r)$$$$\hspace{0,2 cm}$$[$$27,28$$ MP] (30) (4) $$((q\to r)\to (p\wedge q \to r)) \to ((p\to (q\to r)) \to (p\to (p\wedge q \to r)))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$q/q\to r$$, $$r/p\wedge q \to r$$] (31) $$(p\to (q\to r)) \to (p\to (p\wedge q \to r))$$$$\hspace{0,2 cm}$$[$$30,29$$ MP] (32) $$(p\to (p\wedge q\to r))\to (p\wedge q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$7$$ Sub: $$q/p\wedge q$$] (33) (8) $$(p\to (q\to r))\to (p\wedge q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$31,32$$ HS] (34) (4) $$(p\wedge q \to (p \to r)) \to ((p\wedge q \to p) \to (p\wedge q \to r))$$$$\hspace{0,2 cm}$$[$$2$$ SSub: $$p/p\wedge q, q/p$$] (35) 51$$((p\to (q\to r))\to (p\wedge q\to (p\to r)))\to (((p\wedge q\to (p\to r))\to ((p\wedge q\to p)\to(p\wedge q\to r))) \to ((p\to (q\to r))\to ((p\wedge q\to p)\to (p\wedge q\to r))))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/p\to (q\to r), q/p\wedge q\to (p\to r), r/(p\wedge q\to p)\to (p\wedge q\to r)$$] This formula is ‘hidden’ in the following application of HS: (36) (8) $$(p\to (q\to r))\to ((p\wedge q\to p)\to(p\wedge q\to r))$$$$\hspace{0,2 cm}$$[$$33,34$$ HS] (37) (7) $$(p\wedge q\to p)\to ((p\to (q\to r))\to (p\wedge q\to r))$$$$\hspace{0,2 cm}$$[$$36$$ Com] (38) $$p\wedge q\to p$$$$\hspace{0,2 cm}$$[Ax $$6.$$] (39) $$(p\to (q\to r))\to (p\wedge q\to r)$$$$\hspace{0,2 cm}$$[$$37,38$$ MP] [the law of importation] At this point we arrive at the proof of the law of importation with $$l_{imp}=165$$ and $$w_{imp}=51$$. Similarly as in the case of the law of commutation, we define a derived rule, Imp, based on the law of importation. Its application yields $$7$$ lines of a proof (plus 165 for the law of importation, if this is absent in the proof). The formula in line no. 37 results from the previous one by the derived rule Com. We calculate only 7 lines, since the law of commutation is already a part of this proof. (40) $$p \to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$1$$ Sub: $$q/\neg q$$] (41) (4) $$(\neg q \to p)\to (\neg p \to q)$$$$\hspace{0,2 cm}$$[$$14$$ SSub: $$p/q, q/p$$] (42) (8) $$p\to (\neg p\to q)$$$$\hspace{0,2 cm}$$[$$40,41$$ HS] [the Duns Scotus law] (43) (7) $$p\wedge \neg p\to q$$$$\hspace{0,2 cm}$$[$$42$$ Imp] (44) (6) $$(\neg \neg p\to p)\to ((p\to \neg q) \to (\neg \neg p\to \neg q))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/\neg \neg p, q/p, r/\neg q$$] (45) $$(p\to \neg q) \to (\neg \neg p\to \neg q)$$$$\hspace{0,2 cm}$$[$$44,10$$ MP] (46) (6) $$(q\to \neg \neg q)\to ((\neg \neg q\to \neg \neg p) \to (q\to \neg \neg p))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q, q/\neg \neg q, r/\neg \neg p$$] (47) $$p \to \neg \neg p$$$$\hspace{0,2 cm}$$[Ax $$5.$$] (48) $$q \to \neg \neg q$$$$\hspace{0,2 cm}$$[$$47$$ Sub: $$p/q$$] (49) $$(\neg \neg q\to \neg \neg p) \to (q\to \neg \neg p)$$$$\hspace{0,2 cm}$$[$$46,48$$ MP] (50) (6) $$(r\to s) \to ((q\to r) \to (q\to s))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$p/q, q/r, r/s$$] (51) (4) $$((q\to r)\to (q\to s)) \to ((p\to (q\to r)) \to (p\to (q\to s)))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$q/q\to r, r/q\to s$$] (52) (8) $$(r\to s)\to ((p\to (q\to r))\to (p\to (q\to s)))$$$$\hspace{0,2 cm}$$[$$50,51$$ HS] (53) (6) $$(\neg \neg p\to p)\to (((\neg \neg q\to \neg \neg p)\to (q\to \neg \neg p))\to ((\neg \neg q\to \neg \neg p)\to (q\to p)))$$ $$\hspace{0,2 cm}$$[$$52$$ SSub: $$p/\neg \neg q\to \neg \neg p, r/\neg \neg p, s/p$$] (54) $$((\neg \neg q\to \neg \neg p)\to (q\to \neg \neg p))\to ((\neg \neg q\to \neg \neg p)\to (q\to p))$$$$\hspace{0,2 cm}$$[$$53,10$$ MP] (55) $$(\neg \neg q\to \neg \neg p)\to (q\to p)$$$$\hspace{0,2 cm}$$[$$54,49$$ MP] (56) (4) $$(\neg p \to \neg q)\to (\neg \neg q \to \neg \neg p)$$$$\hspace{0,2 cm}$$ [$$12$$ SSub: $$p/\neg p, q/\neg q$$] (57) (8) $$(\neg p\to \neg q)\to (q\to p)$$$$\hspace{0,2 cm}$$[$$56,55$$ HS] (58) $$(\neg \neg p\to \neg q)\to (q\to \neg p)$$$$\hspace{0,2 cm}$$[$$57$$ Sub: $$p/\neg p$$] (59) (8) $$(p\to \neg q)\to (q\to \neg p)$$$$\hspace{0,2 cm}$$[$$45,58$$ HS] (60) (4) $$(p\wedge \neg p\to \neg (p\to p))\to ((p\to p)\to \neg (p\wedge \neg p))$$$$\hspace{0,2 cm}$$[$$59$$ SSub: $$p/p\wedge \neg p, q/p\to p$$] (61) $$p\wedge \neg p\to \neg (p\to p)$$$$\hspace{0,2 cm}$$[$$43$$ Sub: $$q/\neg (p\to p)$$] (62) $$(p\to p)\to \neg (p\wedge \neg p)$$$$\hspace{0,2 cm}$$[$$60,61$$ MP] (63) $$(p \to (q \to p)) \to ((p \to q) \to( p \to p))$$$$\hspace{0,2 cm}$$[$$2$$ Sub: $$r/p$$] (64) $$(p \to q) \to( p \to p)$$$$\hspace{0,2 cm}$$[$$63,1$$ MP] (65) $$(p \to (q\to p)) \to( p \to p)$$$$\hspace{0,2 cm}$$[$$64$$ Sub: $$q/q\to p$$] (66) $$p \to p$$$$\hspace{0,2 cm}$$[$$65,1$$ MP] [the identity law] (67) $$\neg (p\wedge \neg p)$$$$\hspace{0,2 cm}$$[$$62,66$$ MP] (68) $$\neg (\neg p\wedge \neg \neg p)$$$$\hspace{0,2 cm}$$[$$67$$ Sub: $$p/\neg p$$] (69) $$\neg (\neg p \wedge \neg \neg p) \to p\vee \neg p$$$$\hspace{0,2 cm}$$[$$26$$ Sub: $$q/\neg p$$] (70) $$p\lor \neg p$$$$\hspace{0,2 cm}$$[$$69,68$$ MP] The length of the above proof, $$l_{ex~mid}$$, is estimated at $$262$$. Actually, it is easy to see that this number is overestimated, as after adding the lines of the proof of the hypothetical syllogism some of the lines would be repeated. As a matter of fact, we can never exclude the existence of a shorter proof. The width of the proof, $$w_{ex~mid}$$ in symbols, equals $$51$$. Observe also that in line no. 42 we have the Duns Scotus law with $$l_{DSl}=178$$, $$w_{DSl}=51$$, and in line no. 66 — the identity law with $$l_{id}=258$$ and $$w_{id}=51$$. The reversed law of excluded middle (1) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (2) $$(p \to q\vee p) \to ((q \to q\vee p) \to (p \vee q \to q\vee p))$$$$\hspace{0,2 cm}$$ [1 Sub: $$r/q\vee p$$] (3) $$q \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 10.] (4) (4) $$p \to q \vee p$$$$\hspace{0,2 cm}$$[3 SSub: $$q/p, p/q$$] (5) $$(q \to q\vee p) \to (p \vee q \to q\vee p)$$$$\hspace{0,2 cm}$$ [2,4 MP] (6) $$p \to p \vee q$$$$\hspace{0,2 cm}$$ [Ax 9.] (7) (4) $$q \to q \vee p$$$$\hspace{0,2 cm}$$ [6 SSub: $$p/q, q/p$$] (8) $$p \vee q \to q\vee p$$$$\hspace{0,2 cm}$$ [5,7 MP] (9) $$p \vee \neg p \to \neg p\vee p$$$$\hspace{0,2 cm}$$ [8 Sub: $$q/\neg p$$] (10) (262) 51$$p\vee \neg p$$$$\hspace{0,2 cm}$$[the law of excluded middle] (11) $$\neg p\vee p$$$$\hspace{0,2 cm}$$ [9,10 MP] The length of the proof, $$l_{rev~ex~mid}$$, is $$278$$. The width, $$w_{rev~ex~mid}$$, equals 51. Formula: $$p\vee (q\vee \neg p)$$ (1) $$p \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 9.] (2) $$p \to p \vee (q \vee \neg p)$$$$\hspace{0,2 cm}$$[1 Sub: $$q/q\vee \neg p$$] (3) $$q \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 10.] (4) (4) $$\neg p \to (p\vee q) \vee \neg p$$$$\hspace{0,2 cm}$$[3 SSub: $$p/p\vee q, q/\neg p$$] (5) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (6) (4) $$(p \to p\vee (q \vee \neg p)) \to ((\neg p \to p\vee (q \vee \neg p)) \to (p \vee \neg p \to p\vee (q \vee \neg p)))$$ $$\quad$$ [5 SSub: $$q/\neg p, r/p\vee (q \vee \neg p)$$] (7) $$(\neg p \to p\vee (q \vee \neg p)) \to (p \vee \neg p \to p\vee (q \vee \neg p))$$$$\hspace{0,2 cm}$$[6,2 MP] (8) (262) 51$$p \vee \neg p$$$$\hspace{0,2 cm}$$ [the law of excluded middle] (9) (86) $$(p\lor q) \lor r \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[associativity of disjunction 1] (10) $$(p\vee q) \vee \neg p \to p \lor(q \lor \lnot p)$$$$\hspace{0,2 cm}$$[9 Sub: $$r/\lnot p$$] (11) (8) $$\lnot p\to p\lor (q\lor \lnot p)$$$$\hspace{0,2 cm}$$[4,10 HS] (12) $$p \vee \neg p \to p\vee (q \vee \neg p)$$$$\hspace{0,2 cm}$$[7,11 MP] (13) $$p\vee (q \vee \neg p)$$$$\hspace{0,2 cm}$$[12,8 MP] Lines 8–11 need $$262+86+1+8=357$$ lines of proof (the law of the hypothetical syllogism is used in the proof of the law of excluded middle, therefore we do not add the value 46 in line no. 11). Thus the length of the proof, $$l_{p\lor(q\lor\lnot p)}$$ in symbols, is estimated at $$372$$ lines. The width, $$w_{p\lor(q\lor\lnot p)}$$, is 51. Formula: $$\neg p \lor (q\lor p)$$ (1) $$p \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 9.] (2) (4) $$\neg p \to \neg p \vee (q \vee p)$$$$\hspace{0,2 cm}$$[1 SSub: $$p/\neg p, q/q\vee p$$] (3) $$q \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 10.] (4) (4) $$p \to (\neg p\vee q) \vee p$$$$\hspace{0,2 cm}$$[3 SSub: $$p/\neg p\vee q, q/p$$] (5) (262) 51$$p \vee \neg p$$$$\hspace{0,2 cm}$$ [the law of excluded middle] (6) (86) $$(p\lor q) \lor r \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[associativity of disjunction 1] (7) (4) $$(\lnot p\lor q) \lor p \to \lnot p\lor(q\lor p)$$$$\hspace{0,2 cm}$$[6 SSub: $$p/\lnot p,r/p$$] (8) (8) $$p\to \lnot p\lor(q\lor p)$$$$\hspace{0,2 cm}$$[4,7 HS] (9) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (10) (4) $$(p \to \neg p\vee (q \vee p)) \to ((\neg p \to \neg p\vee (q \vee p)) \to (p \vee \neg p \to \neg p\vee (q \vee p)))$$ $$\quad$$ [9 Sub: $$q/\neg p, r/\neg p\vee (q \vee p)$$] (11) $$(\neg p \to \neg p \lor (q \lor p)) \to (p \vee \neg p \to \neg p\vee (q \vee p))$$$$\hspace{0,2 cm}$$[10,8 MP] (12) $$p \vee \neg p \to \neg p \vee (q \lor p)$$$$\hspace{0,2 cm}$$[11,2 MP] (13) $$\neg p\vee (q \vee p)$$$$\hspace{0,2 cm}$$[12,5 MP] Counting similarly as before, we have $$l_{\lnot p\lor(q\lor p)}=378$$ and $$w_{\lnot p\lor(q\lor p)}=51$$. Formula $$(p\to q)\to (r\lor p \to r\lor q)$$ (disjunct added to the left) (1) (19) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[thesis, see the proof of the hypothetical syllogism, line 8] (2) $$q\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 10.] (3) $$q\to r\lor q$$$$\hspace{0,2 cm}$$[2 Sub: $$p/r$$] (4) $$(q\to r\lor q) \to ((p\to q) \to (p\to r\lor q))$$$$\hspace{0,2 cm}$$[1 Sub: $$r/r\lor q$$] (5) $$(p\to q)\to (p\to r\lor q)$$$$\hspace{0,2 cm}$$[4,3 MP] (6) $$(p\to r)\to((q\to r)\to(p\lor q\to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (7) (6) $$(r\to r\lor q)\to((p\to r\lor q)\to(r\lor p \to r\lor q))$$$$\hspace{0,2 cm}$$[6 SSub: $$p/r,q/p,r/r\lor q$$] (8) $$p\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 9.] (9) $$r\to r\lor q$$$$\hspace{0,2 cm}$$[8 Sub: $$p/r$$] (10) $$(p\to r\lor q)\to(r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[7,9 MP] (11) (46+8) 39$$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[5,10 HS] The first formula occurs in the proof of hypothetical syllogism, so the value 19 is included in the value 46 added in line 11. Therefore we may calculate: $$l_{dis~l}=l_{hyp~syl}+22=68$$ and $$w_{dis~l}=39$$. Formula $$(p\to q)\to (p\lor r \to q\lor r)$$ (disjunct added to the right): (1) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[thesis, see the proof of the hypothetical syllogism, line 8] (2) $$p\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 9.] (3) (4) $$q\to q\lor r$$$$\hspace{0,2 cm}$$[2 SSub: $$p/q,q/r$$] (4) $$(q\to q\lor r) \to ((p\to q) \to (p\to q\lor r))$$$$\hspace{0,2 cm}$$[1 Sub: $$r/q\lor r$$] (5) $$(p\to q)\to (p\to q\lor r)$$$$\hspace{0,2 cm}$$[4,3 MP] (6) $$(p\to r)\to((q\to r)\to(p\lor q\to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (7) (4) $$(p\to q\lor r)\to((r\to q\lor r)\to(p\lor r \to q\lor r))$$$$\hspace{0,2 cm}$$[6 SSub: $$q/r,r/q\lor r$$] (8) (74) 39$$(r\to q\lor r) \to ((p\to q\lor r)\to (p\lor r \to q\lor r))$$$$\hspace{0,2 cm}$$[7 Com] (9) $$q\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 10.] (10) (4) $$r\to q\lor r$$$$\hspace{0,2 cm}$$[9 SSub: $$p/q,q/r$$] (11) $$(p\to q\lor r)\to(p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[8,10 MP] (12) (8) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[5,11 HS] The formula in line 1, as well as the hypothetical syllogism, occurs in the proof of the law of commutation, so the cost (74) of the application of Com includes the two. Therefore we have $$l_{dis~r} = 100$$ and $$w_{dis~r}=39$$. Now we are in a position to present: Proof of Theorem 4 Proof. Suppose that $$\vdash A_1, \ldots, A_n$$, where $$n \geq 2$$, is a base sequent. Assume also that it is of the form:   $$ \vdash A_1, \ldots, A_{i-1}, B, A_{i+1}, \ldots, A_{k-1}, \lnot B, A_{k+1}, \ldots, A_n, $$ where $$1 \leq i < k \leq n$$. If $$i+1 < k$$, then we start with the proof of formula $$p \lor (q \lor \lnot p)$$ ($$l_{p \lor (q \lor \lnot p)}=372, w_{p \lor (q \lor \lnot p)}=51$$) and substitute formula $$B$$ for $$p$$ and formula $$A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor A_{k-1})\ldots)$$ for $$q$$. We arrive at: $$(a)$$. $$D = B \lor ((A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor A_{k-1})\ldots)) \lor \lnot B)$$ with the length of the proof $$372+4$$ (substitutions) $$=376$$ and the width equal to $$\max\{51, |D|\}$$. If $$k=i+2$$, i.e., there is only one formula $$A_{i+1}$$ between $$B$$ and $$\lnot B$$, then $$D$$ takes the form: $$B \lor (A_{i+1} \lor \lnot B)$$ and at this level we are done. Now suppose that $$k\geq i+3$$ and let $$C=(A_{i+2}\lor\ldots\lor(A_{k-1}\lor A_k)\ldots)$$. Then we use the law $$(p\to q)\to (r\lor p \to r\lor q)$$ (which adds another $$l_{dis~l}=68$$ lines), we also make use of one of the laws of associativity of disjunction, which is already present among the 372 lines of the proof of $$p\lor(q\lor \lnot p)$$, and we continue as below: $$(b)$$. $$(p\lor q) \lor r \to p\lor(q\lor r)$$ $$(c)$$. (6) $$(A_{i+1} \lor C) \lor \lnot B \to A_{i+1} \lor(C\lor \lnot B)$$$$\hspace{0,2 cm}$$[$$(b)$$ SSub: $$p/A_{i+1}, q/C, r/\lnot B$$] $$(d)$$. (68) $$(p\to q)\to (r\lor p \to r\lor q)$$ $$(e)$$. (6) $$((A_{i+1} \lor C) \lor \lnot B \to A_{i+1} \lor(C\lor \lnot B)) \to (B\lor ((A_{i+1} \lor C) \lor \lnot B) \to B \lor (A_{i+1} \lor(C\lor \lnot B)))$$$$\hspace{0,2 cm}$$[$$(d)$$ SSub: $$p/(A_{i+1} \lor C) \lor \lnot B$$, $$q/A_{i+1} \lor(C\lor \lnot B)$$, $$r/B$$] $$(f)$$. $$B\lor ((A_{i+1} \lor C) \lor \lnot B) \to B \lor (A_{i+1} \lor(C\lor \lnot B))$$$$\hspace{0,2 cm}$$[$$(e),(c)$$ MP] $$(g)$$. $$B \lor (A_{i+1} \lor(C\lor \lnot B))$$$$\hspace{0,2 cm}$$[$$(f), (a)$$ MP]. Again, if $$k=i+3$$, then at this level we are done. But if $$C$$ is a disjunction, then we are forced to continue, in order to rearrange the parentheses inside formula $$(C\lor \lnot B)$$. Let $$k=i+l$$ and suppose that $$k=i+l>i+3$$. Then after reaching line no $$(g)$$. we need another $$6(l-2)$$ substitutions to the associativity law at $$(b)$$ with respect to consecutive inner disjunctions, that is, we will get: $$(h)$$. $$(A_{i+2}\lor C^*)\lor \lnot B \to A_{i+2}\lor(C^*\lor \lnot B)$$ where $$C^*=(A_{i+3}\lor\ldots\lor(A_{k-1}\lor A_k)\ldots)$$, then $$(i)$$. $$(A_{i+3}\lor C^{**})\lor \lnot B \to A_{i+3}\lor(C^{**}\lor \lnot B)$$, where $$C^{**}=(A_{i+4}\lor\ldots\lor(A_{k-1}\lor A_k)\ldots)$$, and so on. At the last step we arrange the parentheses around formulas $$A_{i+l-2}, A_{i+l-1}$$ and $$\lnot B$$ as follows: $$(j)$$. $$(A_{i+l-2}\lor A_{i+l-1})\lor \lnot B \to A_{i+l-2}\lor(A_{i+l-1}\lor \lnot B).$$ However, now we must ‘climb back’ to the more complex disjunctions by adding the consecutive disjuncts to the left. For example, for adding formula $$A_{i+l-3}$$ we need one simultaneous substitution (i.e. 6 substitutions) to $$(d)$$ and one application of MP (we detach $$(j)$$) to obtain: $$(k).$$$$A_{i+l-3} \lor ((A_{i+l-2} \lor A_{i+l-1}) \lor \lnot B) \to A_{i+l-3} \lor (A_{i+l-2} \lor (A_{i+l-1} \lor \lnot B))$$ and further one application of the derived rule HS (8 lines) to obtain:11 $$(l).$$$$(A_{i+l-3} \lor (A_{i+l-2} \lor A_{i+l-1})) \lor \lnot B \to A_{i+l-3} \lor (A_{i+l-2} \lor (A_{i+l-1} \lor \lnot B))$$. Each ‘level up’ adds another 15 lines to the proof, i.e., 7 lines to reach $$(k)$$ and 8 lines to reach $$(l)$$. To sum up, before we add formula $$A_i$$, that is, $$B$$, we need $$15\cdot(l-3)=15\cdot(k-i-3)$$ lines after line number $$(j)$$. Moreover, when for the last time we obtain a formula analogous to that in $$(l)$$, the formula’s length equals   $$\mathbf{L} = \left(\sum_{l=i}^k |A_l|\right) + \left(k-(i-1)-1\right).$$ Recall that $$B$$ is $$A_i$$ and $$\lnot B$$ is $$A_k$$. The value $$(k-(i-1)-1)$$ counts the number of occurrences of connectives (disjunctions) between the formulas. The formula is obtained by an application of HS, therefore the longest formula (at this step of the proof) is the appropriate substitution to the law of hypothetical syllogism; and this formula’s length is: $$3\cdot\mathbf{L}+2$$. After another application of MP to $$(l)$$ (we detach $$(a)$$) we end up with a ready proof of the formula: $$(m).$$$$B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots)).$$ If $$i=1$$ and $$k=n$$, then this is the whole proof. Suppose that $$k<n$$. Now we need to add the formulas to the right of $$\lnot B$$. Thus let $$D^\star = (A_{k+1}\lor(A_{k+2}\lor\ldots\lor(A_{n-1}\lor A_n)\ldots))$$. We start with a simultaneous substitution to Ax 9. (4 substitutions) and MP: $$(n).$$ (4) $$B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots)) \to (B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor$$ $$\quad (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots))) \lor D^\star$$ $$(o).$$$$(B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots))) \lor D^\star$$$$\hspace{0,2 cm}$$[MP] and we have to rearrange the parentheses again. There is $$k-(i-1)$$ formulas from $$B$$ to $$\lnot B$$. As before, we need to add $$k-(i-1)$$ simultaneous substitutions to the law of associativity $$(d)$$ and 15 additional lines for each of the $$k-(i-1)$$ formulas, in order to obtain: $$(p).$$$$B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor (\lnot B \lor (A_{k+1}\lor(A_{k+2}\lor\ldots\lor(A_{n-1}\lor A_n)\ldots)))))\ldots))$$ Finally, if $$i > 1$$, then we add the formulas to the left. For each of $$i-1$$ formulas we add a simultaneous substitution to Ax 10. (4 substitutions) and an application of MP starting with $$A_{i-1}$$ and ending with $$A_1$$. If $$i+1 = k$$, then we start with the proof of $$p\lor\lnot p$$ and substitute $$B$$ for $$p$$. For the cases of $$i>1$$ and/or $$n>k$$ we proceed as above. If $$i=1$$ and/or $$k=n$$, then we omit, respectively, the steps connected with adding the outer formulas. Finally, if the sequent is of the form:   $$\vdash A_1, \ldots, A_{i-1}, \lnot B, A_{i+1}, \ldots, A_{k-1}, B, A_{k+1}, \ldots, A_n$$ then we start with the proof of formula $$\lnot p \lor (q \lor p)$$, if $$i+1 < k$$, or with the proof of $$\lnot p \lor p$$ otherwise. ■ Let $$l_{base~sequent}$$ stand for the length of the proof of the formula corresponding to the base sequent $$\vdash A_1, \ldots, A_n$$. We estimate $$l_{base~sequent}$$ in the worst case:   $$l_{base~sequent}\leq 458+6\cdot(k-i-2)+15\cdot(k-i-3)+6+$$   $$+6\cdot(k-(i-1))+15\cdot(k-(i-1))+5\cdot(i-1)=423+5\cdot k+37\cdot(k-i).$$ Since $$k\leq n$$ and $$k-i< n$$, where $$n$$ is the number of terms of the initial base sequent, we have $$l_{base~sequent}\leq 423+42\cdot n$$, which means that: Corollary 1 The length of the proof of the formula corresponding to a base sequent depends linearly on the number of terms of the sequent, i.e.  $$l_{base~sequent}=\mathcal{O}(n)$$ As to the width, the longest formula that occurs in the above proof is either the one in the proof of formula $$p \lor (q \lor \lnot p)$$ ($$w_{p \lor (q \lor \lnot p)}=51$$), or the one whose length is $$3\cdot\mathbf{L}+2$$ (see above), or, finally, the following one (the last substitution to Ax 10.):   $$D^{\star\star} = A_2 \lor \ldots \lor A_{i-1} \lor \lnot B \lor A_{i+1} \lor \ldots \lor A_{k-1} \lor B \lor A_{k+1} \lor \ldots \lor A_n \to$$   $$A_1 \lor \ldots \lor A_{i-1} \lor \lnot B \lor A_{i+1} \lor \ldots \lor A_{k-1} \lor B \lor A_{k+1} \lor \ldots \lor A_n.$$ We omit parentheses, since $$|D^{\star\star}|$$ does not depend on them. For $$\phi =\ \vdash A_1, \ldots, A_n$$, let $$|\phi| = \sum^{n}_{i=1} |A_i|$$. Then $$|D^{\star\star}|=|\phi|-|A_1|+(n-2)+|\phi|+1+(n-1) \leq 2\cdot(|\phi|+n)$$, whereas $$3\cdot\mathbf{L}+2 \leq 3\cdot(|\phi|+n)+2$$. Since $$2\cdot(|\phi|+n) < 3\cdot(|\phi|+n)+2$$, we have:   $$w_{base~sequent} \leq \max\{51,3\cdot(|\phi|+n)+2\},$$ where for long formulas we will always have $$3\cdot(|\phi|+n)+2 > 51$$. This entails: Corollary 2   $$w_{base~sequent} = \mathcal{O}(|\phi|).$$ Since the number $$n$$ of terms of the base sequent is not greater than $$|\phi|$$, we finally obtain: Corollary 3   $$s_{base~sequent} = \mathcal{O}(|\phi|^2).$$ where $$s_{base~sequent}$$ stands for the size of the proof of the formula corresponding to a base sequent. 4.2 The applications of rules of calculus $$\mathbb{E^{**}}$$ Definition 4 Let $$S = \langle A_{1}, A_{2}, \ldots, A_{n-1}, A_{n} \rangle$$ be a finite, non-empty sequence of formulas of $$\mathcal{L}$$. Then by:   $$\bigvee S$$ we shall mean the following disjunction:   $$A_{1}\vee (A_{2}\vee ...(A_{n-1} \vee A_{n})...).$$ The aim of this subsection is to prove two theorems stated below. Theorem 5 If the following schema:   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E^{**}}$$, and there exists a proof of formula $$\bigvee T$$ in system $$\mathbb{A}$$, then the proof may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. Theorem 6 If the following schema:   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T_{1} \ ; \ \vdash T_{2} \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E^{**}}$$, and there exist a proof of formula $$\bigvee T_{1}$$ and a proof of formula $$\bigvee T_{2}$$ in system $$\mathbb{A}$$, then the proofs may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. The proofs will be conducted with respect to each rule schema of $$\mathbb{E^{**}}$$. Since the rules of $$\mathbb{E^{**}}$$ are given in the form of schemas with metavariables, we will present schemas of the proofs in $$\mathbb{A}$$ also with the use of metavariables. 4.2.1 Rule $$\mathbf{R}_{\lnot\lnot}$$   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \ ' \ A \ ' \ S_{2} \ ; \ \Psi)} $$ Suppose that $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$ (but recall that $$S_1$$ may be empty). We show that if there exists a proof of formula:   \begin{equation}\label{A} C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor (\bigvee S_{2}))))\ldots) \end{equation} (8) in system $$\mathbb{A}$$, then there exists also a proof of:   \begin{equation}\label{not-not-A} C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2}))))\ldots) \end{equation} (9) in system $$\mathbb{A}$$. We show how to obtain one from the other. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor (\bigvee S_{2}))))\ldots)$$$$\hspace{0,2cm}$$[proved by assumption] (2) $$p\to\lnot\lnot p$$$$\hspace{0,2cm}$$[Ax 5.] (3) $$A\to\lnot\lnot A$$$$\hspace{0,2cm}$$[2 Sub: $$p/A$$] (4) (100) 39$$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[disjunct added to the right] (5) (6) $$(A\to \lnot\lnot A)\to (A\lor (\bigvee S_{2}) \to \lnot\lnot A\lor (\bigvee S_{2}))$$$$\hspace{0,2 cm}$$[4 SSub: $$p/A,q/\lnot\lnot A,r/\bigvee S_{2}$$] (6) $$A\lor (\bigvee S_{2}) \to \lnot\lnot A\lor (\bigvee S_{2})$$$$\hspace{0,2 cm}$$[5,3 MP] (7) (68) 39$$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] (8) (6) $$(A\lor (\bigvee S_{2}) \to \lnot\lnot A\lor (\bigvee S_{2}))\to(C_n\lor(A\lor (\bigvee S_{2})) \to C_n\lor(\lnot\lnot A\lor (\bigvee S_{2})))$$ $$\quad$$ [4 SSub: $$p/A\lor (\bigvee S_{2}),q/ \lnot\lnot A\lor (\bigvee S_{2}),r/C_n$$] (9) $$C_n\lor(A\lor (\bigvee S_{2})) \to C_n\lor(\lnot\lnot A\lor (\bigvee S_{2}))$$ [8,6 MP] $$\vdots$$ Steps in lines 8,9 repeated with respect to $$C_{n-1},\ldots,C_1$$ (10) $$\Large($$$$(C_2\lor\ldots\lor (C_n\lor (A \lor (\bigvee S_{2})))\ldots) \to (C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$$$\Large)$$$$\to$$$$\Large($$$$C_1\lor(C_2\lor\ldots\lor (C_n\lor (A \lor (\bigvee S_{2})))\ldots) \to C_1\lor(C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$$$\Large)$$ (11) $$C_1\lor(C_2\lor\ldots\lor (C_n\lor (A \lor (\bigvee S_{2})))\ldots) \to C_1\lor(C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$ (12) $$C_1\lor(C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$$$\hspace{0,2 cm}$$[10,1 MP] Let $$l_C$$ stand for the length of the proof of formula $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor (\bigvee S_{2}))))\ldots)$$ and $$w_C$$ for its width. Then the length of the above proof may be estimated as $$l_C+2+l_{dis~r}+7+l_{dis~l}+7\cdot n+1=l_C+178+7\cdot n$$, where $$n$$ is the number of the terms of $$S_1$$. Therefore the length of the above proof, $$l_{\lnot\lnot}$$ in symbols, depends linearly on the length of $$l_C$$ and the number $$n$$ of terms of $$S_1$$:   \begin{equation}\label{podwojna negacja} l_{\lnot\lnot}= l_C + 7\cdot n + c \end{equation} (10) for a constant $$c$$. The length of the formula in line 10, $$l(10)$$ for short, is bounded by   $$l(10) \leq 4\cdot|\vdash S_{1} \ ' \ A \ ' \ S_{2}|+ 4\cdot|\vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2}|.$$ The coefficients ‘4’ may seem too large, but this overestimation is to include the occurrences of connectives between the formulas. Since $$|\vdash S_{1} \ ' \ A \ ' \ S_{2}| < |\vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2}|$$, the width of the above proof, $$w_{\lnot\lnot}$$, may be estimated as follows:   \begin{equation}\label{w podwojna negacja} w_{\lnot\lnot} \leq \max\{w_C, 39, 8\cdot|\vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2}|\}. \end{equation} (11) 4.2.2 Rule $$\mathbf{R}_\to$$   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \ ' \ A\to B \ ' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \ ' \ \neg A \ ' \ B \ ' \ S_{2} \ ; \ \Psi)}. $$ Let $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$. We show that if there exists a proof of formula:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots)$$ in system $$\mathbb{A}$$, then there exists also a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\to B) \lor (\bigvee S_{2}))))\ldots)$$ in system $$\mathbb{A}$$. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots)$$$$\hspace{0,2 cm}$$[proved by assumption] (2) (179) 51$$p\to (\neg p\to q)$$$$\hspace{0,2 cm}$$ [Duns Scotus law] (3) (4) $$A\to (\neg A\to B)$$$$\hspace{0,2 cm}$$ [2 SSub: $$p/A, q/B$$] (4) (74) $$\neg A\to (A\to B)$$$$\;\;$$ [3 Com] (5) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (6) (6) $$(\neg A\to (A\to B))\to ((B\to (A\to B))\to (\neg A\vee B\to(A\to B)))$$ $$\quad$$ [5 SSub: $$p/\neg A, q/B, r/A\to B$$] (7) $$(B\to (A\to B))\to (\neg A\vee B\to(A\to B))$$$$\;\;$$ [6,4 MP] (8) $$p\to(q\to p)$$$$\hspace{0,2 cm}$$[Ax 1.] (9) (4) $$B\to (A\to B)$$$$\;\;$$ [8 SSub: $$p/B, q/A$$] (10) $$\neg A\vee B\to(A\to B)$$$$\;\;$$ [7,9 MP] (11) (100) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[disjunct added to the right] (12) (6) $$(\lnot A\lor B\to (A\to B))\to ((\lnot A\lor B)\lor (\bigvee S_2) \to (A\to B)\lor (\bigvee S_2))$$ $$\hspace{0,2 cm}$$[11 SSub: $$p/\lnot A\lor B,q/A\to B,r/\bigvee S_2$$] (13) $$(\lnot A\lor B)\lor (\bigvee S_2) \to (A\to B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[12,10 MP] (14) (87) $$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[associativity of disjunction 2] (15) (6) $$\lnot A\lor (B \lor (\bigvee S_2)) \to (\lnot A\lor B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[14 SSub: $$p/\lnot A,q/B,r/\bigvee S_2$$] (16) (8) $$\lnot A\lor(B\lor(\bigvee S_2))\to (A\to B)\lor(\bigvee S_2)$$$$\hspace{0,2 cm}$$[15,13 HS] (17) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] (18) (6) $$(\lnot A\lor(B\lor(\bigvee S_2)) \to (A\to B)\lor(\bigvee S_2))\to (C_n\lor (\lnot A\lor(B\lor(\bigvee S_2))) \to C_n\lor ((A\to B)\lor(\bigvee S_2)))$$$$\hspace{0,2 cm}$$[17 SSub: $$p/\lnot A\lor(B\lor(\bigvee S_2)),q/(A\to B)\lor(\bigvee S_2),r/C_n$$] (19) $$C_n\lor (\lnot A\lor(B\lor(\bigvee S_2))) \to C_n\lor ((A\to B)\lor(\bigvee S_2))$$$$\hspace{0,2 cm}$$[18,16 MP] $$\vdots$$ steps 18, 19 (7 lines) repeated with respect to $$C_{n-1},\ldots,C_1$$ (20) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots) \to$$$$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\to B) \lor (\bigvee S_{2}))))\ldots)$$ (21) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\to B) \lor (\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$[20,1 MP] Let $$l_C$$, $$w_C$$ stands for the length and width of the proof of formula $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots)$$. Then the length of the above proof, $$l_\to$$ in symbols, equals $$480+l_C+7\cdot n$$, thus   \begin{equation}\label{implikacja} l_\to = l_C + 7\cdot n + c \end{equation} (12) for a constant $$c$$. For the width, let $$\psi$$ stand for ‘$$\vdash S_{1} \ ' \ \lnot A \ ' \ B \ ' \ S_{2}$$’. Probably, the longest formula in the proof is the one analogous to that in line no. (18), obtained after adding $$C_1$$. Its length can be estimated by the (upper bound) value $$8\cdot|\psi|$$. Therefore:   \begin{equation}\label{w implikacja} w_\to \leq \max\{w_C, 51, 8\cdot|\psi|\}. \end{equation} (13) 4.2.3 Rule $$\mathbf{R}_{\lnot \wedge}$$ Now we shall consider:   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ \neg (A\wedge B) \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ \neg A \ ' \ \neg B \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)}. $$ We show that a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\lnot B \lor (\bigvee S_{2})))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot (A\land B) \lor (\bigvee S_{2}))))\ldots)$$ (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\lnot B \lor (\bigvee S_{2})))))\ldots)$$$$\hspace{0,2 cm}$$[proved by assumption] (2) $$(p\to q)\to (\neg q\to \neg p)$$$$\;\;$$ [Ax 3.] (3) (4) $$(A\wedge B\to A)\to (\neg A\to \neg(A\wedge B))$$$$\;\;$$ [2 SSub: $$p/A\wedge B, q/A$$] (4) $$p\wedge q\to p$$$$\;\;$$ [Ax 6.] (5) (4) $$A\wedge B\to A$$$$\;\;$$ [4 SSub: $$p/A, q/B$$] (6) $$\neg A\to \neg(A\wedge B)$$$$\;\;$$ [3,5 MP] (7) (4) $$(A\wedge B\to B)\to (\neg B\to \neg(A\wedge B))$$$$\;\;$$ [2 Sub: $$p/A\wedge B, q/B$$] (8) $$p\wedge q\to q$$$$\;\;$$ [Ax 7.] (9) (4) $$A\wedge B\to B$$$$\;\;$$ [8 SSub: $$p/A, q/B$$] (10) $$\neg B\to \neg(A\wedge B)$$$$\;\;$$ [7,9 MP] (11) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (12) (6) $$(\neg A\to \neg(A\wedge B)) \to ((\neg B\to \neg(A\wedge B)) \to (\neg A\vee \neg B\to \neg(A\wedge B)))$$ $$\quad$$ [11 SSub: $$p/\neg A, q/\neg B, r/\neg(A\wedge B)$$] (13) $$(\neg B\to \neg(A\wedge B)) \to (\neg A\vee \neg B\to \neg(A\wedge B))$$$$\;\;$$ [12,6 MP] (14) $$\neg A\vee \neg B\to \neg(A\wedge B)$$$$\;\;$$ [13,10 MP] (15) (100) 39$$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[disjunct added to the right] (16) (6) $$(\lnot A\lor \lnot B\to \lnot(A\land B))\to ((\lnot A\lor \lnot B)\lor (\bigvee S_2) \to \lnot(A \land B)\lor (\bigvee S_2))$$ $$\quad$$ [15 SSub: $$p/\lnot A\lor \lnot B,q/\lnot(A\land B),r/\bigvee S_2$$] (17) $$(\lnot A\lor \lnot B)\lor (\bigvee S_2) \to \lnot(A\land B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[16,14 MP] (18) (87) $$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[associativity of disjunction 2] (19) (6) $$\lnot A\lor (\lnot B \lor (\bigvee S_2)) \to (\lnot A\lor \lnot B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[18 SSub: $$p/\lnot A,q/\lnot B,r/\bigvee S_2$$] (20) (8) $$\lnot A\lor(\lnot B\lor(\bigvee S_2))\to \lnot(A\land B)\lor(\bigvee S_2)$$$$\hspace{0,2 cm}$$[19,17 HS] (21) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] (22) (6) $$\Large($$$$\lnot A\lor(\lnot B\lor(\bigvee S_2)) \to \lnot(A\land B)\lor(\bigvee S_2)$$$$\Large)$$$$\to$$$$\Large($$$$C_n\lor (\lnot A\lor(\lnot B\lor(\bigvee S_2))) \to$$ $$\quad C_n\lor (\lnot(A\land B)\lor(\bigvee S_2))$$$$\Large)$$$$\hspace{0,2 cm}$$[21 SSub: $$p/\lnot A\lor(\lnot B\lor(\bigvee S_2)),q/\lnot(A\land B)\lor(\bigvee S_2),r/C_n$$] (23) $$C_n\lor (\lnot A\lor(\lnot B\lor(\bigvee S_2))) \to C_n\lor (\lnot(A\land B)\lor(\bigvee S_2))$$$$\hspace{0,2 cm}$$[22,20 MP] $$\vdots$$ steps 22,23 (7 lines) repeated with respect to $$C_{n-1},\ldots,C_1$$ (24) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\lnot B \lor (\bigvee S_{2})))))\ldots) \to$$$$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\land B) \lor (\bigvee S_{2}))))\ldots)$$ (25) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\land B) \lor (\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$[24,1 MP] As in the previous cases, we assume that $$l_C$$ and $$w_C$$ stand for the length and width (respectively) of the formula in the first line. The length, $$l_{\lnot\land}$$, of the above proof equals $$l_C+307+7\cdot n$$, thus   \begin{equation}\label{nie i} l_{\lnot\land} = l_C+7\cdot n + c \end{equation} (14) for a constant, $$c$$, and   \begin{equation}\label{w nie i} w_{\lnot\land} \leq \max\{w_C, 39, 8\cdot|\psi|\}, \end{equation} (15) where $$\psi = ~ \vdash S_{1} \hspace{0,1 cm}' \ \neg A \ ' \ \neg B \hspace{0,1 cm}' \ S_{2}$$. 4.2.4 Rule $$\mathbf{R}_\lor$$ To prove Theorem 5, we still need to consider:   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ A\lor B \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ A \ ' \ B \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)}. $$ Now we need to show that a proof of formula:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(B \lor (\bigvee S_{2})))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\lor B) \lor (\bigvee S_{2}))))\ldots).$$ For this purpose we start with: (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(B \lor (\bigvee S_{2})))))\ldots)$$ [proved by assumption] (2) (87) 39$$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[associativity of disjunction 2] (3) (6) $$A\lor (B \lor (\bigvee S_2)) \to (A\lor B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[2 SSub: $$p/A,q/B,r/\bigvee S_2$$] (4) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] and then we add formulas $$C_{n},\ldots,C_1$$ to the left, as in the above cases. This time we have $$l_\lor = l_C+162+7\cdot n$$, thus   \begin{equation}\label{alternatywa} l_\lor = l_C+7\cdot n + c \end{equation} (16) for a constant, $$c$$. Again, as previously,   \begin{equation}\label{w alternatywa} w_\lor \leq \max\{w_C, 39, 8\cdot|\psi|\}, \end{equation} (17) where $$\psi ~=~ \vdash S_{1} \hspace{0,1 cm}' \ A \ ' \ B \hspace{0,1 cm}' \ S_{2}$$. Proof of Theorem 5 The schemas of proofs presented in subsections 4.2.1–4.2.4 supply us with the proof of Theorem 5. We have shown that if   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E^{**}}$$, and there exists a proof of formula $$\bigvee T$$ in system $$\mathbb{A}$$, then the proof may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. Moreover, it is important that the schemas show us how to transform one proof into another. We have also seen that if $$l_C, l_P$$ and $$w_C, w_P$$ stand for the lengths and widths of proofs of the formulas corresponding to, respectively, ‘sequent-conclusion’ $$\vdash T$$ and ‘sequent-premise’ $$\vdash S$$, then by (10), (12), (14), (16): Corollary 4   $$l_P \leq l_C + f(n),$$ where $$n$$ stands for the number of terms of sequence $$T$$, and $$f$$ is a linear function. Observe that, strictly speaking, $$n$$ could be defined as the number of terms of an initial subsequence of $$T$$, not of the whole sequence. But this is enough for our purposes. On the other hand, we can estimate $$f$$ very precisely, as in the case of each rule we have obtained $$f(n)=7\cdot n + c$$ for some constant $$c \in \text{N}_+$$. In turn, by (11), (13), (15), (17): Corollary 5   $$w_P \leq \max\{w_C,c,8\cdot|\vdash T|\}$$ for a constant $$c$$$$(c = 39)$$. 4.2.5 Rule $$\mathbf{R}_\land$$ We proceed to the proof of Theorem 6.   $$ \frac {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' A\wedge B \hspace{0,1 cm}' ~S_{2}~;~ \Psi~)} {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' ~ A \hspace{0,1 cm}'~S_{2}~;~ \vdash S_{1} \hspace{0,1 cm}' ~ B \hspace{0,1 cm}'~S_{2}~;~\Psi~)}. $$ We will show that if there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, and there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots)$$ then the two may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\land B) \lor (\bigvee S_{2}))))\ldots).$$ We start with the proof of the law of conjunction, and distributivity of disjunction with respect to conjunction. The law of conjunction (1) $$p\to (q\to p)$$$$\;\;$$ [Ax 1.] (2) $$(p\to q)\to ((p\to r)\to (p\to q\wedge r))$$$$\;\;$$ [Ax 8.] (3) (6) $$(q\to p)\to ((q\to q)\to (q\to p\wedge q))$$$$\;\;$$ [2 SSub: $$p/q, q/p, r/q$$] (4) (8) $$p\to ((q\to q)\to (q\to p\wedge q))$$$$\;\;$$ [1,3 HS] (5) (7) $$(q\to q)\to (p\to (q\to p\wedge q))$$$$\;\;$$ [4 Com] (6) (258) 51$$p\to p$$$$\;\;$$ [identity law] (7) $$q\to q$$$$\;\;$$ [5 Sub: $$p/q$$] (8) $$p\to (q\to p\wedge q)$$$$\;\;$$ [4,6 MP] In counting the length of the above proof we omit the costs of deriving the laws of hypothetical syllogism and commutation, as they are already counted in the 258 lines for the identity law. Thus we have $$l_{conj} = 283$$ and $$w_{conj}=51$$. The laws of distributivity of disjunction with respect to conjunction (1) $$p\to(q\to p)$$$$\;\;$$ [Ax 1.] (2) $$q\to(p\lor q)$$$$\;\;$$ [Ax 10.] (3) (4) $$r\to((p\land q)\lor r)$$$$\;\;$$ [2 SSub: $$p/p\land q,q/r$$] (4) (4) $$(r\to((p\land q)\lor r)) \to (q\lor r \to (r\to((p\land q)\lor r)))$$$$\;\;$$ [1 SSub: $$p/r\to((p\land q)\lor r),q/q\lor r$$] (5) $$q\lor r \to (r\to((p\land q)\lor r))$$$$\;\;$$ [4,3 MP] (6) (7) $$r\to (q\lor r \to (p\land q)\lor r)$$$$\;\;$$ [5 Com] (7) (283) 51$$p\to(q\to p\wedge q)$$$$\;\;$$ [law of conjunction] (8) (100) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\;\;$$ [disjunct added to the right] (9) (4) $$(q\to p\land q)\to (q\lor r \to (p\land q)\lor r)$$$$\;\;$$ [8 SSub: $$p/q,q/p\land q$$] (10) (8) $$p\to (q\lor r\to (p\land q)\lor r)$$$$\;\;$$ [7,9 HS] (11) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (12) (4) $$(p\to (q\lor r\to (p\land q)\lor r))\to ((r\to (q\lor r\to (p\land q)\lor r))\to (p\vee r\to (q\lor r\to (p\land q)\lor r)))$$ $$\quad$$ [11 SSub: $$q/r,r/q\lor r\to (p\land q)\lor r$$] (13) $$(r\to (q\lor r\to (p\land q)\lor r))\to (p\vee r\to (q\lor r\to (p\land q)\lor r))$$$$\;\;$$ [12,10 MP] (14) $$p\vee r\to (q\lor r\to (p\land q)\lor r)$$$$\;\;$$ [13,6 MP] (15) (7) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$ [14 Imp] Since the proof of the identity law goes through the laws of commutation and importation, and the proof of the law of conjunction goes through the identity law, we may count the length of the above proof, $$l_{distr~1}$$, as equal to $$427$$. In turn, $$w_{distr~1}=51$$. Another law of distributivity: (1) $$p\to(q\to p)$$$$\;\;$$ [Ax 1.] (2) $$p\to p\lor q$$$$\;\;$$ [Ax 9.] (3) $$p\to p\lor (q\land r)$$$$\;\;$$ [2 Sub: $$q/q\land r$$] (4) (4) $$(p\to p\lor (q\land r)) \to (p\lor r \to (p\to p\lor (q\land r)))$$$$\;\;$$ [1 SSub: $$p/r\to((p\land q)\lor r),q/p\lor r$$] (5) $$p\lor r \to (p\to p\lor (q\land r)))$$$$\;\;$$ [4,3 MP] (6) (7) $$p\to (p\lor r \to p\lor (q\land r))$$$$\;\;$$ [5 Com] (7) (283) 51$$p\to(q\to p\wedge q)$$$$\;\;$$ [law of conjunction] (8) (4) $$q\to(r\to q\wedge r)$$$$\;\;$$ [7 SSub: $$p/q,q/r$$] (9) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\;\;$$ [disjunct added to the left] (10) (6) $$(r\to q\land r)\to (p\lor r \to p\lor(q\land r))$$$$\;\;$$ [9 SSub: $$p/r,q/q\land r,r/p$$] (11) (8) $$q\to (p\lor r \to p\lor(q\land r))$$$$\;\;$$ [8,10 HS] (12) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (13) (4) $$(p\to (p\lor r \to p\lor (q\land r)))\to ((q\to (p\lor r \to p\lor(q\land r)))\to (p\vee q\to (p\lor r\to p\lor (q\land r))))$$ $$\quad$$ [12 SSub: $$q/r,r/p\lor r\to p\lor (q\land r)$$] (14) $$(q\to (p\lor r \to p\lor(q\land r)))\to (p\vee q\to (p\lor r\to p\lor (q\land r)))$$$$\;\;$$ [13,6 MP] (15) $$p\vee q\to (p\lor r\to p\lor (q\land r))$$$$\;\;$$ [14,11 MP] (16) (7) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [15 Imp] Similarly as before, the length of the above proof, $$l_{distr~2}$$, is estimated at $$398$$, whereas the width, $$w_{distr~2}$$, equals 51. We go back to the analysis of rule $$\mathbf{R}_\land$$. In line no. 3 below we use the derived rule based on the law of conjunction, Conj for short. Its cost would be 283+6, but the 283 lines are included in the cost of the first law of distributivity. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (2) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (3) (6) 51$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)) ~\land $$$$~~~~~~~~ (C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots))$$$$\;\;$$ [1,2 Conj] (4) (427) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$ [distributivity 1] (5) (6) $$(A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2))\to (A\land B)\lor (\bigvee S_2)$$$$\;\;$$ [4 Sub: $$p/A, q/B,r/\bigvee S_2$$] (6) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\;\;$$ [disjunct added to the left] (7) (6) $$\Large($$$$(A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2)) \to (A\land B)\lor (\bigvee S_2)$$$$\Large)$$$$\to$$$$~~~~$$$$\Large($$$$C_n \lor ((A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2))) \to C_n \lor ((A\land B)\lor (\bigvee S_2))$$$$\Large)$$$$\;\;$$ [6 Sub: $$p/(A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2)),q/(A\land B)\lor (\bigvee S_2),r/C_n$$] (8) $$C_n \lor ((A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2))) \to C_n \lor ((A\land B)\lor (\bigvee S_2))$$$$\;\;$$ [7,5 MP] (9) (398) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [distributivity 2] (10) (6) $$(C_n\lor (A\lor (\bigvee S_2)))\land (C_n\lor (B\lor (\bigvee S_2)))\to C_n\lor ((A\lor (\bigvee S_2))\land (B\lor (\bigvee S_2)))$$ $$\quad$$ [9 Sub: $$p/C_n, q/ A\lor (\bigvee S_2), r/B\lor (\bigvee S_2)$$] (11) (8) $$(C_n\lor (A\lor (\bigvee S_2)))\land (C_n\lor (B\lor (\bigvee S_2))) \to C_n \lor ((A\land B)\lor (\bigvee S_2))$$ $$\quad$$ [10,8 HS] $$\vdots$$ steps 7, 8, 10, 11 (21 lines) repeated with respect to $$C_{n-1},\ldots,C_1$$ (12) $$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)) ~\land $$$$~~~~~~~~ (C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots)) ~~\to$$ $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\land B) \lor (\bigvee S_{2}))))\ldots)$$ (13) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\land B) \lor (\bigvee S_{2}))))\ldots)$$ [12,3 MP] Let $$l_{C1}$$ and $$l_{C2}$$ stand for the length of proofs of formulas in lines 1 and 2, respectively; similarly for the widths $$w_{C1}$$, $$w_{C2}$$. The length of the above proof, $$l_\land$$, may be estimated by $$l_{C1}+l_{C2}+906+21\cdot n$$, thus   \begin{equation}\label{koniunkcja} l_\land = l_{C1}+l_{C2}+21\cdot n+c \end{equation} (18) for a constant, $$c$$. To estimate the width, $$w_\land$$, of the above proof, observe that the longest formula is either that of length $$w_{C1}$$, $$w_{C2}$$, 51, or it is the result of the substitution to the law of hypothetical syllogism, which is necessary to go through line no. 11 for the last time. More specifically, we mean the result of the simultaneous substitution of ‘$$(C_1 \lor \ldots \lor (C_n\lor (A\lor (\bigvee S_2)))\ldots)\land (C_1 \lor \ldots \lor (C_n\lor (B\lor (\bigvee S_2)))\ldots)$$’ for $$p$$, ‘$$C_1 \lor ((C_2 \lor \ldots \lor (A\lor (\bigvee S_2)))\land (C_2 \lor \ldots \lor (B\lor (\bigvee S_2))))$$’ for $$q$$ and ‘$$C_1 \lor (C_2 \lor \ldots \lor ((A\land B)\lor (\bigvee S_2)))$$’ for $$r$$. It is easy to see that the length of this formula is bounded by $$8\cdot|\psi_1|+8\cdot|\psi_2|+4\cdot|\phi|$$, where $$\psi_1 = \ \vdash S_1 \ ' A \ ' S_2$$, $$\psi_2 = \ \vdash S_1 \ ' B \ ' S_2$$, $$\phi = \ \vdash S_1 \ ' A\land B \ ' S_2$$. Since, moreover, $$|\psi_1| < |\phi|$$ and $$|\psi_2| < |\phi|$$, we can estimate $$w_\land$$ as follows   \begin{equation}\label{w koniunkcja} w_\land \leq \max\{w_{C1}, w_{C2}, c, 20\cdot|\phi|\} \end{equation} (19) for a constant, $$c$$ ($$c=51$$). 4.2.6 Rule $$\mathbf{R}_{\lnot\lor}$$   $$ \frac {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg (A\vee B) \hspace{0,1 cm}'~ S_{2}~;~ \Psi~)} {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' ~\neg A \hspace{0,1 cm}'~S_{2}~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg B \hspace{0,1 cm}'~S_{2}~;~\Psi)} .$$ We will show that if there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\bigvee S_{2}))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, and there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$ then the two may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\lor B) \lor (\bigvee S_{2}))))\ldots).$$ For that, we will need the proof of one of the De Morgan’s laws. De Morgan’s law (1) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (2) $$(p\to (\neg p\to q)) \to ((q \to (\neg p\to q)) \to (p\vee q\to (\neg p\to q)))$$$$\;\;$$ [1 Sub: $$r/\neg p\to q$$] (3) (178) 51$$p\to (\neg p\to q)$$$$\;\;$$ [Duns Scotus law] (4) $$(q\to (\neg p\to q)) \to (p\vee q\to (\neg p\to q))$$$$\;\;$$ [2,3 MP] (5) $$p\to (q\to p)$$$$\;\;$$ [Ax 1.] (6) (4) $$q\to (\neg p\to q)$$$$\;\;$$ [5 SSub: $$p/q, q/\neg p$$] (7) $$p\vee q\to (\neg p\to q)$$$$\;\;$$ [4,6 MP] (8) (7) $$\neg p\to (p\vee q\to q)$$$$\;\;$$ [7 Com] (9) $$(p\to q)\to (\neg q\to \neg p)$$$$\;\;$$ [Ax 3.] (10) $$((p\vee q)\to q)\to (\neg q\to \neg (p\vee q))$$$$\;\;$$ [9 Sub: $$p/p\vee q$$] (11) (8) $$\neg p \to (\neg q \to \neg(p\vee q))$$$$\;\;$$ [8,10 HS] (12) (7) $$\neg p\wedge \neg q\to \neg (p\vee q)$$$$\;\;$$[11 Imp] Counting similarly as before, the length of the above proof, $$l_{DM}$$, is bounded by $$211$$, and the width, $$w_{DM}=51$$. We go back to the analysis of rule $$R_{\lnot\lor}$$. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (2) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (3) (6) 51$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\bigvee S_{2}))))\ldots)) ~\land$$$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots))$$$$\;\;$$ [1,2 Conj] (4) (211) $$\neg p\wedge \neg q\to \neg (p\vee q)$$$$\;\;$$[De Morgan’s law] (5) (4) $$\neg A\wedge \neg B\to \neg (A\vee B)$$$$\;\;$$[4 SSub: $$p/A,q/B$$] (6) (100) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\;\;$$[disjunct added to the right] (7) (6) $$(\neg A\wedge \neg B \to \neg (A\vee B))\to ((\neg A\wedge \neg B)\lor (\bigvee S_2) \to \neg (A\vee B)\lor (\bigvee S_2))$$ $$\quad$$ [6 SSub: $$p/\neg A\wedge \neg B,q/\neg (A\vee B),r/\bigvee S_2$$] (8) $$(\neg A\wedge \neg B)\lor (\bigvee S_2) \to \neg (A\vee B)\lor (\bigvee S_2)$$$$\;\;$$[7,5 MP] (9) (427) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$[distributivity 1] (10) (6) $$(\neg A\lor(\bigvee S_2))~ \land~ (\neg B\lor (\bigvee S_2)) \to (\neg A\wedge \neg B)\lor (\bigvee S_2)$$$$\;\;$$ [9 SSub: $$p/\lnot A$$, $$q/\lnot B$$, $$r/\bigvee S_2$$] (11) (8) $$(\neg A\lor(\bigvee S_2)) \land (\neg B\lor (\bigvee S_2)) \to \neg (A\vee B)\lor (\bigvee S_2)$$$$\;\;$$ [10,8 HS] (12) (398) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [distributivity 2] $$\vdots$$ After this step we apply the same mechanism of adding (to the left) formulas $$C_n, \ldots,C_1$$ which was presented in the analysis of rule $$\mathbf{R}_{\land}$$ (it yields $$68$$ lines of the proof of ‘adding to the left’, $$21\cdot n$$ lines $$+ 1$$ line for the final MP). By arguments similar to those presented in the previous case we have   \begin{equation}\label{nie lub} l_{\lnot\lor} = l_{C1}+l_{C2}+21\cdot n+c \end{equation} (20) for $$c=1236$$. For the width we observe that the length of the formula which may be the longest in the proof is bounded by $$8\cdot|\psi_1|+8\cdot|\psi_2|+4\cdot|\phi|$$, where $$\psi_1 = \ \vdash S_1 \ ' \lnot A \ ' S_2$$, $$\psi_2 = \ \vdash S_1 \ ' \lnot B \ ' S_2$$, $$\phi = \ \vdash S_1 \ ' \lnot(A\lor B) \ ' S_2$$. For the same reason as in the previous case, we have   \begin{equation}\label{w nie lub} w_{\lnot\lor} \leq \max\{w_{C1}, w_{C2}, c, 20\cdot|\phi|\} \end{equation} (21) for a constant, $$c$$ ($$c=51$$). 4.2.7 Rule $$\mathbf{R}_{\neg \to}$$ The last rule to consider is:   $$ \frac {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg(A\to B) \hspace{0,1 cm}'~ S_{2}~;~ \Psi~)} {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' ~ A \hspace{0,1 cm}'~S_{2}~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg B \hspace{0,1 cm}'~S_{2}~;~\Psi~)} .$$ For the last time, we show that if there exists a proof of formula:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, and there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$ then the two may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\to B) \lor (\bigvee S_{2}))))\ldots).$$ Again, we will need to prove one additional law. Formula $$p\land\lnot q\to\lnot(p\to q)$$ (1) $$p\to (q\to p)$$$$\;\;$$ [Ax 1.] (2) (4) $$\neg p\vee p\to ((p\to q)\to \neg p\vee p)$$$$\;\;$$ [1 SSub: $$p/\neg p\vee p, q/p\to q$$] (3) (278) 51$$\neg p \vee p$$$$\;\;$$ [reversed law of excluded middle] (4) $$(p\to q)\to \neg p\vee p$$$$\;\;$$ [2,3 MP] (5) $$(p\to (q\to r))\to((p\to q)\to (p\to r))$$$$\;\;$$ [Ax 2.] (6) (6) $$((p\to q)\to (\neg p\vee p\to \neg p\vee q))\to(((p\to q)\to \neg p\vee p)\to ((p\to q)\to \neg p\vee q))$$ $$\quad$$ [5 SSub: $$p/p\to q, q/\neg p\vee p, r/\neg p\vee q$$] (7) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\;\;$$[disjunct added to the left] (8) $$(p\to q)\to (\neg p\vee p\to \neg p\vee q)$$$$\;\;$$ [7 Sub: $$r/ \neg p$$] (9) $$((p\to q)\to \neg p\vee p)\to ((p\to q)\to \neg p\vee q)$$$$\;\;$$ [6,8 MP] (10) $$(p\to q)\to \neg p\vee q$$$$\;\;$$ [9,4 MP] (11) $$(p \to q) \to (\lnot q \to \lnot p)$$$$\;\;$$[Ax 3.] (12) (4) $$((p\to q) \to \lnot p \lor q) \to (\lnot (\lnot p \lor q) \to \lnot (p\to q))$$$$\;\;$$[11 SSub: $$p/p\to q,q/\lnot p \lor q$$] (13) $$\lnot(\lnot p \lor q)\to\lnot(p\to q)$$$$\;\;$$ [12,10 MP] (14) (211) $$\lnot p \land \lnot q\to \lnot(p\lor q)$$$$\;\;$$[De Morgan’s law] (15) $$\lnot\lnot p \land \lnot q\to \lnot(\lnot p\lor q)$$$$\;\;$$[14 Sub: $$p/\lnot p$$] (16) $$p\to \lnot\lnot p$$$$\;\;$$[Ax 5.] (17) (4) $$(p\to q)\to (r\to (p\to q))$$$$\;\;$$ [1 SSub: $$p/p\to q, q/r$$] (18) (0) $$(p\to(q\to r))\to(q\to(p\to r))$$$$\hspace{0,2 cm}$$[the law of commutation, included in line 3] (19) (6) $$(r\to(p\to q))\to(p\to(r\to q))$$$$\;\;$$[18 SSub: $$p/r,q/p,r/q$$] (20) (8) $$(p\to q)\to (p\to (r\to q))$$$$\;\;$$[17,19 HS] (21) (0) $$(p\to(q\to r))\to(p\land q\to r)$$$$\;\;$$ [the law of importation, included in line 3] (22) (4) $$(p\to(r\to q))\to(p\land r\to q)$$$$\;\;$$ [21 SSub: $$q/r,r/q$$] (23) (8) $$(p\to q)\to (p\land r\to q)$$$$\;\;$$[20,22 HS] (24) $$(p\to q) \to ((p \to r) \to (p \to q \wedge r))$$$$\;\;$$[Ax 8.] (25) $$(p\land r\to q) \to ((p\land r \to r) \to (p\land r \to q \wedge r))$$$$\;\;$$[24 Sub: $$p/p\land r$$] (26) (6) $$(p\land r \to r) \to ((p\land r\to q) \to (p\land r \to q \wedge r))$$$$\;\;$$[25 Com] (27) $$p \land q \to q$$$$\;\;$$[Ax 7.] (28) $$p \land r \to r$$$$\;\;$$[27 Sub: $$q/r$$] (29) $$(p\land r\to q) \to (p\land r \to q \wedge r)$$$$\;\;$$[26,28 MP] (30) (8) $$(p\to q)\to(p\land r\to q\land r)$$$$\;\;$$[23,29 HS] (31) (4) $$(p\to \lnot\lnot p)\to(p\land \lnot q \to \lnot\lnot p\land \lnot q)$$$$\;\;$$[30 SSub: $$q/\lnot\lnot p,r/\lnot q$$] (32) $$p\land \lnot q \to \lnot\lnot p \land \lnot q$$ [31,16 MP] (33) (8) $$p\land \lnot q \to \lnot(\lnot p\lor q)$$$$\;\;$$[32,15 HS] (34) (8) $$p\land\lnot q\to\lnot(p\to q)$$$$\;\;$$[33,13 HS] We have $$l_{p\land\lnot q\to\lnot(p\to q)}=651$$ and $$w_{p\land\lnot q\to\lnot(p\to q)}=51$$. We proceed with the following derivation: (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (2) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (3) (6) 51$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)) ~\land$$$$~~~~(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots))$$$$\;\;$$ [1,2 Conj] (4) (651) $$p\land\lnot q\to\lnot(p\to q)$$$$\;\;$$[proved above] (5) (4) $$A\land\lnot B\to\lnot(A\to B)$$$$\;\;$$[4 SSub: $$p/A, q/B$$] (6) (100) $$(p\to q)\to(p\lor r\to q\lor r)$$$$\;\;$$[disjunct added to the right] (7) (6) $$(A\land\lnot B\to \lnot(A\to B))\to((A\land\lnot B)\lor (\bigvee S_2) \to \lnot(A\to B)\lor (\bigvee S_2))$$ $$\quad$$ [6 SSub: $$p/A\land\lnot B,q/\lnot(A\to B),r/\bigvee S_2$$] (8) $$(A\land\lnot B)\lor (\bigvee S_2) \to \lnot(A\to B)\lor (\bigvee S_2)$$$$\;\;$$[7,5 MP] (9) (427) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$ [distributivity 1] (10) $$(A\lor(\bigvee S_2))\land(\lnot B\lor(\bigvee S_2)) \to (A\land\lnot B)\lor (\bigvee S_2)$$$$\;\;$$[9 SSub: $$p/A$$, $$q/\lnot B$$, $$r/\bigvee S_2$$] (11) (8) $$(A\lor(\bigvee S_2))\land(\lnot B\lor(\bigvee S_2)) \to \lnot(A\to B)\lor (\bigvee S_2)$$$$\;\;$$[10,8 HS] (12) (398) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [distributivity 2] $$\vdots$$ After this step we apply, again, the same mechanism of adding formulas $$C_n, \ldots$$, $$C_1$$ which was presented in the analysis of rule $$\mathbf{R}_{\land}$$. We calculate the length and the width as in the former cases and arrive at   \begin{equation}\label{nie implikacja} l_{\lnot\to} = l_{C1}+l_{C2}+21\cdot n+c \end{equation} (22) for $$c=1671$$, and   \begin{equation}\label{w nie implikacja} w_{\lnot\to} \leq \max\{w_{C1}, w_{C2}, c, 20\cdot|\phi|\}, \end{equation} (23) where $$\phi = \ \vdash S_1 \ ' \lnot(A\to B) \ ' S_2$$, $$c=51$$. Proof of Theorem 6 The schemas of proofs presented in subsections 4.2.1–4.2.4 supply us with the proof of Theorem 6. We have shown that if   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T_{1} \ ; \ \vdash T_{2} \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E}^\star$$ and there exist proofs of formulas $$\bigvee T_{1}$$ and $$\bigvee T_{2}$$ in system $$\mathbb{A}$$, then the proofs may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. We have also shown how to transform the proofs of $$\bigvee T_{1}$$ and $$\bigvee T_{2}$$ into a proof of $$\bigvee S$$. We have also seen that if $$l_{C1}, l_{C2}, l_P$$ and $$w_{C1}, w_{C2}, w_P$$ stand for the lengths and widths of the proofs of the formulas corresponding to, respectively, the ‘sequents-conclusions’ and the ‘sequent-premise’, then by (18), (20), (22): Corollary 6   $$l_p \leq l_{C1}+l_{C2}+f(n),$$ where $$n$$ stands for the number of terms of sequence $$T_i$$ ($$i=1$$ or $$i=2$$) and $$f$$ is a linear function of the form $$f(n)=21\cdot n + c$$, for some constant $$c\in\text{N}_+$$. Observe that $$T_1$$ and $$T_2$$ have the same number of terms, thus it is inessential whether $$i=1$$ or $$i=2$$. In turn, by (19), (21), (23): Corollary 7   $$w_p \leq \max\{w_{C1}, w_{C2}, 51,20\cdot|\vdash T_i|\}$$ 5 The algorithm and complexity issues Let $$\mathbf{s} = \langle Q_1,\ldots, Q_m \rangle$$ stand for a Socratic proof of sequent $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$. Then we use $$\mathbf{s}$$ according to the steps described below and ‘produce’ a proof of formula $$A$$ in system $$\mathbb{A}$$. (1) For each base sequent $$\phi$$ in the last question $$Q_m$$ of $$\mathbf{s}$$: a proof of the formula corresponding to $$\phi$$ is generated according to the schema presented in the proof of Theorem 4 (see subsection 4.1). (2) For $$i = m-1, \ldots, 1$$, (a) If the rule applied to question $$Q_i$$ is one of $$\mathbf{R}_{\lnot\lnot}$$, $$\mathbf{R}_{\to}$$, $$\mathbf{R}_{\lnot\land}$$, $$\mathbf{R}_{\lor}$$, then we use the proof of formula corresponding to the sequent-conclusion and transform it into a proof of formula corresponding to the sequent-premise according to the relevant schema presented in one of subsections 4.2.1–4.2.4. (b) If the rule applied to question $$Q_i$$ is one of $$\mathbf{R}_{\land}$$, $$\mathbf{R}_{\lnot\lor}$$, $$\mathbf{R}_{\lnot\to}$$, then we use the two proofs of formulas corresponding to the sequents-conclusions and transform them into a proof of formula corresponding to the sequent-premise according to the relevant schema presented in one of subsections 4.2.1–4.2.4. After the last iteration of for-loop (b) we arrive at a proof of the formula corresponding to sequent $$\vdash A$$ (i.e., formula $$A$$). We can see clearly that step 1. involves generating $$k$$ proofs, where $$k$$ is the number of different base sequents of question $$Q_m$$. (Observe that $$k$$ may be less than the number of constituents of the last question, since there may be repetitions among the constituents.) Step 2 is repeated $$m-1$$ times, where $$m$$ is the number of questions in $$\mathbf{s}$$. Using the algorithm sketched above, to each Socratic proof $$\mathbf{s}$$ in $$\mathbb{E}^{\star\star}$$ of sequent $$\vdash A$$ we can assign a proof $$\mathbf{p}$$ of formula $$A$$ in $$\mathbb{A}$$. This assignment can be formally expressed as a function $$\mathcal{F}$$ with the set of Socratic proofs as the domain and the set of axiomatic proofs as the codomain. Therefore by $$\mathcal{F}(\mathbf{s})$$, where $$\mathbf{s}$$ is a Socratic proof of $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$, we shall mean the proof of $$A$$ in $$\mathbb{A}$$ assigned to $$\mathbf{s}$$ by the above algorithm. Now we shall prove what follows. Lemma 1 Suppose that $$\mathbf{s}$$ is a Socratic proof of sequent $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$ and let $$\mathbf{p}=\mathcal{F}(\mathbf{s})$$. Then: (1) $$l_\mathbf{p}=\mathcal{O}(|A|)$$, where $$l_\mathbf{p}$$ is the length of $$\mathbf{p}$$ (2) $$w_\mathbf{p}=\mathcal{O}(|A|)$$, where $$w_\mathbf{p}$$ is the width of $$\mathbf{p}$$ (3) $$s_\mathbf{p}=\mathcal{O}(|A|^2)$$, where $$s_\mathbf{p}$$ is the size of $$\mathbf{p}$$ In the proof of Lemma 1 we will use the notion of level of Socratic proof, which is understood as follows. Let $$\mathbf{s}$$ be a Socratic transformation (recall that $$\mathbf{s}$$ is a finite sequence of questions which does not have to end with a success, as a Socratic proof does, see Definitions 2, 3). Further, let $$\underline{\phi}$$ stand for an occurrence of sequent $$\phi$$ in $$\mathbf{s}$$. Suppose also that $$\underline{\phi}$$ occurs in $$k$$-th question of $$\mathbf{s}$$. Then we will say that $$\underline{\phi}$$is at$$k$$-th level of$$\mathbf{s}$$. If it does not lead to any confusion, we will neglect the distinction between a sequent and its occurrence and will say that sequent $$\phi$$is at$$k$$-th level of$$\mathbf{s}$$ or that $$\phi$$occurs at$$k$$-th level of$$\mathbf{s}$$. In the proof of Lemma 1 we will also make use of the following fact: Fact 1 Let $$\mathbf{s}$$ be a Socratic proof of sequent $$\vdash A$$ and suppose that sequent $$\phi$$ occurs in $$\mathbf{s}$$ and that it has $$n$$ terms. Then   $$n \leq |\phi| \leq |A|.$$ The first inequality is obvious and the second may be proved by inspection of the rules of $$\mathbb{E}^{\star\star}$$ (and by induction). The point is that if $$\psi$$ is a sequent-premise of a rule and $$\chi$$ is its sequent-conclusion (or one of the two sequents-conclusions), then $$|\chi| \leq |\psi|$$. Proof of Lemma 1: Suppose, as in the statement of Lemma 1, that $$\mathbf{s}$$ is a Socratic proof of sequent $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$ and that $$\mathbf{p}=\mathcal{F}(\mathbf{s})$$. Moreover, let $$\phi$$ be a sequent occurring in $$\mathbf{s}$$. When the proof $$\mathbf{p}$$ is generated, the proof of the formula corresponding to sequent $$\phi$$ must be generated as well. Let $$\mathbf{p}_\phi$$ stand for this proof, $$l_\phi$$ and $$w_\phi$$ will symbolize its length and width, respectively. We will show that   \begin{equation}\label{clause 1 length} l_{\phi} = \mathcal{O}(|\phi|). \end{equation} (24) Once this fact is proved, it will follow that for $$\phi \ = \ \vdash A$$ we have $$l_\mathbf{p}=\mathcal{O}(|A|)$$, as required by clause 1. of Lemma 1. We prove (24) by (reversed) induction with respect to the level at which sequent $$\phi$$ occurs in $$\mathbf{s}$$. Thus suppose that $$\mathbf{s}=\langle Q_1,\ldots, Q_m\rangle$$, i.e. $$\mathbf{s}$$ has $$m$$ terms (questions), and that $$\phi$$ occurs at level $$m$$ in $$\mathbf{s}$$. Then $$\phi$$ is a base sequent and by Corollary 1, $$l_{\phi} = \mathcal{O}(|\phi|)$$. Suppose that $$\phi$$ occurs at level $$k$$, where $$1\leq k < m$$. Then we consider two cases. First, it may happen that there is no application of an erotetic rule to questions $$Q_k,\ldots, Q_{m-1}$$ that is performed with respect to an occurrence of $$\phi$$ (here we mean arbitrary occurrence of $$\phi$$). In other words, $$\phi$$ is then a base sequent which is rewritten from question to question as a part of the context, until the last question is arrived at. Then we repeat the above argument concerning base sequents in the last question. In the second case a rule must be applied with respect to sequent $$\phi$$. To be precise, it may happen that the rule applied to $$k$$-th question of $$\mathbf{s}$$ acts on some other sequent, and $$\phi$$ is a part of the context which is rewritten in the next question. However, a rule must be finally applied with respect to an occurrence of $$\phi$$ in one of the consecutive questions. Suppose that the rule is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi \ ; \ \Psi)}. $$ Sequent $$\psi$$ occurs in $$\mathbf{s}$$ at a level not less than $$k+1$$. Therefore (induction hypothesis):   $$l_\psi = \mathcal{O}(|\psi|),$$ where $$l_\psi$$ is the length of the proof of the formula corresponding to $$\psi$$ (again, the proof must be generated when $$\mathcal{F}(\mathbf{s})$$ is produced). In other words, the induction hypothesis yields that $$l_\psi \leq g(|\psi|)$$ for a certain linear function $$g$$. On the other hand, by Corollary 4, $$l_\phi \leq l_\psi + f(n)$$, where $$n$$ is the number of terms of $$\psi$$ and $$f$$ is a linear function. Finally, we know that $$|\psi| \leq |\phi|$$. To sum up:   $$ l_\phi \leq l_\psi + f(n) \leq g(|\psi|) + f(n) \leq g(|\psi|) + f(|\psi|) \leq g(|\phi|) + f(|\phi|). $$ The two inequalities from the right hold since $$n\leq|\psi|$$ (see Fact 1) and both $$f$$ and $$g$$ are growing functions. Thus we finally arrive at the conclusion that:   $$ l_\phi \leq h(|\phi|), $$ where $$h=g+f$$ is a linear function.12 If the rule applied with respect to $$\phi$$ is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi_1 \ ; \ \psi_2 \ ; \ \Psi)}, $$ then by induction hypothesis we have:   $$l_{\psi_1} \leq g_1(|\psi_1|), l_{\psi_2} \leq g_2(|\psi_2|),$$ where $$l_{\psi_1}$$ and $$l_{\psi_2}$$ are the lengths of the proofs of formulas corresponding to $$\psi_1$$ and $$\psi_2$$, respectively, and $$g_1$$, $$g_2$$ are certain linear and growing functions. By Corollary 6, $$l_\phi \leq l_{\psi_1}+l_{\psi_2}+f(n)$$ where $$n$$ is the number of terms of $$\psi_1$$ (or $$\psi_2$$) and $$f$$ is a linear (and growing) function, therefore:   $$l_\phi \leq l_{\psi_1}+l_{\psi_2}+f(n) \leq g_1(|\psi_1|) + g_2(|\psi_2|) + f(|\psi_1|).$$ Since also $$|\psi_1| \leq |\phi|$$ and $$|\psi_2| \leq |\phi|$$, defining $$h$$ as $$(g_1 + g_2)+f$$ we finally obtain:   $$l_\phi \leq h(|\phi|),$$ where $$h$$ is a linear function. All this entails that (24) is true, and therefore $$l_\mathbf{p}=\mathcal{O}(|A|)$$. The second clause of Lemma 1, concerning width, will be proved in an analogous way. We show that   \begin{equation}\label{clause 2 width} w_{\phi} = \mathcal{O}(|\phi|), \end{equation} (25) holds for each $$\phi$$ occurring in the Socratic proof $$\mathbf{s}$$, and we do it by (reversed) induction with respect to the level at which sequent $$\phi$$ occurs in $$\mathbf{s}$$. If $$\phi$$ is a base sequent at level $$m$$ (where $$\mathbf{s}$$ has $$m$$ terms), then, by Corollary 2, $$w_{\phi} = \mathcal{O}(|\phi|)$$. In the inductive part we analyse, again, two cases: when no rule is applied with respect to $$\phi$$, and when there is one. Here we consider only the second case. Suppose that the rule is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi \ ; \ \Psi)}. $$ Sequent $$\psi$$ occurs in $$\mathbf{s}$$ at a level not less than $$k+1$$, thus by induction hypothesis:   $$w_\psi \leq g(|\psi|),$$ where $$w_\psi$$ is the width of the proof of the formula corresponding to $$\psi$$, and $$g$$ is a linear function. By Corollary 5, $$w_\phi \leq \max\{w_\psi,c,8\cdot|\psi|\}$$, where $$c\leq 51$$. This obviously yields that $$w_\phi \leq h(|\psi|)$$ for some linear (growing) function $$h$$.13 Finally, since $$|\psi| \leq |\phi|$$ and $$h$$ is a growing function:   $$w_\phi \leq h(|\phi|).$$ If the rule applied with respect to $$\phi$$ is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi_1 \ ; \ \psi_2 \ ; \ \Psi)} $$ then by induction hypothesis we have:   $$w_{\psi_1} \leq g_1(|\psi_1|), w_{\psi_2} \leq g_2(|\psi_2|),$$ where $$w_{\psi_1}$$ and $$w_{\psi_2}$$ are the widths of the proofs of the formulas corresponding to $$\psi_1$$ and $$\psi_2$$, respectively, and $$g_1$$, $$g_2$$ are certain linear and growing functions. By Corollary 6 and by the fact that $$|\psi_i| \leq |\phi|$$ for $$i=1,2$$, $$w_\phi \leq \max\{w_{\psi_1}, w_{\psi_2}, 51, 20\cdot|\phi|\}$$. Again, this yields that:   $$w_\phi \leq h(|\phi|),$$ where $$h$$ is a linear function. All this entails that (24) is true, and therefore $$w_\mathbf{p}=\mathcal{O}(|A|)$$. Clause 3 follows from the previous two, as $$s_\mathbf{p}=l_\mathbf{p}\cdot w_\mathbf{p}$$. ■ We did not calculate the size of Socratic proof $$\mathbf{s}$$ of $$\vdash A$$ and we shall not do this. For the purpose of this article it is enough to observe that the size of $$\mathbf{s}$$, understood as the number of all occurrences of signs in $$\mathbf{s}$$, is clearly a function of $$|A|$$. As we have seen, there exists a function, $$\mathcal{F}$$, which assigns to $$\mathbf{s}$$ a proof of formula $$A$$ in $$\mathbb{A}$$ and which is such that the size of $$\mathbf{p}$$ is $$\mathcal{O}(|A|^2)$$. This means that: Theorem 7 Axiomatic system $$\mathbb{A}$$ simulates polynomially erotetic calculus $$\mathbb{E}^{\star\star}$$. 5.1 Back to the R-S system As we have pointed out in the Introduction and in Section 3, there is a close affinity between the right-sided version of the method of Socratic proofs and the R-S method. The affinity follows from the fact that the two methods are based on deriving a Conjunctive Normal Form. The Reader could see this on the examples presented in Sections 2 and 3. The consequence of the affinity of systems is that the erotetic one could be easily replaced with the R-S system in deriving the results presented in this article: the fundamental sequences would replace the base sequents and the rules of the R-S system would replace the rules of $$\mathbb{E^{**}}$$. This means that the algorithm presented in Section 5 works equally well for the case of the R-S system. It follows that axiomatic system $$\mathbb{A}$$ can be shown to simulate polynomially the R-S system. 6 Conclusions and further work The aim of this work was to automatize the process of proof-search in axiomatic system $$\mathbb{A}$$ for CPL. The results presented in this article allow us to transform any algorithm of proof-search by the method of Socratic proofs (in calculus $$\mathbb{E^{**}}$$) into an algorithm of proof-search in axiomatic system $$\mathbb{A}$$. Moreover, the estimations of the lengths and widths of proofs show that axiomatic system $$\mathbb{A}$$ simulates polynomially erotetic calculus $$\mathbb{E}^{\star\star}$$. It is a common perception that proving by means of axioms and rules is a difficult process, in which intuitions and high logical skills are necessarily involved. The results presented here neither prove nor disprove this common belief. We are convinced, however, that the detailed constructions and calculations presented here may shed a new light on the process of proof-search in axiomatic systems. We especially hope that the algorithm sketched in Section 5 will be implemented. Certainly, this is one of the perspectives for the future. The first author admits that the meticulous process of preparing the proofs in the axiomatic system of Tadeusz Batóg was the most difficult but also the most satisfying part of this work. The second author must admit that it would be a shame if we automatized it all. Funding The work of the second author was supported by the Polish National Science Center, grant no [2012/04/A/HS1/00715]. This work originated as a master’s thesis [16] (in Polish) defended by the first author under the supervision of the second author in the Department of Logic and Cognitive Science, Institute of Psychology, Adam Mickiewicz University. Part of the proofs have been changed and the issues concerning complexity have been added in this version. We wish to thank the Referee of the master’s thesis, Professor Andrzej Wiśniewski, and the anonymous Referees of the article for their detailed corrections, valuable comments and suggestions, which helped us to improve this article. Footnotes 1 We will use the term ‘axiomatic system’. 2 To get an idea of the progress in this field see e.g. [12]. Since 1992, the Workshop on Theorem Proving with Analytic Tableaux and Related Methods, and since 1997, the conference TABLEAUX Automated Reasoning with Analytic Tableaux and Related Methods have gathered together researchers in this discipline and published their proceedings. 3 Here by ‘tableau methods’ we mean analytic tableaux like those presented originally by Smullyan [32]. In [9] the authors propose an improved — among others, in terms of efficiency — version of analytic tableaux named the KE system, which was also popularized in [10] and [11]. 4 The acronym ‘RPQ’ is for ‘Right’, ‘Pure’, ‘Quantifier’, where ‘Pure’ is to indicate the fact that the proofs start with pure sentences of the language of $$\mathsf{FOL}$$, that is, sentences containing no parameters. 5 A terminological remark is in order here. What we mean by simulate here is that we can show that there is a function which to every Socratic proof assigns a proof in a sequent calculus, or a function which to every Socratic proof assigns a closed analytic tableau, etc. In terms of simulations used to examine the relative complexity of proof systems this means that the sequent calculus simulates erotetic calculus, or that the tableau system simulates erotetic calculus, where the question remains how fast it does it. 6 For an example of proof-search algorithm written in pseudocode see [26, Appendix]. Admittedly, the algorithm is written for the both-sided version of erotetic calculus $$\mathbb{E}^*$$, but the variant for $$\mathbb{E}^{**}$$ is easily obtained from it. 7 In the details concerning language and axiomatic system this work is based on [1]. 8 This is based on the system by Hilbert and Bernays presented in 1934 (see [17, p. 65]). There are differences, however, between the systems in [1] and [17]. In the latter the laws $$(A \rightarrow (A \rightarrow B)) \rightarrow (A \rightarrow B)$$ and $$(A \rightarrow B) \rightarrow ((B \rightarrow C) \rightarrow (A \rightarrow C))$$ are present instead of the Frege law $$(p \to (q \to r)) \to ((p \to q) \to (p \to r))$$, so each connective is defined by a group of three axioms and the system has 15 axioms. 9 The axiomatic system in [31] is based on axiom schemas and the solely rule modus ponens. In the quotation we have used ‘$$A$$’ instead of ‘$$\beta$$’, to avoid a mix-up of notation. A ‘fundamental’ formula is a disjunction of the terms of a fundamental sequence. 10 The word ‘erotetic’ comes from the Greek word ‘erotema’ which means ‘question’. 11 Formula $$(A_{i+l-3} \lor (A_{i+l-2} \lor A_{i+l-1})) \lor \lnot B\to A_{i+l-3} \lor ((A_{i+l-2} \lor (A_{i+l-1}) \lor \lnot B)$$ has been reached before line $$(j)$$. 12 By $$h=g+f$$ we mean that $$h$$ is defined on the domain of $$g$$ and $$f$$, here $$\text{N}_+$$, and $$h(n)=g(n)+f(n)$$. 13 Let $$h(n)=a\cdot n + b$$. For $$a$$ we take the maximum of the two values: the first coefficient of $$g$$, $$8$$. For $$b$$ we take the maximum of the two values: the second coefficient of $$g$$, $$51$$. References [1] Batóg. T. Podstawy Logiki . AMU Press, 2003. [2] Buss. R. S. Polynomial size proofs of the pigeonhole principle. Journal of Symbolic Logic , 52, 916– 927, 1987. Google Scholar CrossRef Search ADS   [3] Buss. R. S. An introduction to proof theory. In Handbook of Proof Theory , Buss S. R. ed., chapter I, pp. 1– 78. 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Socratic proofs for quantifiers. Journal of Philosophical Logic , 35, 147– 178, 2006. Google Scholar CrossRef Search ADS   [41] Wiśniewski A. Vanackere G. and Leszczyńska. D. Socratic proofs and paraconsistency: a case study. Studia Logica , 80, 433– 468, 2004. © The Author 2017. Published by Oxford University Press. All rights reserved. For Permissions, please email: journals.permissions@oup.com http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Logic Journal of the IGPL Oxford University Press

Automatic proof generation in an axiomatic system for $$\mathsf{CPL}$$ by means of the method of Socratic proofs

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Abstract

Abstract The aim of this article is to describe an algorithm for the automatic generation of proofs in an axiomatic system for Classical Propositional Logic. The idea of the algorithm was taken from a book by Helena Rasiowa and Roman Sikorski [31], where the authors suggest using the method of diagrams to automatically obtain proofs in Classical Propositional Logic. However, in this article the method of diagrams developed by Rasiowa and Sikorski was replaced by a right-sided erotetic calculus developed by Andrzej Wiśniewski. Erotetic calculi have been used before in designing similar algorithms. The proofs are presented together with the estimations of their lengths, widths and sizes — measures introduced for the purposes of the article. The estimations are then used to derive the conclusion that the axiomatic system simulates polynomially the erotetic calculus. 1 Introduction Despite the intensive development of the discipline known as proof-theory, axiomatic systems, also called ‘Frege systems’ or ‘Hilbert systems’,1 are still one of the most universal proof systems, capable of capturing a very wide range of logics for which no other proof systems are known (see [19, p.26]). Although it is acknowledged that in the case of Classical Propositional Logic ($$\mathsf{CPL}$$ for short) purely automatic proof generation is obtainable, e.g. by translating natural deduction proofs into axiomatic system proofs (see e.g. [33, p.34], more references are given below), axiomatic systems are often regarded as difficult to implement or resulting in inefficient algorithms (see [3, p.18]). Probably for this reason there is little interest in the subject, contrary to proof-search by — the still unquestionable leader in this field — resolution, or by tableau and related methods.2 On the other hand, an important motivation to study axiomatic systems comes from theory of computational complexity. During the last few decades, researchers focusing on the computational complexity of proof systems for $$\mathsf{CPL}$$ have shown, int.al., that in the case of $$\mathsf{CPL}$$ axiomatic systems, d.a.g.-like (from directed acyclic graph) natural deduction systems, and sequent calculi with cut belong to the same complexity class ([9], [7], [8], [29]). Moreover, it is known that these systems for $$\mathsf{CPL}$$ are essentially more efficient than sequent systems without cut, resolution or tableau methods3 for $$\mathsf{CPL}$$ (except for [9], [7], [8], [29] see also [2], [34], [6]). After the first presentation of an axiomatic system by Frege in 1879, the discussed results make axiomatic systems for $$\mathsf{CPL}$$ — again — a highly attractive research subject. Translating natural deduction proofs into axiomatic system proofs is actually the content of the proof of Deduction Theorem. The reader will find many constructions of this kind in logic textbooks such as [33] (where the construction is described for the case of intuitionistic logic with the classical case left as an exercise), [13] or the classic textbook [20]. The book [33, p. 34] also mentions a possible translation from natural deduction proofs to axiomatic proofs via$$\lambda$$-terms of type theory. Obviously, a proof-search algorithm for natural deduction must be added to these results to achieve a proof-search algorithm for axiomatic systems. This is not a problem (see e.g. [19]), but algorithms based on a version of the method of diagrams may seem more straightforward. In their joint paper [30], and later in the monograph [31] published in 1963, the authors — Helena Rasiowa and Roman Sikorski — present the method of diagrams of formulas, which nowadays is variously called the Rasiowa-Sikorski method, Rasiowa-Sikorski diagrams, or simply the R-S system. The R-S system for $$\mathsf{CPL}$$ may be used, as the authors claim [31, p. 269], to generate a proof in an axiomatic system for $$\mathsf{CPL}$$ in a fully automatic way. It is worth noticing that the R-S systems have been developed for many logics and found various important applications, among others, in the area of computer science — see e.g. [28], [15], [21], [22], [14]. The researchers working with R-S systems have introduced the term ‘dual tableaux’ for it, to emphasize the duality of R-S systems with respect to analytic tableau systems. The former may be interpreted as deriving the Conjunctive Normal Form of a formula $$A$$, and is a ‘validity checker’, whereas the latter attempts to build a proof of $$A$$ by deriving the Disjunctive Normal Form of $$\lnot A$$ and is thus an ‘unsatisfiability checker’. The idea of exploring the relation between R-S systems and axiomatic systems has been undertaken by Ewa Orłowska and Joanna Golińska-Pilarek in [28] on the level of the First-Order Logic ($$\mathsf{FOL}$$ for short). In the book, the authors present an effective procedure ([28, pp. 22–24]) transforming an RS-proof (i.e., one conducted in the framework of the R-S system) of a formula of the language of $$\mathsf{FOL}$$ into a proof of the formula in an axiomatic system. The focus of the approach in [28] is on the first-order level, therefore the propositional level is, in a way, neglected, as it is assumed that every formula of the language of $$\mathsf{FOL}$$ which falls under a tautological schema is an axiom of the axiomatic system used in their work. Moreover, the authors introduce into their axiomatic system derived rules which are ‘induced by’ the rules of the R-S system, i.e., which map the applications of the R-S rules. On the propositional level, the effective procedure then becomes a quite straightforward imitation of the applications of R-S rules inside the axiomatic system. Thus we have not found in [28] the answer to the question of how to generate a proof in an axiomatic system for $$\mathsf{CPL}$$ with a finite number of axioms and with the rule of substitution — the book was not intended to answer the question; this article is. The right-sided erotetic calculus $$\mathbb{E}^{**}$$ is a proof system grounded in Inferential Erotetic Logic, developed by Andrzej Wiśniewski. $$\mathbb{E}^{**}$$ is the propositional part of the calculus $$\mathbf{E^{RPQ}}$$ for $$\mathsf{FOL}$$ described in [38].4 As we will show in this article, calculus $$\mathbb{E}^{**}$$ may be interpreted as an erotetic version of the R-S system. In fact, the results obtained in this article with the use of the erotetic calculus can be easily reformulated for the case of the R-S system. We emphasize this fact again at the end of this article. The reason to choose the method of Socratic proofs is that it has been used successfully in designing similar algorithms as the one presented in this article, hence the results presented here constitute a part of a wider research project. In [26] the method of Socratic proofs in its both-sided version has been used to simulate the derivation process in sequent calculi both for $$\mathsf{CPL}$$ and $$\mathsf{FOL}$$. In [25] (unpublished) this result is extended to propositional modal logics. Moreover, in [25] the left-sided version of the method of Socratic proofs is used to simulate the derivation process in an analytic tableau system for $$\mathsf{CPL}$$ and $$\mathsf{FOL}$$. There is also another variant of the method of Socratic proofs, based on the so-called reversed sequents, which has been used in [5] to simulate the method of resolution for $$\mathsf{CPL}$$ and some non-classical propositional logics.5 The summarized results show that due to the use of various sequents — or in fact, due to the potential of the structure of a sequent and the numerous interplays between its sides — it is possible to obtain many results concerning the relations between different proof methods, and we think it is important to obtain the results in a uniform manner, i.e., with the use of one proof method. Taking this all into account, as well as the fact that we have at our disposal some algorithms of proof-search by means of the method of Socratic proofs together with an implementation in Prolog,6 we have decided to use $$\mathbb{E}^{**}$$ instead of the R-S system. The formal result of this article is a procedure which transforms a Socratic proof of a sequent $$\vdash A$$ in erotetic calculus $$\mathbb{E}^{**}$$ into a proof of a formula $$A$$ in the axiomatic system $$\mathbb{A}$$ for $$\mathsf{CPL}$$. We introduce three measures to estimate the complexity of the presented proofs: length (the number of lines of a proof), width (the maximum number of occurrences of symbols in a line of a proof), and size (the product of the previous two). Apart from the procedure itself and the detailed calculations which may probably be useful in the implementation of the procedure, the calculations presented in this article yield that the axiomatic system $$\mathbb{A}$$polynomially simulates the erotetic calculus $$\mathbb{E}^{**}$$. Taking into account what we know about the relative complexity of various proof systems this result does not seem surprising. Nevertheless, there are no such detailed results concerning the relation between the axiomatic systems and the erotetic calculi. To the best of our knowledge, there are also no results concerning the relation between the axiomatic systems and the R-S systems, except for that presented in [28]. The article is structured as follows. In Section 2, we describe an axiomatic system for $$\mathsf{CPL}$$, namely $$\mathbb{A}$$, and we recall the original, neat presentation of the R-S system for $$\mathsf{CPL}$$. In Section 3, we describe the method of Socratic proofs, and in Section 4 — the necessary proofs obtained in the system $$\mathbb{A}$$ together with the estimations of their sizes. Section 5 contains a sketch of the whole procedure and discusses some complexity issues. We end up with some conclusions and perspectives for further research. 2 Classical propositional logic: axiomatic system and the Rasiowa-Sikorski system 2.1 Language and semantics Let $$\mathcal{L}$$ be the language of $$\mathsf{CPL}$$ with countably infinitely many propositional variables: $$p_{1}, p_{2}, p_{3}, \ldots$$, connectives: negation ($$\neg $$), implication ($$\to$$), conjunction ($$\wedge$$), disjunction ($$\vee$$), equivalence ($$\leftrightarrow$$) and brackets: $$(,)$$. By FORM$$_{\mathcal{L}}$$ we mean the set of all formulas of $$\mathcal{L}$$, defined in a usual way.7 Till the end of Section 2, instead of ‘formula of $$\mathcal{L}$$’ we will write simply ‘formula’. For simplicity, the letters $$p, q, r, s$$ will be used as propositional variables, and $$p_i$$ will be used as a metavariable. As metavariables for formulas we will use $$A, B, C, D, F$$, possibly with subscripts. Symbols $$X, X_{1}, X_{2}$$ will stand for sets of formulas. We also adopt the usual conventions concerning saving on parentheses. $$\mathcal{L}$$ is equipped with the standard semantics based on Boolean valuations. 2.2 Axiomatic system $$\mathbb{A}$$ for $$\mathsf{CPL}$$ We use the system presented in [1, pp. 42–43],8 for which we shall use the symbol $$\mathbb{A}$$.    Axioms for implication     Axioms for negation  1.  $$p \to (q \to p)$$  3.  $$(p \to q) \to (\neg q \to \neg p)$$  2.  $$( p \to (q \to r)) \to ((p \to q) \to (p \to r))$$  4.  $$\neg \neg p \to p$$        5.  $$p \to \neg \neg p$$     Axioms for conjunction     Axioms for disjunction  6.  $$p \wedge q \to p$$  9.  $$p \to p \vee q $$  7.  $$p \wedge q \to q$$  10.  $$q \to p \vee q$$  8.  $$(p \to q) \to ((p \to r) \to (p \to q \wedge r))$$  11.  $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$     Axioms for equivalence        12.  $$(p\leftrightarrow q) \to (p \to q)$$        13.  $$(p \leftrightarrow q) \to (q \to p)$$        14.  $$(p\to q)\to ((q \to p)\to (p\leftrightarrow q))$$           Axioms for implication     Axioms for negation  1.  $$p \to (q \to p)$$  3.  $$(p \to q) \to (\neg q \to \neg p)$$  2.  $$( p \to (q \to r)) \to ((p \to q) \to (p \to r))$$  4.  $$\neg \neg p \to p$$        5.  $$p \to \neg \neg p$$     Axioms for conjunction     Axioms for disjunction  6.  $$p \wedge q \to p$$  9.  $$p \to p \vee q $$  7.  $$p \wedge q \to q$$  10.  $$q \to p \vee q$$  8.  $$(p \to q) \to ((p \to r) \to (p \to q \wedge r))$$  11.  $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$     Axioms for equivalence        12.  $$(p\leftrightarrow q) \to (p \to q)$$        13.  $$(p \leftrightarrow q) \to (q \to p)$$        14.  $$(p\to q)\to ((q \to p)\to (p\leftrightarrow q))$$        The set of the 14 axioms presented above will be referred to by $$Apc$$ (this is the English-language version of the symbol $$Arz$$ used in [1]), whereas the set of all valid formulas of $$\mathsf{CPL}$$ will be represented as $$Tpc$$ (again, instead of $$Trz$$ used in [1]). Theorems of $$\mathsf{CPL}$$ are derived in $$\mathbb{A}$$ from the above axioms by the rule of modus ponens (MP) and / or the rule of substitution (Sub):   \[\begin{matrix} \frac{\begin{matrix} \begin{matrix} A\to B \\ A \\ \end{matrix} \\ \end{matrix}}{B} & {} & {} & \frac{B}{S(A,{{p}_{i}},B)}, \\ \end{matrix}\] where the operation of substitution of formula$$A$$for variable$$p_{i}$$in formula$$B$$ is defined in a usual way. Proof of formula$$A$$from set$$X$$of formulas in$$\mathbb{A}$$ is also defined in a standard way: as a finite sequence of formulas of $$\mathcal{L}$$ ending with $$A$$ and such that each formula of the sequence is an element of $$X$$ or results from some previous term(s) of the sequence by MP or Sub. If there exists a proof of $$A$$ from $$X$$ in $$\mathbb{A}$$, then we write $$A\in Cn(X)$$ ($$Cn$$ stands for the consequence operator). Obviously, the following holds: Theorem 1 (Soundness and completeness of $$\mathbb{A}$$ with respect to Boolean semantics)   $$Cn(Apc) = Tpc.$$ 2.3 The Rasiowa and Sikorski system Below we describe the elegant and simple R-S system for $$\mathsf{CPL}$$. This presentation is strictly based on [31, Chapter VII], we adjust the notation, however, to that used in this article. In this subsection, $$\Gamma$$ will be used as a metavariable for (possibly empty) sequences of formulas of the form:   \begin{equation}\label{1} A_{1},...,A_{m}. \end{equation} (1) If $$\Gamma$$ is of the form (1), $$\Gamma^\sharp$$ is a sequence of formulas: $$B_{1},...,B_{n}$$ and $$A$$, $$B$$ are formulas, then the following expressions:   \begin{eqnarray*} &\Gamma \ ' \ \Gamma^\sharp&\nonumber\\[3pt] &\Gamma \ ' \ A \ ' \ \Gamma^\sharp&\nonumber\\[3pt] &\Gamma \ ' \ A \ ' \ B \ ' \ \Gamma^\sharp& \end{eqnarray*} will denote, respectively, the following sequences:   \begin{eqnarray*} &A_{1},...,A_{m}, B_{1},...,B_{n}&\\[3pt] &A_{1},...,A_{m}, A, B_{1},...,B_{n}&\\[3pt] &A_{1},...,A_{m}, A, B, B_{1},...,B_{n}.& \end{eqnarray*} The symbol ‘$$'$$’ is therefore used for the concatenation of finite sequences of formulas, where the cases of single formulas $$A$$, $$B$$ are treated as cases of single-term sequences $$\langle A \rangle$$, $$\langle B \rangle$$. We say that a formula is indecomposable if it is either a propositional variable or the negation of a propositional variable. A sequence (1) is said to be indecomposable, if its terms are indecomposable formulas; it is said to be fundamental if it contains a formula and its negation. By a rule schema we shall understand a pair $$\langle \Gamma,\Gamma^{0} \rangle$$ or a triple $$\langle \Gamma,\Gamma^{0}, \Gamma^{1}\rangle$$ of finite, non-empty sequences of formulas. These are presented as follows:   \begin{equation}\label{2} \frac{\Gamma}{\Gamma^{0}} \end{equation} (2)  \begin{equation}\label{3} \frac{\Gamma}{\Gamma^{0}~;~\Gamma^{1}}. \end{equation} (3) The authors of the method call $$\Gamma$$ the conclusion of the rule schema, and $$\Gamma^{0},\Gamma^{1}$$ — its premises. In the case of schema (3), $$\Gamma^{0},\Gamma^{1}$$ are called the left and the right premise, respectively. Let us observe that, as in sequent calculi, this naming preserves the direction of proving, though not the direction of proof-search; the process of proving starts with the premises (axioms, or fundamental sequences) to reach the final conclusion (a thesis), whereas the process of proof-search goes in the opposite direction. The R-S system for $$\mathsf{CPL}$$ has no axioms and is composed of the following rule schemas (symbol $$\Gamma^*$$ always denotes an indecomposable sequence, which may be empty):   \begin{align*} &(D) ~~~~~~~~& \frac{\Gamma^*,(A \lor B), \Gamma^{**}}{\Gamma^*, A, B, \Gamma^{**}} & & & (\lnot D) ~~~~~~~~& \frac{\Gamma^*,\lnot(A \lor B), \Gamma^{**}}{\Gamma^*,\lnot A,\Gamma^{**}~;~\Gamma^*,\lnot B, \Gamma^{**}}\\ \\ &(\lnot C) & \frac{\Gamma^*,\lnot(A \land B), \Gamma^{**}}{\Gamma^*,\lnot A,\lnot B, \Gamma^{**}} & & & (C) & \frac{\Gamma^*,(A \land B), \Gamma^{**}}{\Gamma^*,A,\Gamma^{**}~;~\Gamma^*,B, \Gamma^{**}}\\ \\ &(I) & \frac{\Gamma^*,(A \rightarrow B), \Gamma^{**}}{\Gamma^*,\lnot A, B, \Gamma^{**}} & & & (\lnot I) & \frac{\Gamma^*,\lnot(A \rightarrow B), \Gamma^{**}}{\Gamma^*,A,\Gamma^{**}~;~\Gamma^*,\lnot B, \Gamma^{**}}\\ \\ &(\lnot N) & \frac{\Gamma^*,\lnot \lnot A, \Gamma^{**}}{\Gamma^*, A, \Gamma^{**}} \end{align*} As we can see, the rules of the forms $$(D), (\lnot C), (I), (\lnot N)$$ are of type (2), whereas the rules of the forms $$(\lnot D), (C), (\lnot I)$$ are of type (3). Observe that if a sequence $$\Gamma$$ of formulas is not indecomposable, then $$\Gamma$$ is the conclusion of one of these schemas. The assumption that $$\Gamma^*$$ denotes exclusively indecomposable sequences warrants that $$\Gamma$$ is the conclusion of exactly one of the rule schemas. As we shall see, the notion of diagram of a formula will be introduced as a certain partial mapping; the mappings will be defined on finite binary sequences. Therefore, the letters i, j will denote finite, possibly empty, sequences:   \begin{equation}\label{4} i_{1},...,i_{n} \end{equation} (4) of integers $$0$$, $$1$$. We will write j$$\le$$i, if j is an initial (proper or improper) segment of i. If i is of the form (4), then i, 0 and i, 1 will denote the sequences:   $$i_{1},...,i_{n}, 0$$ and   $$i_{1},...,i_{n}, 1$$ respectively. The empty sequence will be denoted by $$\mathbf{O}$$. By definition, $$\mathbf{O} \le$$i for each i. Now we are in a position to introduce the definition of diagram. The following is almost an exact quotation from [31, p. 266]. Definition 1 By the diagram of a formula$$A$$ we shall mean any mapping which, with certain sequences $${\boldsymbol{i}}$$, associates non-empty finite sequences $$\Gamma_{\boldsymbol{i}}$$ of formulas, and which is defined by induction as follows: (1) $$\Gamma_{\mathbf{O}}$$ is the sequence formed only of the formula $$A$$. (2) If $$\Gamma_{\boldsymbol{i}}$$ is defined but is either fundamental or indecomposable, then $$\Gamma_{\boldsymbol{i},0}$$ and $$\Gamma_{\boldsymbol{i},1}$$ are not defined. (3) If $$\Gamma_{\boldsymbol{i}}$$ is defined and is neither fundamental nor indecomposable, then $$\Gamma_{\boldsymbol{i}}$$ is the conclusion of exactly one of the schemas of rules. If $$\Gamma_{\boldsymbol{i}}$$ is the conclusion of schema $$(D), (\lnot C), (I)$$ or $$(\lnot N)$$, then $$\Gamma_{\boldsymbol{i},0}$$ is the only premise of this schema and $$\Gamma_{\boldsymbol{i},1}$$ is not defined. If $$\Gamma_{\boldsymbol{i}}$$ is the conclusion of schema $$(\lnot D), (C)$$ or $$(\lnot I)$$, then $$\Gamma_{\boldsymbol{i},0}$$ and $$\Gamma_{\boldsymbol{i},1}$$ are the left and the right premises of this schema. (4) If $$\Gamma_{\boldsymbol{i}}$$ is not defined, then $$\Gamma_{\boldsymbol{i},0}$$ and $$\Gamma_{\boldsymbol{i},1}$$ are not defined. As we have observed above, diagrams are partial mappings defined on the set of all finite binary sequences. Now we may also observe that for each formula $$A$$, the diagram of $$A$$ is uniquely determined by the formula. Moreover, the diagram of any formula is always finite, i.e. $$\Gamma_{{\boldsymbol{i}}}$$ is defined only for a finite number of binary sequences. We will say that $$\Gamma_{{\boldsymbol{i}}}$$ is an end sequence of the diagram of $$A$$ if $$\Gamma_{{\boldsymbol{i}}}$$ is fundamental or indecomposable, i.e. if $$\Gamma_{{\boldsymbol{i}},0}$$ and $$\Gamma_{{\boldsymbol{i}},1}$$ are not defined. Finally: Theorem 2 (Completeness of the R-S system) A formula $$A$$ is valid iff all end sequences in the diagram of $$A$$ are fundamental. The proof of this theorem presented in [31] is based on the fact that the diagram of a formula is uniquely determined by the rules of the R-S system. This means that for an arbitrary formula its diagram may be obtained ‘mechanically’; then one may determine validity of the formula by inspection of the end sequences of its diagram. Here are two examples of diagrams. Example 1 The diagram for formula $$((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r):$$ The formula is valid, and the end sequences of its diagram are fundamental. Example 2 The diagram for formula $$((p \rightarrow q) \land (r \rightarrow q)) \rightarrow \lnot (p \land \lnot r):$$ The formula is not valid, as witnessed by its diagram. 2.4 A task crazy enough to undertake In their summary of the method of diagrams, Rasiowa and Sikorski state that the diagram of a valid formula $$A$$ may be used to yield a construction of a proof of $$A$$ in the axiomatic system presented in their book. Their argumentation runs as follows. By Theorem 2, we know that each end sequence of the diagram of $$A$$ is fundamental. Moreover, if $$\Gamma$$ is a fundamental sequence, then the disjunction of all terms of $$\Gamma$$ is a valid formula. ‘The construction is based on the fact that for every fundamental formula $$A$$ we can effectively give a formal proof of $$A$$ from (…) [axioms] by means of modus ponens.’ [31, p. 269].9 Next, if sequence $$\Gamma$$ results from sequence $$\Gamma_{0}$$ by a rule falling under the schema (2), or results from $$\Gamma_{0}$$ and $$\Gamma_{1}$$ by a rule falling under the schema (3), then the proof of the formula denoting the disjunction of terms of $$\Gamma_{0}$$ (respectively, $$\Gamma_0$$ and $$\Gamma_{1}$$) may be transformed into the proof of disjunction of the terms of $$\Gamma$$. However, the authors state ‘We shall not describe these formal proofs in detail because the description is rather long.’ The task seemed crazy enough to undertake. 3 The method of Socratic proofs The method of Socratic proofs has been created as an explication of the — Socratic indeed — idea of solving questions by pure questioning. In the most basic version, the questions concern the validity of formulas of language of $$\mathsf{CPL}$$. Through progressive simplification of the logical structure of consecutive questions we are led to a question concerning some basic properties of the relation of entailment, like its reflexivity or ex falso quodlibet. The answer to such a question may be considered ‘obvious’. Since questions come into play, the method must be grounded in the logic of questions which analyses inferential relations between questions. This is the basic task of Inferential Erotetic Logic. 3.1 Inferential Erotetic Logic Inferential Erotetic10 Logic (hereafter $$\mathsf{IEL}$$) focuses on the so-called erotetic inferences — inferences in which questions play the role of conclusions and/or premises — and proposes criteria of validity of such inferences. $$\mathsf{IEL}$$ was invented by Andrzej Wiśniewski in the late 1980s (see [35]) and then further developed in various directions (see [36], [37], [40], [39], [27] — to mention the most important publications). It originated as an alternative with respect to the widespread (in that time) understanding of the logic of questions as focusing on analyses of the answerhood relation, and also in opposition to the interrogative model of inquiry developed by Jaakko Hintikka (see [18]). The method of Socratic proofs illustrates the importance of questions in proof theory. It has already been described for Classical Logic and various non-classical logics (see [37], [41], [40], [23], [24], [5], [4]). Below we present the method of Socratic proofs for $$\mathsf{CPL}$$ in the right-sided version (see [38], where $$\mathbf{E^{RPQ}}$$ for $$\mathsf{FOL}$$ has been introduced; as we have pointed out in the introduction, calculus $$\mathbb{E}^{**}$$ is the propositional part of calculus $$\mathbf{E^{RPQ}}$$ for $$\mathsf{FOL}$$). The choice of this version has been motivated by the aims of this article, i.e., by the strict correspondence between the right-sided erotetic calculus $$\mathbb{E^{**}}$$ and the R-S system. We describe formal language $$\mathcal{L_{\vdash}^{?}}$$ equipped with questions, and then we present erotetic calculus $$\mathbb{E^{**}}$$. Since this article concentrates on proof-theoretical issues we do not focus on the erotetic interpretation of the calculus. The Reader will find it in the cited works by Andrzej Wiśniewski. 3.2 Language $$\mathcal{L}_{\vdash}^{?}$$ Right-sided erotetic calculus $$\mathbb{E^{**}}$$ is worded in language $$\mathcal{L}_{\vdash}^{?}$$ and its declarative part is based on right-sided sequents. Formulas of language $$\mathcal{L_{\vdash}^{?}}$$ are built of the formulas of $$\mathcal{L}$$ and the following additional symbols: $$\vdash, ?, \&, \underline{ng}$$ and the comma. Finite, possibly empty sequences of formulas of language $$\mathcal{L}$$ will be represented by capital letters $$S, T, U, W$$. Each formula of language $$\mathcal{L}_{\vdash}^{?}$$ is either a declarative formula, or a question, and each declarative formula of language $$\mathcal{L}_{\vdash}^{?}$$ is either atomic or complex. Atomic declarative formulas of language$$\mathcal{L_{\vdash}^{?}}$$ are right-sided sequents, i.e., expressions of the form:   \begin{equation}\label{sequent} \vdash S, \end{equation} (5) where $$S$$ is a finite and non-empty sequence of formulas of language $$\mathcal{L}$$. If $$S = \langle A_1, \ldots, A_n\rangle$$, then we will represent (5) as   \begin{equation}\label{sequentA} \vdash A_{1},...,A_{n} \end{equation} (6) omitting the angle brackets. In the sequel, by terms of sequent (6) we shall mean formulas $$A_1, \ldots, A_n$$, i.e. the terms of sequence $$S$$ in $$\vdash S$$. We will say that sequent (5) is valid iff there is no valuation $$V$$ such that each term of $$S$$ is false under $$V$$. Thus sequent (6) is valid iff formula $$A_{1} \lor (A_{2}\lor \ldots(A_{n-1} \lor A_{n})\ldots)$$ is valid. As we can see, semantically the right-sided sequents may be interpreted as disjunctions of formulas. Questions of language$$\mathcal{L_{\vdash}^{?}}$$ are expressions of the form $$?(\Phi)$$, where $$\Phi$$ is a finite and non-empty sequence of atomic declarative formulas of language $$\mathcal{L_{\vdash}^{?}}$$ (sequents). Again, if $$\Phi = \langle \vdash~S_1, \ldots, \vdash~S_n \rangle$$, then we represent question $$?(\Phi)$$ as:   \begin{equation}\label{question} ?(\vdash S_1, \ldots, \vdash S_n) \end{equation} (7) and we say that question $$?(\Phi)$$ is based on sequents$$\vdash S_1, \ldots, \vdash S_n$$. We shall use $$Q, Q_1, Q^*$$ as metavariables for questions. Complex declarative formulas of language$$\mathcal{L_{\vdash}^{?}}$$ are built from atomic declarative formulas by connectives ‘$$\&$$’ and/or ‘$$\underline{ng}$$’. The complex formulas are used mainly to express answers to questions of $$\mathcal{L_{\vdash}^{?}}$$. All questions of $$\mathcal{L_{\vdash}^{?}}$$ are polar, and thus an affirmative answer to (7) states, roughly speaking, that all terms of (7) are valid sequents. The negative answer, in turn, states that it is not the case. As we have noted above, in this article we will not consider the strictly ‘erotetic’ issues of the method of Socratic proofs, as all this may be found in the works by Andrzej Wiśniewski. We will use the semicolon ‘;’ for the concatenation of sequences of sequents. We adopt conventions analogous to that pertaining to the concatenation of sequences of formulas of $$\mathcal{L}$$ (see page 5). 3.3 Erotetic calculus $$\mathbb{E^{**}}$$ Calculus $$\mathbb{E^{**}}$$ is a set of erotetic rules, i.e., rules transforming questions. There are no axioms. Table 1 presents the rules’ schemas (in the sequel we use the terms ‘schema of a rule’ and ‘rule’ interchangeably). Table 1 the rules of $$\mathbb{E}^{**}$$.   $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A\to B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A\to B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \land B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \land B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \lor B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \lor B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A\to B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A\to B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\to}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \land B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\vdash S\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \land B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\land}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\: \lnot(A \lor B)\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\: \lnot A\:'\:T\:;\:\vdash S\:'\: \lnot B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lnot\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:A \lor B\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:B\:'\:T\:;\:\Psi)} {\textbf{R}_{\lor}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:\neg\neg A\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\Psi)} {\textbf{R}_{\neg\neg}}$$     $$\frac {?(\Phi\:;\:\vdash S\:'\:\neg\neg A\:'\:T\:;\:\Psi)} {?(\Phi\:;\:\vdash S\:'\:A\:'\:T\:;\:\Psi)} {\textbf{R}_{\neg\neg}}$$   The question in the numerator of an erotetic rule is called the question-premise, and the one in the denominator — the question-conclusion. Thus, contrary to the rules of the R-S system, the direction of proving as defined by the erotetic rules coincides with the direction of proof-search. The sequent which is distinguished in the question-premise of a rule will be called its sequent-premise. The sequent(s) distinguished in the question-conclusion of a rule will be called its sequent(s)-conclusion(s). Finite sequences of questions transformed by the above rules are called Socratic transformations. Formally: Definition 2 (Socratic transformation) Let $$Q$$ be a question of language $$\mathcal{L_{\vdash}^{?}}$$. A finite sequence of questions $$\langle Q_{1}, Q_{2},...,Q_{n} \rangle$$ is a Socratic transformation of question$$Q$$ via the rules of $$\mathbb{E^{**}}$$ iff $$Q = Q_{1}$$ and each question of the sequence, except for the first one, results from the previous question by an application of one of the rules of $$\mathbb{E^{**}}$$. As we have written above, the method of Socratic proofs is an explication of the idea of answering questions by pure questioning. In accordance with this idea we will regard a Socratic transformation as successful, if its last question is, in a way, a trivial one. The notion of Socratic proof embodies this idea in the following way. Definition 3 (Socratic proof) Let $$\ \vdash S$$ be a sequent of language $$\mathcal{L_{\vdash}^{?}}$$. A Socratic proof of sequent$$\vdash S$$in$$\mathbb{E^{**}}$$ is a Socratic transformation of question $$?(\vdash S)$$ such that each constituent of its last question is of one of the following forms: $$\vdash T\: ' \: B\: ' \: U \: ' \: \neg B\: ' \: W,$$ $$\vdash T\: ' \: \neg B\: ' \: U \: ' \: B\: ' \: W.$$ Sequents of one of the two forms indicated in the definition are obviously valid sequents (as far as the law of excluded middle is an obvious one); in this sense the answer to the last question of a Socratic proof is obviously affirmative. Further, we shall say that sequents of one of the two forms are base sequents. The following theorem is true: Theorem 3 There is a Socratic proof of sequent $$\vdash A$$ of language $$\mathcal{L_{\vdash}^{?}}$$ in $$\mathbb{E^{**}}$$ iff formula $$A$$ of language $$\mathcal{L}$$ is valid. The Reader will find the relevant proofs in [38]. Let us now present two examples of Socratic transformations. The first one is a Socratic proof of the corresponding sequent, the second one is not. Example 3 Socratic transformation of question $$?~(~\vdash~ ((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r) ~):$$  $$ \frac {?~(~\vdash~ ((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r)~)} {?~(~\vdash~ \lnot((p \rightarrow q) \land (q \rightarrow r)), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ \lnot(p \rightarrow q), \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot (p \land \lnot r) ~;~ \vdash~ p, \lnot r, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, \lnot (q \rightarrow r), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, q, \lnot (p \land \lnot r) ~;~ \vdash ~ \lnot q, \lnot r, \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, q, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot r, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, q, \lnot (p \land \lnot r) ~;~ \vdash ~ \lnot q, \lnot r, \lnot p, \lnot\lnot r) ~)} . $$ The Socratic transformation is a Socratic proof of sequent $$\vdash~ ((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r)$$. Let us emphasize that the last question is based on sequents containing exactly the same sequences of formulas which occur in the leaves of the R-S diagram for formula $$((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \lnot (p \land \lnot r)$$ (compare Example 1). This example illustrates a close connection between calculus $$\mathbb{E^{**}}$$ and the method of Rasiowa and Sikorski, which justifies us in saying that calculus $$\mathbb{E^{**}}$$ is an erotetic version of the R-S system. The connection, in turn, derives from the fact that the two methods are based on deriving Conjunctive Normal Form of a formula. Example 4 Socratic transformation of question $$?~(~\vdash~ ((p \rightarrow q) \land (r \rightarrow q)) \rightarrow \lnot (p \land \lnot r) ~):$$   $$ \frac{?~(~\vdash~ \lnot((p \rightarrow q) \land (r \rightarrow q)), \lnot (p \land \lnot r) ~)} {?~(~\vdash~ ((p \rightarrow q) \land (r \rightarrow q)) \rightarrow \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ \lnot(p \rightarrow q), \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~) }\\ \frac {} {?~(~\vdash~ p, r, \lnot (p \land \lnot r) ~;~ \vdash~ p, \lnot q, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot (p \land \lnot r)~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, \lnot (r \rightarrow q), \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot (p \land \lnot r) ~;~ \vdash ~ \lnot q, \lnot q, \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, \lnot \lnot r ~;~ \vdash ~ \lnot q, \lnot q, \lnot(p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, r ~;~ \vdash ~ \lnot q, \lnot q, \lnot (p \land \lnot r) ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, r ~;~ \vdash ~ \lnot q, \lnot q, \lnot p, \lnot\lnot r ~)}\\ \frac {} {?~(~\vdash~ p, r, \lnot p, \lnot \lnot r ~;~ \vdash~ p, \lnot q, \lnot p, \lnot \lnot r~; ~\vdash ~ \lnot q, r, \lnot p, r ~;~ \vdash ~ \lnot q, \lnot q, \lnot p, r ~)} .$$ Again, the transformation should be compared with Example 2. 4 From Socratic proofs in $$\mathbb{E^{**}}$$ to proofs in $$\mathbb{A}$$ 4.1 Base sequents This subsection is devoted to proof of the following theorem: Theorem 4 If $$\vdash A_{1},...,A_{n}$$ is a base sequent, then formula: $$(*)$$$$A_{1}\vee (A_{2}\vee ...\vee (A_{n-1} \vee A_{n})...)$$ is a thesis of $$\mathbb{A}$$. Further, we will say that formula $$(*)$$ corresponds to sequent $$\vdash A_{1},...,A_{n}$$. So the above theorem states that a formula which corresponds to a base sequent is provable in $$\mathbb{A}$$. Before we proceed to the proof of Theorem 4, however, we will prove several laws of $$\mathsf{CPL}$$ for further use. In the first account of this work, we wanted to obtain ‘clear’ proofs in $$\mathbb{A}$$ without making use of the derived rules except for the rule of simultaneous substitution — and that is how the work is done in [16]. In this article, however, shortcuts are necessary, so for some of the laws of $$\mathsf{CPL}$$ proved below we will introduce a derived rule based on the law; but then we will also estimate the cost of applying such a rule in a proof. The application of the standard rule of substitution will be indicated by: [no. of line Sub: $$p_i/B_i$$], whereas the application of the rule of simultaneous substitution will be indicated by: [no. of line SSub: $$p_{i_1}/B_1, \ldots, p_{i_n}/B_n$$]. For each proof $$\mathbf{p}$$ in $$\mathbb{A}$$ we calculate (or estimate the upper bound for) the length of proof$$\mathbf{p}$$, symbolically $$l_\mathbf{p}$$, understood as the number of lines of $$\mathbf{p}$$. In the case of the application of a derived rule we want to calculate (or at least estimate) the number of lines that the proof would have, were the applications of the derived rules replaced with applications of the primary rules. As we know, when the application of the rule of simultaneous substitution is replaced with the applications of the standard substitution rule, for some variables an intermediate variable must be introduced to derive the result correctly. So the cost of the application of the simultaneous substitution rule is, in the worst case, twice as large as the cost of the subsequent applications of the standard substitution rule. More specifically, suppose that the rule of simultaneous substitution of formula $$B_1$$ for $$p_{i_1}$$, …, $$B_n$$ for $$p_{i_n}$$ is used to formula $$A$$ (where $$n \geq 2$$). Let $$p_{k_1}, \ldots, p_{k_n}$$ stand for $$n$$ pairwise distinct variables which do not occur in $$A$$ nor in any of $$B_1, \ldots, B_n$$. First, we apply the rule of substitution $$n$$ times:   $$C = S(p_{k_n}, p_{i_n}, S(\ \ldots \ S(p_{k_1}, p_{i_1}, A) \ \ldots ))$$ and then another $$n$$ times starting with formula $$C$$:   $$D = S(B_n, {p_{k_n}}, S(\ \ldots \ S(B_1, p_{k_1}, C) \ \ldots \ ))$$ which gives us $$2\cdot n$$ applications of the rule of substitution — i.e. $$2\cdot n$$ lines in the proof — to get from $$A$$ to $$D$$. Another useful measure will be that of width of proof in$$\mathbb{A}$$. Let $$|A|$$ stand for the number of occurrences of signs in formula $$A$$ without parentheses. $$|A|$$ is called the length of$$A$$. Then the width of proof$$\mathbf{p}$$in$$\mathbb{A}$$, symbolically $$w_\mathbf{p}$$, is understood as the maximum:   \begin{align*} w_\mathbf{p} = \max \{|A|: A~ \text{is a term of}~ \mathbf{p} \}. \end{align*} Given the two measures we may give an upper bound for the total number of signs occurring in a proof. By size of proof$$\mathbf{p}$$in$$\mathbb{A}$$, symbolically $$s_\mathbf{p}$$, we shall mean the product $$l_\mathbf{p}\cdot w_\mathbf{p}$$. This kind of measure corresponds well to time-complexity (see [29, p. 551]). In the proofs presented below we shall indicate, as the first piece of information, the number of lines ‘abbreviated’ whenever a derived rule is used. So the second line of the proof below starts with ‘(4)’, as this application of simultaneous substitution abbreviates 4 applications of Sub. In bold — as in the fifth line — we give the length of formula which is the longest formula in the proof (if there are more formulas of the same length, as it happens in this proof, we indicate only the first one). This number is the width of the proof. On the right side, in square brackets, we leave the usual didascalia. First, the proof of the hypothetical syllogism in $$\mathbb{A}$$: The law of hypothetical syllogism (1) $$p \to (q \to p)$$$$\hspace{0,2 cm}$$[Ax $$1.$$] (2) (4) $$((p\to (q\to r))\to ((p\to q) \to (p\to r)))\to ((q\to r) \to ((p\to (q\to r))\to ((p\to q) \to$$ $$\quad (p\to r))))$$$$\hspace{0,2 cm}$$ [$$1$$ SSub: $$p/(p\to (q\to r))\to ((p\to q) \to (p\to r)),q/ q\to r$$] (3) $$( p \to (q \to r)) \to ((p \to q) \to( p \to r))$$$$\hspace{0,2 cm}$$[Ax $$2.$$] (4) $$(q\to r) \to ((p\to (q\to r))\to ((p\to q) \to (p\to r)))$$$$\hspace{0,2 cm}$$ [$$2,3$$ MP] (5) (6) 39$$( (q\to r) \to ((p\to (q\to r)) \to ((p\to q) \to (p\to r)))) \to (((q\to r) \to (p\to (q\to r))) \to$$ $$\quad((q\to r) \to ((p\to q) \to (p\to r))))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q\to r, q/p\to (q\to r), r/(p\to q) \to (p\to r)$$] (6) $$((q\to r) \to (p\to (q\to r))) \to ((q\to r) \to ((p\to q) \to (p\to r)))$$$$\hspace{0,2 cm}$$[$$5,4$$ MP] (7) (4) $$(q\to r)\to (p \to (q\to r))$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/ q\to r, q/p$$] (8) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[$$6,7$$ MP] (9) (6) $$(\neg \neg p \to p) \to ((\neg q\to \neg \neg p) \to (\neg q \to p))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$p/\neg q, q/\neg\neg p, r/p$$] (10) $$\neg \neg p \to p$$$$\hspace{0,2 cm}$$[Ax $$4.$$] (11) $$(\neg q\to \neg \neg p) \to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$9,10$$ MP] (12) (6) $$((q\to r) \to ((p \to q) \to (p\to r))) \to (((q\to r) \to (p \to q)) \to ((q\to r) \to (p\to r)))$$ $$\quad$$ [$$3$$ SSub: $$p/q\to r, q/ p \to q, r/ p\to r$$] (13) $$((q\to r) \to (p \to q)) \to ((q\to r) \to (p\to r ))$$$$\hspace{0,2 cm}$$[$$12,8$$ MP] (14) (6) $$(((q\to r) \to (p\to q))\to ((q\to r) \to (p \to r))) \to (((p\to q)\to ((q\to r) \to (p\to q)))\to$$ $$\quad((p\to q)\to ((q\to r) \to (p\to r))))$$$$\hspace{0,2 cm}$$ [$$8$$ SSub: $$q/(q\to r) \to (p\to q) , r/(q\to r) \to (p \to r),$$ $$\quad p/p\to q$$] (15) $$((p\to q)\to ((q\to r) \to (p\to q))) \to ((p\to q)\to ((q\to r) \to (p\to r)))$$$$\hspace{0,2 cm}$$[$$14,13$$ MP] (16) (4) $$(p\to q) \to ((q \to r) \to (p\to q))$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/ p\to q, q/ q \to r$$] (17) $$(p\to q)\to ((q\to r) \to (p\to r))$$$$\hspace{0,2 cm}$$[$$15,16$$ MP] We will use $$l_{hyp~syl}$$ for the length of this proof and $$w_{hyp~syl}$$ for its width. As we can see, $$l_{hyp~syl}=46$$ and $$w_{hyp~syl}=39$$. Moreover, in the $$8^{th}$$ line there is a thesis which will be used later, so let us note that the length of its proof equals $$19$$ and the width of its proof equals $$39$$. The use of the derived rule based on the hypothetical syllogism will be indicated by [$$a,~b$$ HS], where line no. $$a$$ contains the result of substitution to formula $$(p \to q)$$, line no. $$b$$ – to $$(q \to r)$$, and the result is the suitable substitution to $$(p\to r)$$. To sum up, the use of HS yields 6 applications of Sub and two applications of MP (i.e. 8 lines). We must remember that the first use of HS in a proof also yields the above 46 lines and a formula of length 39. Two laws of associativity of disjunction The proof of $$(p\lor q) \lor r \to p\lor(q\lor r):$$ (1) $$p \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$9.$$] (2) (4) $$q \rightarrow q \lor r$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/q, q/r$$] (3) $$q \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$10.$$] (4) $$q \lor r \rightarrow p \lor (q \lor r)$$$$\hspace{0,2 cm}$$[$$3$$ Sub: $$q/q\lor r$$] (5) (46+8) 39$$q \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[$$2, 4$$ HS] (6) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax $$11.$$] (7) $$(p \to p\lor (q\lor r)) \to ((q \to p\lor (q\lor r)) \to (p \vee q \to p\lor (q\lor r)))$$$$\hspace{0,2 cm}$$[$$6$$ Sub: $$r/p\lor (q\lor r)$$] (8) $$p \rightarrow p \lor (q\lor r)$$$$\hspace{0,2 cm}$$[$$1$$ Sub: $$q/q\lor r$$] (9) $$(q \to p\lor (q\lor r)) \to (p \vee q \to p\lor (q\lor r))$$$$\hspace{0,2 cm}$$[$$7, 8$$ MP] (10) $$p \vee q \to p\lor (q\lor r)$$$$\hspace{0,2 cm}$$[$$9, 5$$ MP] (11) (4) $$r \rightarrow q \lor r$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q, q/r$$] (12) (8) $$r\to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[$$11, 4$$ HS] (13) (6) $$(p\lor q \to p\lor(q\lor r)) \to ((r \to p\lor(q\lor r)) \to ((p\lor q) \lor r \to p\lor(q\lor r)))$$ $$\quad$$ [$$6$$ SSub: $$p/p\lor q, q/r , r/p\lor(q\lor r)$$] (14) $$(r \to p\lor(q\lor r)) \to ((p\lor q) \lor r \to p\lor(q\lor r))$$$$\hspace{0,2 cm}$$[$$13, 10$$ MP] (15) $$(p\lor q) \lor r \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[$$14, 12$$ MP] Again, we use the self-explanatory notation and write $$l_{as~dis~1}=86$$, $$w_{as~dis~1} = 39$$. The proof of $$p\lor (q \lor r) \to (p\lor q)\lor r$$: (1) $$p \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$9.$$] (2) (4) $$p\lor q \rightarrow (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/p\lor q, q/r$$] (3) (46+8) 39$$p\to (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$1, 2$$ HS] (4) $$q \rightarrow p \lor q$$$$\hspace{0,2 cm}$$[Ax $$10.$$] (5) (8) $$q\to (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$4,2$$ HS] (6) (4) $$r \rightarrow (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$4$$ SSub: $$p/p\lor q, q/r $$] (7) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax $$11.$$] (8) (6) $$(q \to (p\lor q) \lor r) \to ((r \to (p\lor q) \lor r) \to (q \lor r \to (p\lor q) \lor r))$$ $$\quad$$ [$$7$$ SSub: $$p/q, q/r, r/(p\lor q) \lor r$$] (9) $$(r \to (p\lor q) \lor r) \to (q \lor r \to (p\lor q) \lor r)$$$$\hspace{0,2 cm}$$[$$8, 5$$ MP] (10) $$q \lor r \to (p\lor q) \lor r$$$$\hspace{0,2 cm}$$[$$9, 6$$ MP] (11) (4) $$(p \to (p\lor q) \lor r) \to ((q\lor r \to (p\lor q) \lor r) \to (p \vee (q\lor r) \to (p\lor q) \lor r))$$ $$\quad$$ [$$7$$ SSub: $$q/q\lor r, r/(p\lor q) \lor r$$] (12) $$(q\lor r \to (p\lor q) \lor r) \to (p \vee (q\lor r) \to (p\lor q) \lor r)$$$$\hspace{0,2 cm}$$[$$11, 3$$ MP] (13) $$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[$$12, 10$$ MP] We have: $$l_{as~dis~2}=87$$ and $$w_{as~dis~2}=39$$. Further, we will also need the proofs of formulas: $$p\vee \neg p$$, $$\neg p\vee p$$, $$p\vee (q \vee \neg p)$$, $$\neg p\vee (q \vee p)$$. Below we present a proof of the first formula (the law of excluded middle). The proof is long, but it contains some important intermediate steps, like reaching the laws of commutation and importation. Further we will use the derived rules based on these laws. The laws of commutation and importation, the Duns Scotus law, the law of identity, the law of excluded middle (1) $$p \to (q \to p)$$$$\hspace{0,2 cm}$$[Ax $$1.$$] (2) $$(p \to (q \to r)) \to ((p \to q) \to( p \to r))$$$$\hspace{0,2 cm}$$[Ax $$2.$$] (3) (46) 39$$(p\to q)\to ((q\to r) \to (p\to r))$$$$\hspace{0,2 cm}$$[hypothetical syllogism] (4) (6) $$(q\to (p\to q))\to (((p\to q)\to (p\to r)) \to (q\to (p\to r)))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q, q/p\to q, r/p\to r$$] (5) (4) $$q \to (p \to q)$$$$\hspace{0,2 cm}$$[$$1$$ SSub: $$p/q, q/p$$] (6) $$((p\to q)\to (p\to r)) \to (q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$4,5$$ MP] (7) (8) $$(p\to (q\to r))\to (q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$2,6$$ HS] [the law of commutation] The seven lines above constitute a proof of the law of commutation with $$l_{com}=67$$ and $$w_{com}=39$$. Further we will use a derived rule based on the law. The application of the derived rule will be indicated by $$[a$$ Com$$]$$, where line no. $$a$$ contains the result of substitution to $$p\to(q\to r)$$, and the result of the application of Com is the suitable substitution to $$q\to(p\to r)$$. The application of Com yields $$67$$ lines of the proof, one application of the rule of simultaneous substitution (6 lines) and one application of MP, i.e., $$74$$ lines. We continue the proof of the law of excluded middle. (8) (19) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[thesis, see the proof of the hypothetical syllogism, line 8] (9) (6) $$(\neg \neg p \to p) \to ((\neg q\to \neg \neg p) \to (\neg q \to p))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$p/\neg q, q/\neg \neg p, r/p $$] (10) $$\neg \neg p \to p$$$$\hspace{0,2 cm}$$[Ax $$4.$$] (11) $$(\neg q\to \neg \neg p) \to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$9,10$$ MP] (12) $$(p \to q)\to (\neg q \to \neg p)$$$$\hspace{0,2 cm}$$ [Ax $$3.$$] (13) $$(\neg p \to q)\to (\neg q \to \neg \neg p)$$$$\hspace{0,2 cm}$$ [$$12$$ Sub: $$p/ \neg p$$] (14) (8) $$(\neg p \to q)\to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$13,11$$ HS] (15) $$(p \to (p \vee q))\to (\neg (p \vee q) \to \neg p)$$$$\hspace{0,2 cm}$$ [$$12$$ Sub: $$q/ p \vee q$$] (16) $$p \to p \vee q$$$$\hspace{0,2 cm}$$[Ax $$9.$$] (17) $$\neg (p \vee q) \to \neg p$$$$\hspace{0,2 cm}$$[$$15,16$$ MP] (18) (4) $$(q \to p\vee q)\to (\neg (p\vee q) \to \neg q)$$$$\hspace{0,2 cm}$$ [$$12$$ SSub: $$p/q, q/p \lor q$$] (19) $$q\to p \vee q $$$$\hspace{0,2 cm}$$[Ax $$10.$$] (20) $$\neg (p\vee q) \to \neg q$$$$\hspace{0,2 cm}$$[$$18,19$$ MP] (21) $$(p \to q) \to ((p \to r) \to (p \to q \wedge r))$$$$\hspace{0,2 cm}$$ [Ax $$8.$$] (22) (6) $$(\neg (p\vee q) \to \neg p) \to ((\neg (p\vee q) \to \neg q) \to (\neg (p\vee q) \to \neg p \wedge \neg q))$$ $$\quad$$ [$$21$$ SSub: $$p/\neg (p\vee q), q/\neg p, r/\neg q$$] (23) $$(\neg (p\vee q) \to \neg q) \to (\neg (p\vee q) \to \neg p \wedge \neg q)$$$$\hspace{0,2 cm}$$[$$22,17$$ MP] (24) $$\neg (p\vee q) \to \neg p \wedge \neg q$$$$\hspace{0,2 cm}$$[$$23,20$$ MP] (25) (4) $$(\neg (p\vee q) \to \neg p \wedge \neg q)\to (\neg (\neg p \wedge \neg q) \to p\vee q)$$$$\hspace{0,2 cm}$$[$$14$$ SSub: $$p/p\vee q, q/\neg p \wedge \neg q$$] (26) $$\neg (\neg p \wedge \neg q) \to p\vee q$$$$\hspace{0,2 cm}$$[$$25,24$$ MP] (27) $$(p\wedge q\to q)\to ((q\to r) \to (p\wedge q\to r))$$$$\hspace{0,2 cm}$$[$$3$$ Sub: $$p/p\wedge q$$] (28) $$p \wedge q \to q$$$$\hspace{0,2 cm}$$[Ax $$7.$$] (29) $$(q\to r) \to (p\wedge q\to r)$$$$\hspace{0,2 cm}$$[$$27,28$$ MP] (30) (4) $$((q\to r)\to (p\wedge q \to r)) \to ((p\to (q\to r)) \to (p\to (p\wedge q \to r)))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$q/q\to r$$, $$r/p\wedge q \to r$$] (31) $$(p\to (q\to r)) \to (p\to (p\wedge q \to r))$$$$\hspace{0,2 cm}$$[$$30,29$$ MP] (32) $$(p\to (p\wedge q\to r))\to (p\wedge q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$7$$ Sub: $$q/p\wedge q$$] (33) (8) $$(p\to (q\to r))\to (p\wedge q\to (p\to r))$$$$\hspace{0,2 cm}$$[$$31,32$$ HS] (34) (4) $$(p\wedge q \to (p \to r)) \to ((p\wedge q \to p) \to (p\wedge q \to r))$$$$\hspace{0,2 cm}$$[$$2$$ SSub: $$p/p\wedge q, q/p$$] (35) 51$$((p\to (q\to r))\to (p\wedge q\to (p\to r)))\to (((p\wedge q\to (p\to r))\to ((p\wedge q\to p)\to(p\wedge q\to r))) \to ((p\to (q\to r))\to ((p\wedge q\to p)\to (p\wedge q\to r))))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/p\to (q\to r), q/p\wedge q\to (p\to r), r/(p\wedge q\to p)\to (p\wedge q\to r)$$] This formula is ‘hidden’ in the following application of HS: (36) (8) $$(p\to (q\to r))\to ((p\wedge q\to p)\to(p\wedge q\to r))$$$$\hspace{0,2 cm}$$[$$33,34$$ HS] (37) (7) $$(p\wedge q\to p)\to ((p\to (q\to r))\to (p\wedge q\to r))$$$$\hspace{0,2 cm}$$[$$36$$ Com] (38) $$p\wedge q\to p$$$$\hspace{0,2 cm}$$[Ax $$6.$$] (39) $$(p\to (q\to r))\to (p\wedge q\to r)$$$$\hspace{0,2 cm}$$[$$37,38$$ MP] [the law of importation] At this point we arrive at the proof of the law of importation with $$l_{imp}=165$$ and $$w_{imp}=51$$. Similarly as in the case of the law of commutation, we define a derived rule, Imp, based on the law of importation. Its application yields $$7$$ lines of a proof (plus 165 for the law of importation, if this is absent in the proof). The formula in line no. 37 results from the previous one by the derived rule Com. We calculate only 7 lines, since the law of commutation is already a part of this proof. (40) $$p \to (\neg q \to p)$$$$\hspace{0,2 cm}$$[$$1$$ Sub: $$q/\neg q$$] (41) (4) $$(\neg q \to p)\to (\neg p \to q)$$$$\hspace{0,2 cm}$$[$$14$$ SSub: $$p/q, q/p$$] (42) (8) $$p\to (\neg p\to q)$$$$\hspace{0,2 cm}$$[$$40,41$$ HS] [the Duns Scotus law] (43) (7) $$p\wedge \neg p\to q$$$$\hspace{0,2 cm}$$[$$42$$ Imp] (44) (6) $$(\neg \neg p\to p)\to ((p\to \neg q) \to (\neg \neg p\to \neg q))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/\neg \neg p, q/p, r/\neg q$$] (45) $$(p\to \neg q) \to (\neg \neg p\to \neg q)$$$$\hspace{0,2 cm}$$[$$44,10$$ MP] (46) (6) $$(q\to \neg \neg q)\to ((\neg \neg q\to \neg \neg p) \to (q\to \neg \neg p))$$$$\hspace{0,2 cm}$$[$$3$$ SSub: $$p/q, q/\neg \neg q, r/\neg \neg p$$] (47) $$p \to \neg \neg p$$$$\hspace{0,2 cm}$$[Ax $$5.$$] (48) $$q \to \neg \neg q$$$$\hspace{0,2 cm}$$[$$47$$ Sub: $$p/q$$] (49) $$(\neg \neg q\to \neg \neg p) \to (q\to \neg \neg p)$$$$\hspace{0,2 cm}$$[$$46,48$$ MP] (50) (6) $$(r\to s) \to ((q\to r) \to (q\to s))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$p/q, q/r, r/s$$] (51) (4) $$((q\to r)\to (q\to s)) \to ((p\to (q\to r)) \to (p\to (q\to s)))$$$$\hspace{0,2 cm}$$[$$8$$ SSub: $$q/q\to r, r/q\to s$$] (52) (8) $$(r\to s)\to ((p\to (q\to r))\to (p\to (q\to s)))$$$$\hspace{0,2 cm}$$[$$50,51$$ HS] (53) (6) $$(\neg \neg p\to p)\to (((\neg \neg q\to \neg \neg p)\to (q\to \neg \neg p))\to ((\neg \neg q\to \neg \neg p)\to (q\to p)))$$ $$\hspace{0,2 cm}$$[$$52$$ SSub: $$p/\neg \neg q\to \neg \neg p, r/\neg \neg p, s/p$$] (54) $$((\neg \neg q\to \neg \neg p)\to (q\to \neg \neg p))\to ((\neg \neg q\to \neg \neg p)\to (q\to p))$$$$\hspace{0,2 cm}$$[$$53,10$$ MP] (55) $$(\neg \neg q\to \neg \neg p)\to (q\to p)$$$$\hspace{0,2 cm}$$[$$54,49$$ MP] (56) (4) $$(\neg p \to \neg q)\to (\neg \neg q \to \neg \neg p)$$$$\hspace{0,2 cm}$$ [$$12$$ SSub: $$p/\neg p, q/\neg q$$] (57) (8) $$(\neg p\to \neg q)\to (q\to p)$$$$\hspace{0,2 cm}$$[$$56,55$$ HS] (58) $$(\neg \neg p\to \neg q)\to (q\to \neg p)$$$$\hspace{0,2 cm}$$[$$57$$ Sub: $$p/\neg p$$] (59) (8) $$(p\to \neg q)\to (q\to \neg p)$$$$\hspace{0,2 cm}$$[$$45,58$$ HS] (60) (4) $$(p\wedge \neg p\to \neg (p\to p))\to ((p\to p)\to \neg (p\wedge \neg p))$$$$\hspace{0,2 cm}$$[$$59$$ SSub: $$p/p\wedge \neg p, q/p\to p$$] (61) $$p\wedge \neg p\to \neg (p\to p)$$$$\hspace{0,2 cm}$$[$$43$$ Sub: $$q/\neg (p\to p)$$] (62) $$(p\to p)\to \neg (p\wedge \neg p)$$$$\hspace{0,2 cm}$$[$$60,61$$ MP] (63) $$(p \to (q \to p)) \to ((p \to q) \to( p \to p))$$$$\hspace{0,2 cm}$$[$$2$$ Sub: $$r/p$$] (64) $$(p \to q) \to( p \to p)$$$$\hspace{0,2 cm}$$[$$63,1$$ MP] (65) $$(p \to (q\to p)) \to( p \to p)$$$$\hspace{0,2 cm}$$[$$64$$ Sub: $$q/q\to p$$] (66) $$p \to p$$$$\hspace{0,2 cm}$$[$$65,1$$ MP] [the identity law] (67) $$\neg (p\wedge \neg p)$$$$\hspace{0,2 cm}$$[$$62,66$$ MP] (68) $$\neg (\neg p\wedge \neg \neg p)$$$$\hspace{0,2 cm}$$[$$67$$ Sub: $$p/\neg p$$] (69) $$\neg (\neg p \wedge \neg \neg p) \to p\vee \neg p$$$$\hspace{0,2 cm}$$[$$26$$ Sub: $$q/\neg p$$] (70) $$p\lor \neg p$$$$\hspace{0,2 cm}$$[$$69,68$$ MP] The length of the above proof, $$l_{ex~mid}$$, is estimated at $$262$$. Actually, it is easy to see that this number is overestimated, as after adding the lines of the proof of the hypothetical syllogism some of the lines would be repeated. As a matter of fact, we can never exclude the existence of a shorter proof. The width of the proof, $$w_{ex~mid}$$ in symbols, equals $$51$$. Observe also that in line no. 42 we have the Duns Scotus law with $$l_{DSl}=178$$, $$w_{DSl}=51$$, and in line no. 66 — the identity law with $$l_{id}=258$$ and $$w_{id}=51$$. The reversed law of excluded middle (1) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (2) $$(p \to q\vee p) \to ((q \to q\vee p) \to (p \vee q \to q\vee p))$$$$\hspace{0,2 cm}$$ [1 Sub: $$r/q\vee p$$] (3) $$q \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 10.] (4) (4) $$p \to q \vee p$$$$\hspace{0,2 cm}$$[3 SSub: $$q/p, p/q$$] (5) $$(q \to q\vee p) \to (p \vee q \to q\vee p)$$$$\hspace{0,2 cm}$$ [2,4 MP] (6) $$p \to p \vee q$$$$\hspace{0,2 cm}$$ [Ax 9.] (7) (4) $$q \to q \vee p$$$$\hspace{0,2 cm}$$ [6 SSub: $$p/q, q/p$$] (8) $$p \vee q \to q\vee p$$$$\hspace{0,2 cm}$$ [5,7 MP] (9) $$p \vee \neg p \to \neg p\vee p$$$$\hspace{0,2 cm}$$ [8 Sub: $$q/\neg p$$] (10) (262) 51$$p\vee \neg p$$$$\hspace{0,2 cm}$$[the law of excluded middle] (11) $$\neg p\vee p$$$$\hspace{0,2 cm}$$ [9,10 MP] The length of the proof, $$l_{rev~ex~mid}$$, is $$278$$. The width, $$w_{rev~ex~mid}$$, equals 51. Formula: $$p\vee (q\vee \neg p)$$ (1) $$p \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 9.] (2) $$p \to p \vee (q \vee \neg p)$$$$\hspace{0,2 cm}$$[1 Sub: $$q/q\vee \neg p$$] (3) $$q \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 10.] (4) (4) $$\neg p \to (p\vee q) \vee \neg p$$$$\hspace{0,2 cm}$$[3 SSub: $$p/p\vee q, q/\neg p$$] (5) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (6) (4) $$(p \to p\vee (q \vee \neg p)) \to ((\neg p \to p\vee (q \vee \neg p)) \to (p \vee \neg p \to p\vee (q \vee \neg p)))$$ $$\quad$$ [5 SSub: $$q/\neg p, r/p\vee (q \vee \neg p)$$] (7) $$(\neg p \to p\vee (q \vee \neg p)) \to (p \vee \neg p \to p\vee (q \vee \neg p))$$$$\hspace{0,2 cm}$$[6,2 MP] (8) (262) 51$$p \vee \neg p$$$$\hspace{0,2 cm}$$ [the law of excluded middle] (9) (86) $$(p\lor q) \lor r \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[associativity of disjunction 1] (10) $$(p\vee q) \vee \neg p \to p \lor(q \lor \lnot p)$$$$\hspace{0,2 cm}$$[9 Sub: $$r/\lnot p$$] (11) (8) $$\lnot p\to p\lor (q\lor \lnot p)$$$$\hspace{0,2 cm}$$[4,10 HS] (12) $$p \vee \neg p \to p\vee (q \vee \neg p)$$$$\hspace{0,2 cm}$$[7,11 MP] (13) $$p\vee (q \vee \neg p)$$$$\hspace{0,2 cm}$$[12,8 MP] Lines 8–11 need $$262+86+1+8=357$$ lines of proof (the law of the hypothetical syllogism is used in the proof of the law of excluded middle, therefore we do not add the value 46 in line no. 11). Thus the length of the proof, $$l_{p\lor(q\lor\lnot p)}$$ in symbols, is estimated at $$372$$ lines. The width, $$w_{p\lor(q\lor\lnot p)}$$, is 51. Formula: $$\neg p \lor (q\lor p)$$ (1) $$p \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 9.] (2) (4) $$\neg p \to \neg p \vee (q \vee p)$$$$\hspace{0,2 cm}$$[1 SSub: $$p/\neg p, q/q\vee p$$] (3) $$q \to p \vee q$$$$\hspace{0,2 cm}$$[Ax 10.] (4) (4) $$p \to (\neg p\vee q) \vee p$$$$\hspace{0,2 cm}$$[3 SSub: $$p/\neg p\vee q, q/p$$] (5) (262) 51$$p \vee \neg p$$$$\hspace{0,2 cm}$$ [the law of excluded middle] (6) (86) $$(p\lor q) \lor r \to p\lor(q\lor r)$$$$\hspace{0,2 cm}$$[associativity of disjunction 1] (7) (4) $$(\lnot p\lor q) \lor p \to \lnot p\lor(q\lor p)$$$$\hspace{0,2 cm}$$[6 SSub: $$p/\lnot p,r/p$$] (8) (8) $$p\to \lnot p\lor(q\lor p)$$$$\hspace{0,2 cm}$$[4,7 HS] (9) $$(p \to r) \to ((q \to r) \to (p \vee q \to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (10) (4) $$(p \to \neg p\vee (q \vee p)) \to ((\neg p \to \neg p\vee (q \vee p)) \to (p \vee \neg p \to \neg p\vee (q \vee p)))$$ $$\quad$$ [9 Sub: $$q/\neg p, r/\neg p\vee (q \vee p)$$] (11) $$(\neg p \to \neg p \lor (q \lor p)) \to (p \vee \neg p \to \neg p\vee (q \vee p))$$$$\hspace{0,2 cm}$$[10,8 MP] (12) $$p \vee \neg p \to \neg p \vee (q \lor p)$$$$\hspace{0,2 cm}$$[11,2 MP] (13) $$\neg p\vee (q \vee p)$$$$\hspace{0,2 cm}$$[12,5 MP] Counting similarly as before, we have $$l_{\lnot p\lor(q\lor p)}=378$$ and $$w_{\lnot p\lor(q\lor p)}=51$$. Formula $$(p\to q)\to (r\lor p \to r\lor q)$$ (disjunct added to the left) (1) (19) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[thesis, see the proof of the hypothetical syllogism, line 8] (2) $$q\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 10.] (3) $$q\to r\lor q$$$$\hspace{0,2 cm}$$[2 Sub: $$p/r$$] (4) $$(q\to r\lor q) \to ((p\to q) \to (p\to r\lor q))$$$$\hspace{0,2 cm}$$[1 Sub: $$r/r\lor q$$] (5) $$(p\to q)\to (p\to r\lor q)$$$$\hspace{0,2 cm}$$[4,3 MP] (6) $$(p\to r)\to((q\to r)\to(p\lor q\to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (7) (6) $$(r\to r\lor q)\to((p\to r\lor q)\to(r\lor p \to r\lor q))$$$$\hspace{0,2 cm}$$[6 SSub: $$p/r,q/p,r/r\lor q$$] (8) $$p\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 9.] (9) $$r\to r\lor q$$$$\hspace{0,2 cm}$$[8 Sub: $$p/r$$] (10) $$(p\to r\lor q)\to(r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[7,9 MP] (11) (46+8) 39$$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[5,10 HS] The first formula occurs in the proof of hypothetical syllogism, so the value 19 is included in the value 46 added in line 11. Therefore we may calculate: $$l_{dis~l}=l_{hyp~syl}+22=68$$ and $$w_{dis~l}=39$$. Formula $$(p\to q)\to (p\lor r \to q\lor r)$$ (disjunct added to the right): (1) $$(q\to r) \to ((p\to q) \to (p\to r))$$$$\hspace{0,2 cm}$$[thesis, see the proof of the hypothetical syllogism, line 8] (2) $$p\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 9.] (3) (4) $$q\to q\lor r$$$$\hspace{0,2 cm}$$[2 SSub: $$p/q,q/r$$] (4) $$(q\to q\lor r) \to ((p\to q) \to (p\to q\lor r))$$$$\hspace{0,2 cm}$$[1 Sub: $$r/q\lor r$$] (5) $$(p\to q)\to (p\to q\lor r)$$$$\hspace{0,2 cm}$$[4,3 MP] (6) $$(p\to r)\to((q\to r)\to(p\lor q\to r))$$$$\hspace{0,2 cm}$$[Ax 11.] (7) (4) $$(p\to q\lor r)\to((r\to q\lor r)\to(p\lor r \to q\lor r))$$$$\hspace{0,2 cm}$$[6 SSub: $$q/r,r/q\lor r$$] (8) (74) 39$$(r\to q\lor r) \to ((p\to q\lor r)\to (p\lor r \to q\lor r))$$$$\hspace{0,2 cm}$$[7 Com] (9) $$q\to p\lor q$$$$\hspace{0,2 cm}$$[Ax 10.] (10) (4) $$r\to q\lor r$$$$\hspace{0,2 cm}$$[9 SSub: $$p/q,q/r$$] (11) $$(p\to q\lor r)\to(p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[8,10 MP] (12) (8) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[5,11 HS] The formula in line 1, as well as the hypothetical syllogism, occurs in the proof of the law of commutation, so the cost (74) of the application of Com includes the two. Therefore we have $$l_{dis~r} = 100$$ and $$w_{dis~r}=39$$. Now we are in a position to present: Proof of Theorem 4 Proof. Suppose that $$\vdash A_1, \ldots, A_n$$, where $$n \geq 2$$, is a base sequent. Assume also that it is of the form:   $$ \vdash A_1, \ldots, A_{i-1}, B, A_{i+1}, \ldots, A_{k-1}, \lnot B, A_{k+1}, \ldots, A_n, $$ where $$1 \leq i < k \leq n$$. If $$i+1 < k$$, then we start with the proof of formula $$p \lor (q \lor \lnot p)$$ ($$l_{p \lor (q \lor \lnot p)}=372, w_{p \lor (q \lor \lnot p)}=51$$) and substitute formula $$B$$ for $$p$$ and formula $$A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor A_{k-1})\ldots)$$ for $$q$$. We arrive at: $$(a)$$. $$D = B \lor ((A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor A_{k-1})\ldots)) \lor \lnot B)$$ with the length of the proof $$372+4$$ (substitutions) $$=376$$ and the width equal to $$\max\{51, |D|\}$$. If $$k=i+2$$, i.e., there is only one formula $$A_{i+1}$$ between $$B$$ and $$\lnot B$$, then $$D$$ takes the form: $$B \lor (A_{i+1} \lor \lnot B)$$ and at this level we are done. Now suppose that $$k\geq i+3$$ and let $$C=(A_{i+2}\lor\ldots\lor(A_{k-1}\lor A_k)\ldots)$$. Then we use the law $$(p\to q)\to (r\lor p \to r\lor q)$$ (which adds another $$l_{dis~l}=68$$ lines), we also make use of one of the laws of associativity of disjunction, which is already present among the 372 lines of the proof of $$p\lor(q\lor \lnot p)$$, and we continue as below: $$(b)$$. $$(p\lor q) \lor r \to p\lor(q\lor r)$$ $$(c)$$. (6) $$(A_{i+1} \lor C) \lor \lnot B \to A_{i+1} \lor(C\lor \lnot B)$$$$\hspace{0,2 cm}$$[$$(b)$$ SSub: $$p/A_{i+1}, q/C, r/\lnot B$$] $$(d)$$. (68) $$(p\to q)\to (r\lor p \to r\lor q)$$ $$(e)$$. (6) $$((A_{i+1} \lor C) \lor \lnot B \to A_{i+1} \lor(C\lor \lnot B)) \to (B\lor ((A_{i+1} \lor C) \lor \lnot B) \to B \lor (A_{i+1} \lor(C\lor \lnot B)))$$$$\hspace{0,2 cm}$$[$$(d)$$ SSub: $$p/(A_{i+1} \lor C) \lor \lnot B$$, $$q/A_{i+1} \lor(C\lor \lnot B)$$, $$r/B$$] $$(f)$$. $$B\lor ((A_{i+1} \lor C) \lor \lnot B) \to B \lor (A_{i+1} \lor(C\lor \lnot B))$$$$\hspace{0,2 cm}$$[$$(e),(c)$$ MP] $$(g)$$. $$B \lor (A_{i+1} \lor(C\lor \lnot B))$$$$\hspace{0,2 cm}$$[$$(f), (a)$$ MP]. Again, if $$k=i+3$$, then at this level we are done. But if $$C$$ is a disjunction, then we are forced to continue, in order to rearrange the parentheses inside formula $$(C\lor \lnot B)$$. Let $$k=i+l$$ and suppose that $$k=i+l>i+3$$. Then after reaching line no $$(g)$$. we need another $$6(l-2)$$ substitutions to the associativity law at $$(b)$$ with respect to consecutive inner disjunctions, that is, we will get: $$(h)$$. $$(A_{i+2}\lor C^*)\lor \lnot B \to A_{i+2}\lor(C^*\lor \lnot B)$$ where $$C^*=(A_{i+3}\lor\ldots\lor(A_{k-1}\lor A_k)\ldots)$$, then $$(i)$$. $$(A_{i+3}\lor C^{**})\lor \lnot B \to A_{i+3}\lor(C^{**}\lor \lnot B)$$, where $$C^{**}=(A_{i+4}\lor\ldots\lor(A_{k-1}\lor A_k)\ldots)$$, and so on. At the last step we arrange the parentheses around formulas $$A_{i+l-2}, A_{i+l-1}$$ and $$\lnot B$$ as follows: $$(j)$$. $$(A_{i+l-2}\lor A_{i+l-1})\lor \lnot B \to A_{i+l-2}\lor(A_{i+l-1}\lor \lnot B).$$ However, now we must ‘climb back’ to the more complex disjunctions by adding the consecutive disjuncts to the left. For example, for adding formula $$A_{i+l-3}$$ we need one simultaneous substitution (i.e. 6 substitutions) to $$(d)$$ and one application of MP (we detach $$(j)$$) to obtain: $$(k).$$$$A_{i+l-3} \lor ((A_{i+l-2} \lor A_{i+l-1}) \lor \lnot B) \to A_{i+l-3} \lor (A_{i+l-2} \lor (A_{i+l-1} \lor \lnot B))$$ and further one application of the derived rule HS (8 lines) to obtain:11 $$(l).$$$$(A_{i+l-3} \lor (A_{i+l-2} \lor A_{i+l-1})) \lor \lnot B \to A_{i+l-3} \lor (A_{i+l-2} \lor (A_{i+l-1} \lor \lnot B))$$. Each ‘level up’ adds another 15 lines to the proof, i.e., 7 lines to reach $$(k)$$ and 8 lines to reach $$(l)$$. To sum up, before we add formula $$A_i$$, that is, $$B$$, we need $$15\cdot(l-3)=15\cdot(k-i-3)$$ lines after line number $$(j)$$. Moreover, when for the last time we obtain a formula analogous to that in $$(l)$$, the formula’s length equals   $$\mathbf{L} = \left(\sum_{l=i}^k |A_l|\right) + \left(k-(i-1)-1\right).$$ Recall that $$B$$ is $$A_i$$ and $$\lnot B$$ is $$A_k$$. The value $$(k-(i-1)-1)$$ counts the number of occurrences of connectives (disjunctions) between the formulas. The formula is obtained by an application of HS, therefore the longest formula (at this step of the proof) is the appropriate substitution to the law of hypothetical syllogism; and this formula’s length is: $$3\cdot\mathbf{L}+2$$. After another application of MP to $$(l)$$ (we detach $$(a)$$) we end up with a ready proof of the formula: $$(m).$$$$B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots)).$$ If $$i=1$$ and $$k=n$$, then this is the whole proof. Suppose that $$k<n$$. Now we need to add the formulas to the right of $$\lnot B$$. Thus let $$D^\star = (A_{k+1}\lor(A_{k+2}\lor\ldots\lor(A_{n-1}\lor A_n)\ldots))$$. We start with a simultaneous substitution to Ax 9. (4 substitutions) and MP: $$(n).$$ (4) $$B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots)) \to (B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor$$ $$\quad (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots))) \lor D^\star$$ $$(o).$$$$(B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor \lnot B))\ldots))) \lor D^\star$$$$\hspace{0,2 cm}$$[MP] and we have to rearrange the parentheses again. There is $$k-(i-1)$$ formulas from $$B$$ to $$\lnot B$$. As before, we need to add $$k-(i-1)$$ simultaneous substitutions to the law of associativity $$(d)$$ and 15 additional lines for each of the $$k-(i-1)$$ formulas, in order to obtain: $$(p).$$$$B \lor (A_{i+1} \lor(A_{i+2}\lor \ldots \lor (A_{k-2} \lor (A_{k-1} \lor (\lnot B \lor (A_{k+1}\lor(A_{k+2}\lor\ldots\lor(A_{n-1}\lor A_n)\ldots)))))\ldots))$$ Finally, if $$i > 1$$, then we add the formulas to the left. For each of $$i-1$$ formulas we add a simultaneous substitution to Ax 10. (4 substitutions) and an application of MP starting with $$A_{i-1}$$ and ending with $$A_1$$. If $$i+1 = k$$, then we start with the proof of $$p\lor\lnot p$$ and substitute $$B$$ for $$p$$. For the cases of $$i>1$$ and/or $$n>k$$ we proceed as above. If $$i=1$$ and/or $$k=n$$, then we omit, respectively, the steps connected with adding the outer formulas. Finally, if the sequent is of the form:   $$\vdash A_1, \ldots, A_{i-1}, \lnot B, A_{i+1}, \ldots, A_{k-1}, B, A_{k+1}, \ldots, A_n$$ then we start with the proof of formula $$\lnot p \lor (q \lor p)$$, if $$i+1 < k$$, or with the proof of $$\lnot p \lor p$$ otherwise. ■ Let $$l_{base~sequent}$$ stand for the length of the proof of the formula corresponding to the base sequent $$\vdash A_1, \ldots, A_n$$. We estimate $$l_{base~sequent}$$ in the worst case:   $$l_{base~sequent}\leq 458+6\cdot(k-i-2)+15\cdot(k-i-3)+6+$$   $$+6\cdot(k-(i-1))+15\cdot(k-(i-1))+5\cdot(i-1)=423+5\cdot k+37\cdot(k-i).$$ Since $$k\leq n$$ and $$k-i< n$$, where $$n$$ is the number of terms of the initial base sequent, we have $$l_{base~sequent}\leq 423+42\cdot n$$, which means that: Corollary 1 The length of the proof of the formula corresponding to a base sequent depends linearly on the number of terms of the sequent, i.e.  $$l_{base~sequent}=\mathcal{O}(n)$$ As to the width, the longest formula that occurs in the above proof is either the one in the proof of formula $$p \lor (q \lor \lnot p)$$ ($$w_{p \lor (q \lor \lnot p)}=51$$), or the one whose length is $$3\cdot\mathbf{L}+2$$ (see above), or, finally, the following one (the last substitution to Ax 10.):   $$D^{\star\star} = A_2 \lor \ldots \lor A_{i-1} \lor \lnot B \lor A_{i+1} \lor \ldots \lor A_{k-1} \lor B \lor A_{k+1} \lor \ldots \lor A_n \to$$   $$A_1 \lor \ldots \lor A_{i-1} \lor \lnot B \lor A_{i+1} \lor \ldots \lor A_{k-1} \lor B \lor A_{k+1} \lor \ldots \lor A_n.$$ We omit parentheses, since $$|D^{\star\star}|$$ does not depend on them. For $$\phi =\ \vdash A_1, \ldots, A_n$$, let $$|\phi| = \sum^{n}_{i=1} |A_i|$$. Then $$|D^{\star\star}|=|\phi|-|A_1|+(n-2)+|\phi|+1+(n-1) \leq 2\cdot(|\phi|+n)$$, whereas $$3\cdot\mathbf{L}+2 \leq 3\cdot(|\phi|+n)+2$$. Since $$2\cdot(|\phi|+n) < 3\cdot(|\phi|+n)+2$$, we have:   $$w_{base~sequent} \leq \max\{51,3\cdot(|\phi|+n)+2\},$$ where for long formulas we will always have $$3\cdot(|\phi|+n)+2 > 51$$. This entails: Corollary 2   $$w_{base~sequent} = \mathcal{O}(|\phi|).$$ Since the number $$n$$ of terms of the base sequent is not greater than $$|\phi|$$, we finally obtain: Corollary 3   $$s_{base~sequent} = \mathcal{O}(|\phi|^2).$$ where $$s_{base~sequent}$$ stands for the size of the proof of the formula corresponding to a base sequent. 4.2 The applications of rules of calculus $$\mathbb{E^{**}}$$ Definition 4 Let $$S = \langle A_{1}, A_{2}, \ldots, A_{n-1}, A_{n} \rangle$$ be a finite, non-empty sequence of formulas of $$\mathcal{L}$$. Then by:   $$\bigvee S$$ we shall mean the following disjunction:   $$A_{1}\vee (A_{2}\vee ...(A_{n-1} \vee A_{n})...).$$ The aim of this subsection is to prove two theorems stated below. Theorem 5 If the following schema:   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E^{**}}$$, and there exists a proof of formula $$\bigvee T$$ in system $$\mathbb{A}$$, then the proof may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. Theorem 6 If the following schema:   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T_{1} \ ; \ \vdash T_{2} \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E^{**}}$$, and there exist a proof of formula $$\bigvee T_{1}$$ and a proof of formula $$\bigvee T_{2}$$ in system $$\mathbb{A}$$, then the proofs may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. The proofs will be conducted with respect to each rule schema of $$\mathbb{E^{**}}$$. Since the rules of $$\mathbb{E^{**}}$$ are given in the form of schemas with metavariables, we will present schemas of the proofs in $$\mathbb{A}$$ also with the use of metavariables. 4.2.1 Rule $$\mathbf{R}_{\lnot\lnot}$$   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \ ' \ A \ ' \ S_{2} \ ; \ \Psi)} $$ Suppose that $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$ (but recall that $$S_1$$ may be empty). We show that if there exists a proof of formula:   \begin{equation}\label{A} C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor (\bigvee S_{2}))))\ldots) \end{equation} (8) in system $$\mathbb{A}$$, then there exists also a proof of:   \begin{equation}\label{not-not-A} C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2}))))\ldots) \end{equation} (9) in system $$\mathbb{A}$$. We show how to obtain one from the other. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor (\bigvee S_{2}))))\ldots)$$$$\hspace{0,2cm}$$[proved by assumption] (2) $$p\to\lnot\lnot p$$$$\hspace{0,2cm}$$[Ax 5.] (3) $$A\to\lnot\lnot A$$$$\hspace{0,2cm}$$[2 Sub: $$p/A$$] (4) (100) 39$$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[disjunct added to the right] (5) (6) $$(A\to \lnot\lnot A)\to (A\lor (\bigvee S_{2}) \to \lnot\lnot A\lor (\bigvee S_{2}))$$$$\hspace{0,2 cm}$$[4 SSub: $$p/A,q/\lnot\lnot A,r/\bigvee S_{2}$$] (6) $$A\lor (\bigvee S_{2}) \to \lnot\lnot A\lor (\bigvee S_{2})$$$$\hspace{0,2 cm}$$[5,3 MP] (7) (68) 39$$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] (8) (6) $$(A\lor (\bigvee S_{2}) \to \lnot\lnot A\lor (\bigvee S_{2}))\to(C_n\lor(A\lor (\bigvee S_{2})) \to C_n\lor(\lnot\lnot A\lor (\bigvee S_{2})))$$ $$\quad$$ [4 SSub: $$p/A\lor (\bigvee S_{2}),q/ \lnot\lnot A\lor (\bigvee S_{2}),r/C_n$$] (9) $$C_n\lor(A\lor (\bigvee S_{2})) \to C_n\lor(\lnot\lnot A\lor (\bigvee S_{2}))$$ [8,6 MP] $$\vdots$$ Steps in lines 8,9 repeated with respect to $$C_{n-1},\ldots,C_1$$ (10) $$\Large($$$$(C_2\lor\ldots\lor (C_n\lor (A \lor (\bigvee S_{2})))\ldots) \to (C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$$$\Large)$$$$\to$$$$\Large($$$$C_1\lor(C_2\lor\ldots\lor (C_n\lor (A \lor (\bigvee S_{2})))\ldots) \to C_1\lor(C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$$$\Large)$$ (11) $$C_1\lor(C_2\lor\ldots\lor (C_n\lor (A \lor (\bigvee S_{2})))\ldots) \to C_1\lor(C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$ (12) $$C_1\lor(C_2\lor\ldots\lor (C_n\lor (\lnot\lnot A \lor (\bigvee S_{2})))\ldots)$$$$\hspace{0,2 cm}$$[10,1 MP] Let $$l_C$$ stand for the length of the proof of formula $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor (\bigvee S_{2}))))\ldots)$$ and $$w_C$$ for its width. Then the length of the above proof may be estimated as $$l_C+2+l_{dis~r}+7+l_{dis~l}+7\cdot n+1=l_C+178+7\cdot n$$, where $$n$$ is the number of the terms of $$S_1$$. Therefore the length of the above proof, $$l_{\lnot\lnot}$$ in symbols, depends linearly on the length of $$l_C$$ and the number $$n$$ of terms of $$S_1$$:   \begin{equation}\label{podwojna negacja} l_{\lnot\lnot}= l_C + 7\cdot n + c \end{equation} (10) for a constant $$c$$. The length of the formula in line 10, $$l(10)$$ for short, is bounded by   $$l(10) \leq 4\cdot|\vdash S_{1} \ ' \ A \ ' \ S_{2}|+ 4\cdot|\vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2}|.$$ The coefficients ‘4’ may seem too large, but this overestimation is to include the occurrences of connectives between the formulas. Since $$|\vdash S_{1} \ ' \ A \ ' \ S_{2}| < |\vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2}|$$, the width of the above proof, $$w_{\lnot\lnot}$$, may be estimated as follows:   \begin{equation}\label{w podwojna negacja} w_{\lnot\lnot} \leq \max\{w_C, 39, 8\cdot|\vdash S_{1} \ ' \ \lnot\lnot A \ ' \ S_{2}|\}. \end{equation} (11) 4.2.2 Rule $$\mathbf{R}_\to$$   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \ ' \ A\to B \ ' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \ ' \ \neg A \ ' \ B \ ' \ S_{2} \ ; \ \Psi)}. $$ Let $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$. We show that if there exists a proof of formula:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots)$$ in system $$\mathbb{A}$$, then there exists also a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\to B) \lor (\bigvee S_{2}))))\ldots)$$ in system $$\mathbb{A}$$. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots)$$$$\hspace{0,2 cm}$$[proved by assumption] (2) (179) 51$$p\to (\neg p\to q)$$$$\hspace{0,2 cm}$$ [Duns Scotus law] (3) (4) $$A\to (\neg A\to B)$$$$\hspace{0,2 cm}$$ [2 SSub: $$p/A, q/B$$] (4) (74) $$\neg A\to (A\to B)$$$$\;\;$$ [3 Com] (5) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (6) (6) $$(\neg A\to (A\to B))\to ((B\to (A\to B))\to (\neg A\vee B\to(A\to B)))$$ $$\quad$$ [5 SSub: $$p/\neg A, q/B, r/A\to B$$] (7) $$(B\to (A\to B))\to (\neg A\vee B\to(A\to B))$$$$\;\;$$ [6,4 MP] (8) $$p\to(q\to p)$$$$\hspace{0,2 cm}$$[Ax 1.] (9) (4) $$B\to (A\to B)$$$$\;\;$$ [8 SSub: $$p/B, q/A$$] (10) $$\neg A\vee B\to(A\to B)$$$$\;\;$$ [7,9 MP] (11) (100) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[disjunct added to the right] (12) (6) $$(\lnot A\lor B\to (A\to B))\to ((\lnot A\lor B)\lor (\bigvee S_2) \to (A\to B)\lor (\bigvee S_2))$$ $$\hspace{0,2 cm}$$[11 SSub: $$p/\lnot A\lor B,q/A\to B,r/\bigvee S_2$$] (13) $$(\lnot A\lor B)\lor (\bigvee S_2) \to (A\to B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[12,10 MP] (14) (87) $$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[associativity of disjunction 2] (15) (6) $$\lnot A\lor (B \lor (\bigvee S_2)) \to (\lnot A\lor B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[14 SSub: $$p/\lnot A,q/B,r/\bigvee S_2$$] (16) (8) $$\lnot A\lor(B\lor(\bigvee S_2))\to (A\to B)\lor(\bigvee S_2)$$$$\hspace{0,2 cm}$$[15,13 HS] (17) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] (18) (6) $$(\lnot A\lor(B\lor(\bigvee S_2)) \to (A\to B)\lor(\bigvee S_2))\to (C_n\lor (\lnot A\lor(B\lor(\bigvee S_2))) \to C_n\lor ((A\to B)\lor(\bigvee S_2)))$$$$\hspace{0,2 cm}$$[17 SSub: $$p/\lnot A\lor(B\lor(\bigvee S_2)),q/(A\to B)\lor(\bigvee S_2),r/C_n$$] (19) $$C_n\lor (\lnot A\lor(B\lor(\bigvee S_2))) \to C_n\lor ((A\to B)\lor(\bigvee S_2))$$$$\hspace{0,2 cm}$$[18,16 MP] $$\vdots$$ steps 18, 19 (7 lines) repeated with respect to $$C_{n-1},\ldots,C_1$$ (20) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots) \to$$$$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\to B) \lor (\bigvee S_{2}))))\ldots)$$ (21) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\to B) \lor (\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$[20,1 MP] Let $$l_C$$, $$w_C$$ stands for the length and width of the proof of formula $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(B \lor (\bigvee S_{2})))))\ldots)$$. Then the length of the above proof, $$l_\to$$ in symbols, equals $$480+l_C+7\cdot n$$, thus   \begin{equation}\label{implikacja} l_\to = l_C + 7\cdot n + c \end{equation} (12) for a constant $$c$$. For the width, let $$\psi$$ stand for ‘$$\vdash S_{1} \ ' \ \lnot A \ ' \ B \ ' \ S_{2}$$’. Probably, the longest formula in the proof is the one analogous to that in line no. (18), obtained after adding $$C_1$$. Its length can be estimated by the (upper bound) value $$8\cdot|\psi|$$. Therefore:   \begin{equation}\label{w implikacja} w_\to \leq \max\{w_C, 51, 8\cdot|\psi|\}. \end{equation} (13) 4.2.3 Rule $$\mathbf{R}_{\lnot \wedge}$$ Now we shall consider:   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ \neg (A\wedge B) \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ \neg A \ ' \ \neg B \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)}. $$ We show that a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\lnot B \lor (\bigvee S_{2})))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot (A\land B) \lor (\bigvee S_{2}))))\ldots)$$ (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\lnot B \lor (\bigvee S_{2})))))\ldots)$$$$\hspace{0,2 cm}$$[proved by assumption] (2) $$(p\to q)\to (\neg q\to \neg p)$$$$\;\;$$ [Ax 3.] (3) (4) $$(A\wedge B\to A)\to (\neg A\to \neg(A\wedge B))$$$$\;\;$$ [2 SSub: $$p/A\wedge B, q/A$$] (4) $$p\wedge q\to p$$$$\;\;$$ [Ax 6.] (5) (4) $$A\wedge B\to A$$$$\;\;$$ [4 SSub: $$p/A, q/B$$] (6) $$\neg A\to \neg(A\wedge B)$$$$\;\;$$ [3,5 MP] (7) (4) $$(A\wedge B\to B)\to (\neg B\to \neg(A\wedge B))$$$$\;\;$$ [2 Sub: $$p/A\wedge B, q/B$$] (8) $$p\wedge q\to q$$$$\;\;$$ [Ax 7.] (9) (4) $$A\wedge B\to B$$$$\;\;$$ [8 SSub: $$p/A, q/B$$] (10) $$\neg B\to \neg(A\wedge B)$$$$\;\;$$ [7,9 MP] (11) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (12) (6) $$(\neg A\to \neg(A\wedge B)) \to ((\neg B\to \neg(A\wedge B)) \to (\neg A\vee \neg B\to \neg(A\wedge B)))$$ $$\quad$$ [11 SSub: $$p/\neg A, q/\neg B, r/\neg(A\wedge B)$$] (13) $$(\neg B\to \neg(A\wedge B)) \to (\neg A\vee \neg B\to \neg(A\wedge B))$$$$\;\;$$ [12,6 MP] (14) $$\neg A\vee \neg B\to \neg(A\wedge B)$$$$\;\;$$ [13,10 MP] (15) (100) 39$$(p\to q)\to (p\lor r \to q\lor r)$$$$\hspace{0,2 cm}$$[disjunct added to the right] (16) (6) $$(\lnot A\lor \lnot B\to \lnot(A\land B))\to ((\lnot A\lor \lnot B)\lor (\bigvee S_2) \to \lnot(A \land B)\lor (\bigvee S_2))$$ $$\quad$$ [15 SSub: $$p/\lnot A\lor \lnot B,q/\lnot(A\land B),r/\bigvee S_2$$] (17) $$(\lnot A\lor \lnot B)\lor (\bigvee S_2) \to \lnot(A\land B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[16,14 MP] (18) (87) $$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[associativity of disjunction 2] (19) (6) $$\lnot A\lor (\lnot B \lor (\bigvee S_2)) \to (\lnot A\lor \lnot B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[18 SSub: $$p/\lnot A,q/\lnot B,r/\bigvee S_2$$] (20) (8) $$\lnot A\lor(\lnot B\lor(\bigvee S_2))\to \lnot(A\land B)\lor(\bigvee S_2)$$$$\hspace{0,2 cm}$$[19,17 HS] (21) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] (22) (6) $$\Large($$$$\lnot A\lor(\lnot B\lor(\bigvee S_2)) \to \lnot(A\land B)\lor(\bigvee S_2)$$$$\Large)$$$$\to$$$$\Large($$$$C_n\lor (\lnot A\lor(\lnot B\lor(\bigvee S_2))) \to$$ $$\quad C_n\lor (\lnot(A\land B)\lor(\bigvee S_2))$$$$\Large)$$$$\hspace{0,2 cm}$$[21 SSub: $$p/\lnot A\lor(\lnot B\lor(\bigvee S_2)),q/\lnot(A\land B)\lor(\bigvee S_2),r/C_n$$] (23) $$C_n\lor (\lnot A\lor(\lnot B\lor(\bigvee S_2))) \to C_n\lor (\lnot(A\land B)\lor(\bigvee S_2))$$$$\hspace{0,2 cm}$$[22,20 MP] $$\vdots$$ steps 22,23 (7 lines) repeated with respect to $$C_{n-1},\ldots,C_1$$ (24) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\lnot B \lor (\bigvee S_{2})))))\ldots) \to$$$$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\land B) \lor (\bigvee S_{2}))))\ldots)$$ (25) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\land B) \lor (\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$[24,1 MP] As in the previous cases, we assume that $$l_C$$ and $$w_C$$ stand for the length and width (respectively) of the formula in the first line. The length, $$l_{\lnot\land}$$, of the above proof equals $$l_C+307+7\cdot n$$, thus   \begin{equation}\label{nie i} l_{\lnot\land} = l_C+7\cdot n + c \end{equation} (14) for a constant, $$c$$, and   \begin{equation}\label{w nie i} w_{\lnot\land} \leq \max\{w_C, 39, 8\cdot|\psi|\}, \end{equation} (15) where $$\psi = ~ \vdash S_{1} \hspace{0,1 cm}' \ \neg A \ ' \ \neg B \hspace{0,1 cm}' \ S_{2}$$. 4.2.4 Rule $$\mathbf{R}_\lor$$ To prove Theorem 5, we still need to consider:   $$ \frac {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ A\lor B \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)} {?(\Phi \ ; \ \vdash S_{1} \hspace{0,1 cm}' \ A \ ' \ B \hspace{0,1 cm}' \ S_{2} \ ; \ \Psi)}. $$ Now we need to show that a proof of formula:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(B \lor (\bigvee S_{2})))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\lor B) \lor (\bigvee S_{2}))))\ldots).$$ For this purpose we start with: (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(B \lor (\bigvee S_{2})))))\ldots)$$ [proved by assumption] (2) (87) 39$$p\lor (q \lor r) \to (p\lor q)\lor r$$$$\hspace{0,2 cm}$$[associativity of disjunction 2] (3) (6) $$A\lor (B \lor (\bigvee S_2)) \to (A\lor B)\lor (\bigvee S_2)$$$$\hspace{0,2 cm}$$[2 SSub: $$p/A,q/B,r/\bigvee S_2$$] (4) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\hspace{0,2 cm}$$[disjunct added to the left] and then we add formulas $$C_{n},\ldots,C_1$$ to the left, as in the above cases. This time we have $$l_\lor = l_C+162+7\cdot n$$, thus   \begin{equation}\label{alternatywa} l_\lor = l_C+7\cdot n + c \end{equation} (16) for a constant, $$c$$. Again, as previously,   \begin{equation}\label{w alternatywa} w_\lor \leq \max\{w_C, 39, 8\cdot|\psi|\}, \end{equation} (17) where $$\psi ~=~ \vdash S_{1} \hspace{0,1 cm}' \ A \ ' \ B \hspace{0,1 cm}' \ S_{2}$$. Proof of Theorem 5 The schemas of proofs presented in subsections 4.2.1–4.2.4 supply us with the proof of Theorem 5. We have shown that if   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E^{**}}$$, and there exists a proof of formula $$\bigvee T$$ in system $$\mathbb{A}$$, then the proof may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. Moreover, it is important that the schemas show us how to transform one proof into another. We have also seen that if $$l_C, l_P$$ and $$w_C, w_P$$ stand for the lengths and widths of proofs of the formulas corresponding to, respectively, ‘sequent-conclusion’ $$\vdash T$$ and ‘sequent-premise’ $$\vdash S$$, then by (10), (12), (14), (16): Corollary 4   $$l_P \leq l_C + f(n),$$ where $$n$$ stands for the number of terms of sequence $$T$$, and $$f$$ is a linear function. Observe that, strictly speaking, $$n$$ could be defined as the number of terms of an initial subsequence of $$T$$, not of the whole sequence. But this is enough for our purposes. On the other hand, we can estimate $$f$$ very precisely, as in the case of each rule we have obtained $$f(n)=7\cdot n + c$$ for some constant $$c \in \text{N}_+$$. In turn, by (11), (13), (15), (17): Corollary 5   $$w_P \leq \max\{w_C,c,8\cdot|\vdash T|\}$$ for a constant $$c$$$$(c = 39)$$. 4.2.5 Rule $$\mathbf{R}_\land$$ We proceed to the proof of Theorem 6.   $$ \frac {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' A\wedge B \hspace{0,1 cm}' ~S_{2}~;~ \Psi~)} {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' ~ A \hspace{0,1 cm}'~S_{2}~;~ \vdash S_{1} \hspace{0,1 cm}' ~ B \hspace{0,1 cm}'~S_{2}~;~\Psi~)}. $$ We will show that if there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, and there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots)$$ then the two may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\land B) \lor (\bigvee S_{2}))))\ldots).$$ We start with the proof of the law of conjunction, and distributivity of disjunction with respect to conjunction. The law of conjunction (1) $$p\to (q\to p)$$$$\;\;$$ [Ax 1.] (2) $$(p\to q)\to ((p\to r)\to (p\to q\wedge r))$$$$\;\;$$ [Ax 8.] (3) (6) $$(q\to p)\to ((q\to q)\to (q\to p\wedge q))$$$$\;\;$$ [2 SSub: $$p/q, q/p, r/q$$] (4) (8) $$p\to ((q\to q)\to (q\to p\wedge q))$$$$\;\;$$ [1,3 HS] (5) (7) $$(q\to q)\to (p\to (q\to p\wedge q))$$$$\;\;$$ [4 Com] (6) (258) 51$$p\to p$$$$\;\;$$ [identity law] (7) $$q\to q$$$$\;\;$$ [5 Sub: $$p/q$$] (8) $$p\to (q\to p\wedge q)$$$$\;\;$$ [4,6 MP] In counting the length of the above proof we omit the costs of deriving the laws of hypothetical syllogism and commutation, as they are already counted in the 258 lines for the identity law. Thus we have $$l_{conj} = 283$$ and $$w_{conj}=51$$. The laws of distributivity of disjunction with respect to conjunction (1) $$p\to(q\to p)$$$$\;\;$$ [Ax 1.] (2) $$q\to(p\lor q)$$$$\;\;$$ [Ax 10.] (3) (4) $$r\to((p\land q)\lor r)$$$$\;\;$$ [2 SSub: $$p/p\land q,q/r$$] (4) (4) $$(r\to((p\land q)\lor r)) \to (q\lor r \to (r\to((p\land q)\lor r)))$$$$\;\;$$ [1 SSub: $$p/r\to((p\land q)\lor r),q/q\lor r$$] (5) $$q\lor r \to (r\to((p\land q)\lor r))$$$$\;\;$$ [4,3 MP] (6) (7) $$r\to (q\lor r \to (p\land q)\lor r)$$$$\;\;$$ [5 Com] (7) (283) 51$$p\to(q\to p\wedge q)$$$$\;\;$$ [law of conjunction] (8) (100) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\;\;$$ [disjunct added to the right] (9) (4) $$(q\to p\land q)\to (q\lor r \to (p\land q)\lor r)$$$$\;\;$$ [8 SSub: $$p/q,q/p\land q$$] (10) (8) $$p\to (q\lor r\to (p\land q)\lor r)$$$$\;\;$$ [7,9 HS] (11) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (12) (4) $$(p\to (q\lor r\to (p\land q)\lor r))\to ((r\to (q\lor r\to (p\land q)\lor r))\to (p\vee r\to (q\lor r\to (p\land q)\lor r)))$$ $$\quad$$ [11 SSub: $$q/r,r/q\lor r\to (p\land q)\lor r$$] (13) $$(r\to (q\lor r\to (p\land q)\lor r))\to (p\vee r\to (q\lor r\to (p\land q)\lor r))$$$$\;\;$$ [12,10 MP] (14) $$p\vee r\to (q\lor r\to (p\land q)\lor r)$$$$\;\;$$ [13,6 MP] (15) (7) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$ [14 Imp] Since the proof of the identity law goes through the laws of commutation and importation, and the proof of the law of conjunction goes through the identity law, we may count the length of the above proof, $$l_{distr~1}$$, as equal to $$427$$. In turn, $$w_{distr~1}=51$$. Another law of distributivity: (1) $$p\to(q\to p)$$$$\;\;$$ [Ax 1.] (2) $$p\to p\lor q$$$$\;\;$$ [Ax 9.] (3) $$p\to p\lor (q\land r)$$$$\;\;$$ [2 Sub: $$q/q\land r$$] (4) (4) $$(p\to p\lor (q\land r)) \to (p\lor r \to (p\to p\lor (q\land r)))$$$$\;\;$$ [1 SSub: $$p/r\to((p\land q)\lor r),q/p\lor r$$] (5) $$p\lor r \to (p\to p\lor (q\land r)))$$$$\;\;$$ [4,3 MP] (6) (7) $$p\to (p\lor r \to p\lor (q\land r))$$$$\;\;$$ [5 Com] (7) (283) 51$$p\to(q\to p\wedge q)$$$$\;\;$$ [law of conjunction] (8) (4) $$q\to(r\to q\wedge r)$$$$\;\;$$ [7 SSub: $$p/q,q/r$$] (9) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\;\;$$ [disjunct added to the left] (10) (6) $$(r\to q\land r)\to (p\lor r \to p\lor(q\land r))$$$$\;\;$$ [9 SSub: $$p/r,q/q\land r,r/p$$] (11) (8) $$q\to (p\lor r \to p\lor(q\land r))$$$$\;\;$$ [8,10 HS] (12) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (13) (4) $$(p\to (p\lor r \to p\lor (q\land r)))\to ((q\to (p\lor r \to p\lor(q\land r)))\to (p\vee q\to (p\lor r\to p\lor (q\land r))))$$ $$\quad$$ [12 SSub: $$q/r,r/p\lor r\to p\lor (q\land r)$$] (14) $$(q\to (p\lor r \to p\lor(q\land r)))\to (p\vee q\to (p\lor r\to p\lor (q\land r)))$$$$\;\;$$ [13,6 MP] (15) $$p\vee q\to (p\lor r\to p\lor (q\land r))$$$$\;\;$$ [14,11 MP] (16) (7) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [15 Imp] Similarly as before, the length of the above proof, $$l_{distr~2}$$, is estimated at $$398$$, whereas the width, $$w_{distr~2}$$, equals 51. We go back to the analysis of rule $$\mathbf{R}_\land$$. In line no. 3 below we use the derived rule based on the law of conjunction, Conj for short. Its cost would be 283+6, but the 283 lines are included in the cost of the first law of distributivity. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (2) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (3) (6) 51$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)) ~\land $$$$~~~~~~~~ (C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots))$$$$\;\;$$ [1,2 Conj] (4) (427) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$ [distributivity 1] (5) (6) $$(A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2))\to (A\land B)\lor (\bigvee S_2)$$$$\;\;$$ [4 Sub: $$p/A, q/B,r/\bigvee S_2$$] (6) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\;\;$$ [disjunct added to the left] (7) (6) $$\Large($$$$(A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2)) \to (A\land B)\lor (\bigvee S_2)$$$$\Large)$$$$\to$$$$~~~~$$$$\Large($$$$C_n \lor ((A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2))) \to C_n \lor ((A\land B)\lor (\bigvee S_2))$$$$\Large)$$$$\;\;$$ [6 Sub: $$p/(A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2)),q/(A\land B)\lor (\bigvee S_2),r/C_n$$] (8) $$C_n \lor ((A\lor (\bigvee S_2))\land(B\lor (\bigvee S_2))) \to C_n \lor ((A\land B)\lor (\bigvee S_2))$$$$\;\;$$ [7,5 MP] (9) (398) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [distributivity 2] (10) (6) $$(C_n\lor (A\lor (\bigvee S_2)))\land (C_n\lor (B\lor (\bigvee S_2)))\to C_n\lor ((A\lor (\bigvee S_2))\land (B\lor (\bigvee S_2)))$$ $$\quad$$ [9 Sub: $$p/C_n, q/ A\lor (\bigvee S_2), r/B\lor (\bigvee S_2)$$] (11) (8) $$(C_n\lor (A\lor (\bigvee S_2)))\land (C_n\lor (B\lor (\bigvee S_2))) \to C_n \lor ((A\land B)\lor (\bigvee S_2))$$ $$\quad$$ [10,8 HS] $$\vdots$$ steps 7, 8, 10, 11 (21 lines) repeated with respect to $$C_{n-1},\ldots,C_1$$ (12) $$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)) ~\land $$$$~~~~~~~~ (C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (B \lor(\bigvee S_{2}))))\ldots)) ~~\to$$ $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\land B) \lor (\bigvee S_{2}))))\ldots)$$ (13) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor ((A\land B) \lor (\bigvee S_{2}))))\ldots)$$ [12,3 MP] Let $$l_{C1}$$ and $$l_{C2}$$ stand for the length of proofs of formulas in lines 1 and 2, respectively; similarly for the widths $$w_{C1}$$, $$w_{C2}$$. The length of the above proof, $$l_\land$$, may be estimated by $$l_{C1}+l_{C2}+906+21\cdot n$$, thus   \begin{equation}\label{koniunkcja} l_\land = l_{C1}+l_{C2}+21\cdot n+c \end{equation} (18) for a constant, $$c$$. To estimate the width, $$w_\land$$, of the above proof, observe that the longest formula is either that of length $$w_{C1}$$, $$w_{C2}$$, 51, or it is the result of the substitution to the law of hypothetical syllogism, which is necessary to go through line no. 11 for the last time. More specifically, we mean the result of the simultaneous substitution of ‘$$(C_1 \lor \ldots \lor (C_n\lor (A\lor (\bigvee S_2)))\ldots)\land (C_1 \lor \ldots \lor (C_n\lor (B\lor (\bigvee S_2)))\ldots)$$’ for $$p$$, ‘$$C_1 \lor ((C_2 \lor \ldots \lor (A\lor (\bigvee S_2)))\land (C_2 \lor \ldots \lor (B\lor (\bigvee S_2))))$$’ for $$q$$ and ‘$$C_1 \lor (C_2 \lor \ldots \lor ((A\land B)\lor (\bigvee S_2)))$$’ for $$r$$. It is easy to see that the length of this formula is bounded by $$8\cdot|\psi_1|+8\cdot|\psi_2|+4\cdot|\phi|$$, where $$\psi_1 = \ \vdash S_1 \ ' A \ ' S_2$$, $$\psi_2 = \ \vdash S_1 \ ' B \ ' S_2$$, $$\phi = \ \vdash S_1 \ ' A\land B \ ' S_2$$. Since, moreover, $$|\psi_1| < |\phi|$$ and $$|\psi_2| < |\phi|$$, we can estimate $$w_\land$$ as follows   \begin{equation}\label{w koniunkcja} w_\land \leq \max\{w_{C1}, w_{C2}, c, 20\cdot|\phi|\} \end{equation} (19) for a constant, $$c$$ ($$c=51$$). 4.2.6 Rule $$\mathbf{R}_{\lnot\lor}$$   $$ \frac {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg (A\vee B) \hspace{0,1 cm}'~ S_{2}~;~ \Psi~)} {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' ~\neg A \hspace{0,1 cm}'~S_{2}~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg B \hspace{0,1 cm}'~S_{2}~;~\Psi)} .$$ We will show that if there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\bigvee S_{2}))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, and there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$ then the two may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\lor B) \lor (\bigvee S_{2}))))\ldots).$$ For that, we will need the proof of one of the De Morgan’s laws. De Morgan’s law (1) $$(p\to r)\to ((q\to r)\to (p\vee q\to r))$$$$\;\;$$ [Ax 11.] (2) $$(p\to (\neg p\to q)) \to ((q \to (\neg p\to q)) \to (p\vee q\to (\neg p\to q)))$$$$\;\;$$ [1 Sub: $$r/\neg p\to q$$] (3) (178) 51$$p\to (\neg p\to q)$$$$\;\;$$ [Duns Scotus law] (4) $$(q\to (\neg p\to q)) \to (p\vee q\to (\neg p\to q))$$$$\;\;$$ [2,3 MP] (5) $$p\to (q\to p)$$$$\;\;$$ [Ax 1.] (6) (4) $$q\to (\neg p\to q)$$$$\;\;$$ [5 SSub: $$p/q, q/\neg p$$] (7) $$p\vee q\to (\neg p\to q)$$$$\;\;$$ [4,6 MP] (8) (7) $$\neg p\to (p\vee q\to q)$$$$\;\;$$ [7 Com] (9) $$(p\to q)\to (\neg q\to \neg p)$$$$\;\;$$ [Ax 3.] (10) $$((p\vee q)\to q)\to (\neg q\to \neg (p\vee q))$$$$\;\;$$ [9 Sub: $$p/p\vee q$$] (11) (8) $$\neg p \to (\neg q \to \neg(p\vee q))$$$$\;\;$$ [8,10 HS] (12) (7) $$\neg p\wedge \neg q\to \neg (p\vee q)$$$$\;\;$$[11 Imp] Counting similarly as before, the length of the above proof, $$l_{DM}$$, is bounded by $$211$$, and the width, $$w_{DM}=51$$. We go back to the analysis of rule $$R_{\lnot\lor}$$. (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (2) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (3) (6) 51$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot A \lor(\bigvee S_{2}))))\ldots)) ~\land$$$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots))$$$$\;\;$$ [1,2 Conj] (4) (211) $$\neg p\wedge \neg q\to \neg (p\vee q)$$$$\;\;$$[De Morgan’s law] (5) (4) $$\neg A\wedge \neg B\to \neg (A\vee B)$$$$\;\;$$[4 SSub: $$p/A,q/B$$] (6) (100) $$(p\to q)\to (p\lor r \to q\lor r)$$$$\;\;$$[disjunct added to the right] (7) (6) $$(\neg A\wedge \neg B \to \neg (A\vee B))\to ((\neg A\wedge \neg B)\lor (\bigvee S_2) \to \neg (A\vee B)\lor (\bigvee S_2))$$ $$\quad$$ [6 SSub: $$p/\neg A\wedge \neg B,q/\neg (A\vee B),r/\bigvee S_2$$] (8) $$(\neg A\wedge \neg B)\lor (\bigvee S_2) \to \neg (A\vee B)\lor (\bigvee S_2)$$$$\;\;$$[7,5 MP] (9) (427) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$[distributivity 1] (10) (6) $$(\neg A\lor(\bigvee S_2))~ \land~ (\neg B\lor (\bigvee S_2)) \to (\neg A\wedge \neg B)\lor (\bigvee S_2)$$$$\;\;$$ [9 SSub: $$p/\lnot A$$, $$q/\lnot B$$, $$r/\bigvee S_2$$] (11) (8) $$(\neg A\lor(\bigvee S_2)) \land (\neg B\lor (\bigvee S_2)) \to \neg (A\vee B)\lor (\bigvee S_2)$$$$\;\;$$ [10,8 HS] (12) (398) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [distributivity 2] $$\vdots$$ After this step we apply the same mechanism of adding (to the left) formulas $$C_n, \ldots,C_1$$ which was presented in the analysis of rule $$\mathbf{R}_{\land}$$ (it yields $$68$$ lines of the proof of ‘adding to the left’, $$21\cdot n$$ lines $$+ 1$$ line for the final MP). By arguments similar to those presented in the previous case we have   \begin{equation}\label{nie lub} l_{\lnot\lor} = l_{C1}+l_{C2}+21\cdot n+c \end{equation} (20) for $$c=1236$$. For the width we observe that the length of the formula which may be the longest in the proof is bounded by $$8\cdot|\psi_1|+8\cdot|\psi_2|+4\cdot|\phi|$$, where $$\psi_1 = \ \vdash S_1 \ ' \lnot A \ ' S_2$$, $$\psi_2 = \ \vdash S_1 \ ' \lnot B \ ' S_2$$, $$\phi = \ \vdash S_1 \ ' \lnot(A\lor B) \ ' S_2$$. For the same reason as in the previous case, we have   \begin{equation}\label{w nie lub} w_{\lnot\lor} \leq \max\{w_{C1}, w_{C2}, c, 20\cdot|\phi|\} \end{equation} (21) for a constant, $$c$$ ($$c=51$$). 4.2.7 Rule $$\mathbf{R}_{\neg \to}$$ The last rule to consider is:   $$ \frac {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg(A\to B) \hspace{0,1 cm}'~ S_{2}~;~ \Psi~)} {?(~\Phi ~;~ \vdash S_{1} \hspace{0,1 cm}' ~ A \hspace{0,1 cm}'~S_{2}~;~ \vdash S_{1} \hspace{0,1 cm}'~ \neg B \hspace{0,1 cm}'~S_{2}~;~\Psi~)} .$$ For the last time, we show that if there exists a proof of formula:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots),$$ where $$S_1 = \langle C_1,C_2,\ldots,C_{n-1},C_n \rangle$$, and there exists a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$ then the two may be used to obtain a proof of:   $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot(A\to B) \lor (\bigvee S_{2}))))\ldots).$$ Again, we will need to prove one additional law. Formula $$p\land\lnot q\to\lnot(p\to q)$$ (1) $$p\to (q\to p)$$$$\;\;$$ [Ax 1.] (2) (4) $$\neg p\vee p\to ((p\to q)\to \neg p\vee p)$$$$\;\;$$ [1 SSub: $$p/\neg p\vee p, q/p\to q$$] (3) (278) 51$$\neg p \vee p$$$$\;\;$$ [reversed law of excluded middle] (4) $$(p\to q)\to \neg p\vee p$$$$\;\;$$ [2,3 MP] (5) $$(p\to (q\to r))\to((p\to q)\to (p\to r))$$$$\;\;$$ [Ax 2.] (6) (6) $$((p\to q)\to (\neg p\vee p\to \neg p\vee q))\to(((p\to q)\to \neg p\vee p)\to ((p\to q)\to \neg p\vee q))$$ $$\quad$$ [5 SSub: $$p/p\to q, q/\neg p\vee p, r/\neg p\vee q$$] (7) (68) $$(p\to q)\to (r\lor p \to r\lor q)$$$$\;\;$$[disjunct added to the left] (8) $$(p\to q)\to (\neg p\vee p\to \neg p\vee q)$$$$\;\;$$ [7 Sub: $$r/ \neg p$$] (9) $$((p\to q)\to \neg p\vee p)\to ((p\to q)\to \neg p\vee q)$$$$\;\;$$ [6,8 MP] (10) $$(p\to q)\to \neg p\vee q$$$$\;\;$$ [9,4 MP] (11) $$(p \to q) \to (\lnot q \to \lnot p)$$$$\;\;$$[Ax 3.] (12) (4) $$((p\to q) \to \lnot p \lor q) \to (\lnot (\lnot p \lor q) \to \lnot (p\to q))$$$$\;\;$$[11 SSub: $$p/p\to q,q/\lnot p \lor q$$] (13) $$\lnot(\lnot p \lor q)\to\lnot(p\to q)$$$$\;\;$$ [12,10 MP] (14) (211) $$\lnot p \land \lnot q\to \lnot(p\lor q)$$$$\;\;$$[De Morgan’s law] (15) $$\lnot\lnot p \land \lnot q\to \lnot(\lnot p\lor q)$$$$\;\;$$[14 Sub: $$p/\lnot p$$] (16) $$p\to \lnot\lnot p$$$$\;\;$$[Ax 5.] (17) (4) $$(p\to q)\to (r\to (p\to q))$$$$\;\;$$ [1 SSub: $$p/p\to q, q/r$$] (18) (0) $$(p\to(q\to r))\to(q\to(p\to r))$$$$\hspace{0,2 cm}$$[the law of commutation, included in line 3] (19) (6) $$(r\to(p\to q))\to(p\to(r\to q))$$$$\;\;$$[18 SSub: $$p/r,q/p,r/q$$] (20) (8) $$(p\to q)\to (p\to (r\to q))$$$$\;\;$$[17,19 HS] (21) (0) $$(p\to(q\to r))\to(p\land q\to r)$$$$\;\;$$ [the law of importation, included in line 3] (22) (4) $$(p\to(r\to q))\to(p\land r\to q)$$$$\;\;$$ [21 SSub: $$q/r,r/q$$] (23) (8) $$(p\to q)\to (p\land r\to q)$$$$\;\;$$[20,22 HS] (24) $$(p\to q) \to ((p \to r) \to (p \to q \wedge r))$$$$\;\;$$[Ax 8.] (25) $$(p\land r\to q) \to ((p\land r \to r) \to (p\land r \to q \wedge r))$$$$\;\;$$[24 Sub: $$p/p\land r$$] (26) (6) $$(p\land r \to r) \to ((p\land r\to q) \to (p\land r \to q \wedge r))$$$$\;\;$$[25 Com] (27) $$p \land q \to q$$$$\;\;$$[Ax 7.] (28) $$p \land r \to r$$$$\;\;$$[27 Sub: $$q/r$$] (29) $$(p\land r\to q) \to (p\land r \to q \wedge r)$$$$\;\;$$[26,28 MP] (30) (8) $$(p\to q)\to(p\land r\to q\land r)$$$$\;\;$$[23,29 HS] (31) (4) $$(p\to \lnot\lnot p)\to(p\land \lnot q \to \lnot\lnot p\land \lnot q)$$$$\;\;$$[30 SSub: $$q/\lnot\lnot p,r/\lnot q$$] (32) $$p\land \lnot q \to \lnot\lnot p \land \lnot q$$ [31,16 MP] (33) (8) $$p\land \lnot q \to \lnot(\lnot p\lor q)$$$$\;\;$$[32,15 HS] (34) (8) $$p\land\lnot q\to\lnot(p\to q)$$$$\;\;$$[33,13 HS] We have $$l_{p\land\lnot q\to\lnot(p\to q)}=651$$ and $$w_{p\land\lnot q\to\lnot(p\to q)}=51$$. We proceed with the following derivation: (1) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (2) $$C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots)$$$$\hspace{0,2 cm}$$ [proved by assumption] (3) (6) 51$$(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (A \lor(\bigvee S_{2}))))\ldots)) ~\land$$$$~~~~(C_1\lor(C_2\lor\ldots\lor(C_{n-1}\lor (C_n\lor (\lnot B \lor(\bigvee S_{2}))))\ldots))$$$$\;\;$$ [1,2 Conj] (4) (651) $$p\land\lnot q\to\lnot(p\to q)$$$$\;\;$$[proved above] (5) (4) $$A\land\lnot B\to\lnot(A\to B)$$$$\;\;$$[4 SSub: $$p/A, q/B$$] (6) (100) $$(p\to q)\to(p\lor r\to q\lor r)$$$$\;\;$$[disjunct added to the right] (7) (6) $$(A\land\lnot B\to \lnot(A\to B))\to((A\land\lnot B)\lor (\bigvee S_2) \to \lnot(A\to B)\lor (\bigvee S_2))$$ $$\quad$$ [6 SSub: $$p/A\land\lnot B,q/\lnot(A\to B),r/\bigvee S_2$$] (8) $$(A\land\lnot B)\lor (\bigvee S_2) \to \lnot(A\to B)\lor (\bigvee S_2)$$$$\;\;$$[7,5 MP] (9) (427) $$(p\lor r)\land(q\lor r)\to (p\land q)\lor r$$$$\;\;$$ [distributivity 1] (10) $$(A\lor(\bigvee S_2))\land(\lnot B\lor(\bigvee S_2)) \to (A\land\lnot B)\lor (\bigvee S_2)$$$$\;\;$$[9 SSub: $$p/A$$, $$q/\lnot B$$, $$r/\bigvee S_2$$] (11) (8) $$(A\lor(\bigvee S_2))\land(\lnot B\lor(\bigvee S_2)) \to \lnot(A\to B)\lor (\bigvee S_2)$$$$\;\;$$[10,8 HS] (12) (398) $$(p\lor q)\land (p\lor r)\to p\lor (q\land r)$$$$\;\;$$ [distributivity 2] $$\vdots$$ After this step we apply, again, the same mechanism of adding formulas $$C_n, \ldots$$, $$C_1$$ which was presented in the analysis of rule $$\mathbf{R}_{\land}$$. We calculate the length and the width as in the former cases and arrive at   \begin{equation}\label{nie implikacja} l_{\lnot\to} = l_{C1}+l_{C2}+21\cdot n+c \end{equation} (22) for $$c=1671$$, and   \begin{equation}\label{w nie implikacja} w_{\lnot\to} \leq \max\{w_{C1}, w_{C2}, c, 20\cdot|\phi|\}, \end{equation} (23) where $$\phi = \ \vdash S_1 \ ' \lnot(A\to B) \ ' S_2$$, $$c=51$$. Proof of Theorem 6 The schemas of proofs presented in subsections 4.2.1–4.2.4 supply us with the proof of Theorem 6. We have shown that if   $$ \frac {?(\ \Phi \ ; \ \vdash S \ ; \ \Psi \ )} {?(\ \Phi \ ; \ \vdash T_{1} \ ; \ \vdash T_{2} \ ; \ \Psi \ )} $$ is a schema of an erotetic rule of calculus $$\mathbb{E}^\star$$ and there exist proofs of formulas $$\bigvee T_{1}$$ and $$\bigvee T_{2}$$ in system $$\mathbb{A}$$, then the proofs may be used to obtain a proof of formula $$\bigvee S$$ in system $$\mathbb{A}$$. We have also shown how to transform the proofs of $$\bigvee T_{1}$$ and $$\bigvee T_{2}$$ into a proof of $$\bigvee S$$. We have also seen that if $$l_{C1}, l_{C2}, l_P$$ and $$w_{C1}, w_{C2}, w_P$$ stand for the lengths and widths of the proofs of the formulas corresponding to, respectively, the ‘sequents-conclusions’ and the ‘sequent-premise’, then by (18), (20), (22): Corollary 6   $$l_p \leq l_{C1}+l_{C2}+f(n),$$ where $$n$$ stands for the number of terms of sequence $$T_i$$ ($$i=1$$ or $$i=2$$) and $$f$$ is a linear function of the form $$f(n)=21\cdot n + c$$, for some constant $$c\in\text{N}_+$$. Observe that $$T_1$$ and $$T_2$$ have the same number of terms, thus it is inessential whether $$i=1$$ or $$i=2$$. In turn, by (19), (21), (23): Corollary 7   $$w_p \leq \max\{w_{C1}, w_{C2}, 51,20\cdot|\vdash T_i|\}$$ 5 The algorithm and complexity issues Let $$\mathbf{s} = \langle Q_1,\ldots, Q_m \rangle$$ stand for a Socratic proof of sequent $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$. Then we use $$\mathbf{s}$$ according to the steps described below and ‘produce’ a proof of formula $$A$$ in system $$\mathbb{A}$$. (1) For each base sequent $$\phi$$ in the last question $$Q_m$$ of $$\mathbf{s}$$: a proof of the formula corresponding to $$\phi$$ is generated according to the schema presented in the proof of Theorem 4 (see subsection 4.1). (2) For $$i = m-1, \ldots, 1$$, (a) If the rule applied to question $$Q_i$$ is one of $$\mathbf{R}_{\lnot\lnot}$$, $$\mathbf{R}_{\to}$$, $$\mathbf{R}_{\lnot\land}$$, $$\mathbf{R}_{\lor}$$, then we use the proof of formula corresponding to the sequent-conclusion and transform it into a proof of formula corresponding to the sequent-premise according to the relevant schema presented in one of subsections 4.2.1–4.2.4. (b) If the rule applied to question $$Q_i$$ is one of $$\mathbf{R}_{\land}$$, $$\mathbf{R}_{\lnot\lor}$$, $$\mathbf{R}_{\lnot\to}$$, then we use the two proofs of formulas corresponding to the sequents-conclusions and transform them into a proof of formula corresponding to the sequent-premise according to the relevant schema presented in one of subsections 4.2.1–4.2.4. After the last iteration of for-loop (b) we arrive at a proof of the formula corresponding to sequent $$\vdash A$$ (i.e., formula $$A$$). We can see clearly that step 1. involves generating $$k$$ proofs, where $$k$$ is the number of different base sequents of question $$Q_m$$. (Observe that $$k$$ may be less than the number of constituents of the last question, since there may be repetitions among the constituents.) Step 2 is repeated $$m-1$$ times, where $$m$$ is the number of questions in $$\mathbf{s}$$. Using the algorithm sketched above, to each Socratic proof $$\mathbf{s}$$ in $$\mathbb{E}^{\star\star}$$ of sequent $$\vdash A$$ we can assign a proof $$\mathbf{p}$$ of formula $$A$$ in $$\mathbb{A}$$. This assignment can be formally expressed as a function $$\mathcal{F}$$ with the set of Socratic proofs as the domain and the set of axiomatic proofs as the codomain. Therefore by $$\mathcal{F}(\mathbf{s})$$, where $$\mathbf{s}$$ is a Socratic proof of $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$, we shall mean the proof of $$A$$ in $$\mathbb{A}$$ assigned to $$\mathbf{s}$$ by the above algorithm. Now we shall prove what follows. Lemma 1 Suppose that $$\mathbf{s}$$ is a Socratic proof of sequent $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$ and let $$\mathbf{p}=\mathcal{F}(\mathbf{s})$$. Then: (1) $$l_\mathbf{p}=\mathcal{O}(|A|)$$, where $$l_\mathbf{p}$$ is the length of $$\mathbf{p}$$ (2) $$w_\mathbf{p}=\mathcal{O}(|A|)$$, where $$w_\mathbf{p}$$ is the width of $$\mathbf{p}$$ (3) $$s_\mathbf{p}=\mathcal{O}(|A|^2)$$, where $$s_\mathbf{p}$$ is the size of $$\mathbf{p}$$ In the proof of Lemma 1 we will use the notion of level of Socratic proof, which is understood as follows. Let $$\mathbf{s}$$ be a Socratic transformation (recall that $$\mathbf{s}$$ is a finite sequence of questions which does not have to end with a success, as a Socratic proof does, see Definitions 2, 3). Further, let $$\underline{\phi}$$ stand for an occurrence of sequent $$\phi$$ in $$\mathbf{s}$$. Suppose also that $$\underline{\phi}$$ occurs in $$k$$-th question of $$\mathbf{s}$$. Then we will say that $$\underline{\phi}$$is at$$k$$-th level of$$\mathbf{s}$$. If it does not lead to any confusion, we will neglect the distinction between a sequent and its occurrence and will say that sequent $$\phi$$is at$$k$$-th level of$$\mathbf{s}$$ or that $$\phi$$occurs at$$k$$-th level of$$\mathbf{s}$$. In the proof of Lemma 1 we will also make use of the following fact: Fact 1 Let $$\mathbf{s}$$ be a Socratic proof of sequent $$\vdash A$$ and suppose that sequent $$\phi$$ occurs in $$\mathbf{s}$$ and that it has $$n$$ terms. Then   $$n \leq |\phi| \leq |A|.$$ The first inequality is obvious and the second may be proved by inspection of the rules of $$\mathbb{E}^{\star\star}$$ (and by induction). The point is that if $$\psi$$ is a sequent-premise of a rule and $$\chi$$ is its sequent-conclusion (or one of the two sequents-conclusions), then $$|\chi| \leq |\psi|$$. Proof of Lemma 1: Suppose, as in the statement of Lemma 1, that $$\mathbf{s}$$ is a Socratic proof of sequent $$\vdash A$$ in $$\mathbb{E}^{\star\star}$$ and that $$\mathbf{p}=\mathcal{F}(\mathbf{s})$$. Moreover, let $$\phi$$ be a sequent occurring in $$\mathbf{s}$$. When the proof $$\mathbf{p}$$ is generated, the proof of the formula corresponding to sequent $$\phi$$ must be generated as well. Let $$\mathbf{p}_\phi$$ stand for this proof, $$l_\phi$$ and $$w_\phi$$ will symbolize its length and width, respectively. We will show that   \begin{equation}\label{clause 1 length} l_{\phi} = \mathcal{O}(|\phi|). \end{equation} (24) Once this fact is proved, it will follow that for $$\phi \ = \ \vdash A$$ we have $$l_\mathbf{p}=\mathcal{O}(|A|)$$, as required by clause 1. of Lemma 1. We prove (24) by (reversed) induction with respect to the level at which sequent $$\phi$$ occurs in $$\mathbf{s}$$. Thus suppose that $$\mathbf{s}=\langle Q_1,\ldots, Q_m\rangle$$, i.e. $$\mathbf{s}$$ has $$m$$ terms (questions), and that $$\phi$$ occurs at level $$m$$ in $$\mathbf{s}$$. Then $$\phi$$ is a base sequent and by Corollary 1, $$l_{\phi} = \mathcal{O}(|\phi|)$$. Suppose that $$\phi$$ occurs at level $$k$$, where $$1\leq k < m$$. Then we consider two cases. First, it may happen that there is no application of an erotetic rule to questions $$Q_k,\ldots, Q_{m-1}$$ that is performed with respect to an occurrence of $$\phi$$ (here we mean arbitrary occurrence of $$\phi$$). In other words, $$\phi$$ is then a base sequent which is rewritten from question to question as a part of the context, until the last question is arrived at. Then we repeat the above argument concerning base sequents in the last question. In the second case a rule must be applied with respect to sequent $$\phi$$. To be precise, it may happen that the rule applied to $$k$$-th question of $$\mathbf{s}$$ acts on some other sequent, and $$\phi$$ is a part of the context which is rewritten in the next question. However, a rule must be finally applied with respect to an occurrence of $$\phi$$ in one of the consecutive questions. Suppose that the rule is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi \ ; \ \Psi)}. $$ Sequent $$\psi$$ occurs in $$\mathbf{s}$$ at a level not less than $$k+1$$. Therefore (induction hypothesis):   $$l_\psi = \mathcal{O}(|\psi|),$$ where $$l_\psi$$ is the length of the proof of the formula corresponding to $$\psi$$ (again, the proof must be generated when $$\mathcal{F}(\mathbf{s})$$ is produced). In other words, the induction hypothesis yields that $$l_\psi \leq g(|\psi|)$$ for a certain linear function $$g$$. On the other hand, by Corollary 4, $$l_\phi \leq l_\psi + f(n)$$, where $$n$$ is the number of terms of $$\psi$$ and $$f$$ is a linear function. Finally, we know that $$|\psi| \leq |\phi|$$. To sum up:   $$ l_\phi \leq l_\psi + f(n) \leq g(|\psi|) + f(n) \leq g(|\psi|) + f(|\psi|) \leq g(|\phi|) + f(|\phi|). $$ The two inequalities from the right hold since $$n\leq|\psi|$$ (see Fact 1) and both $$f$$ and $$g$$ are growing functions. Thus we finally arrive at the conclusion that:   $$ l_\phi \leq h(|\phi|), $$ where $$h=g+f$$ is a linear function.12 If the rule applied with respect to $$\phi$$ is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi_1 \ ; \ \psi_2 \ ; \ \Psi)}, $$ then by induction hypothesis we have:   $$l_{\psi_1} \leq g_1(|\psi_1|), l_{\psi_2} \leq g_2(|\psi_2|),$$ where $$l_{\psi_1}$$ and $$l_{\psi_2}$$ are the lengths of the proofs of formulas corresponding to $$\psi_1$$ and $$\psi_2$$, respectively, and $$g_1$$, $$g_2$$ are certain linear and growing functions. By Corollary 6, $$l_\phi \leq l_{\psi_1}+l_{\psi_2}+f(n)$$ where $$n$$ is the number of terms of $$\psi_1$$ (or $$\psi_2$$) and $$f$$ is a linear (and growing) function, therefore:   $$l_\phi \leq l_{\psi_1}+l_{\psi_2}+f(n) \leq g_1(|\psi_1|) + g_2(|\psi_2|) + f(|\psi_1|).$$ Since also $$|\psi_1| \leq |\phi|$$ and $$|\psi_2| \leq |\phi|$$, defining $$h$$ as $$(g_1 + g_2)+f$$ we finally obtain:   $$l_\phi \leq h(|\phi|),$$ where $$h$$ is a linear function. All this entails that (24) is true, and therefore $$l_\mathbf{p}=\mathcal{O}(|A|)$$. The second clause of Lemma 1, concerning width, will be proved in an analogous way. We show that   \begin{equation}\label{clause 2 width} w_{\phi} = \mathcal{O}(|\phi|), \end{equation} (25) holds for each $$\phi$$ occurring in the Socratic proof $$\mathbf{s}$$, and we do it by (reversed) induction with respect to the level at which sequent $$\phi$$ occurs in $$\mathbf{s}$$. If $$\phi$$ is a base sequent at level $$m$$ (where $$\mathbf{s}$$ has $$m$$ terms), then, by Corollary 2, $$w_{\phi} = \mathcal{O}(|\phi|)$$. In the inductive part we analyse, again, two cases: when no rule is applied with respect to $$\phi$$, and when there is one. Here we consider only the second case. Suppose that the rule is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi \ ; \ \Psi)}. $$ Sequent $$\psi$$ occurs in $$\mathbf{s}$$ at a level not less than $$k+1$$, thus by induction hypothesis:   $$w_\psi \leq g(|\psi|),$$ where $$w_\psi$$ is the width of the proof of the formula corresponding to $$\psi$$, and $$g$$ is a linear function. By Corollary 5, $$w_\phi \leq \max\{w_\psi,c,8\cdot|\psi|\}$$, where $$c\leq 51$$. This obviously yields that $$w_\phi \leq h(|\psi|)$$ for some linear (growing) function $$h$$.13 Finally, since $$|\psi| \leq |\phi|$$ and $$h$$ is a growing function:   $$w_\phi \leq h(|\phi|).$$ If the rule applied with respect to $$\phi$$ is of the form:   $$ \frac {?(\Phi \ ; \ \phi \ ; \ \Psi)} {?(\Phi \ ; \ \psi_1 \ ; \ \psi_2 \ ; \ \Psi)} $$ then by induction hypothesis we have:   $$w_{\psi_1} \leq g_1(|\psi_1|), w_{\psi_2} \leq g_2(|\psi_2|),$$ where $$w_{\psi_1}$$ and $$w_{\psi_2}$$ are the widths of the proofs of the formulas corresponding to $$\psi_1$$ and $$\psi_2$$, respectively, and $$g_1$$, $$g_2$$ are certain linear and growing functions. By Corollary 6 and by the fact that $$|\psi_i| \leq |\phi|$$ for $$i=1,2$$, $$w_\phi \leq \max\{w_{\psi_1}, w_{\psi_2}, 51, 20\cdot|\phi|\}$$. Again, this yields that:   $$w_\phi \leq h(|\phi|),$$ where $$h$$ is a linear function. All this entails that (24) is true, and therefore $$w_\mathbf{p}=\mathcal{O}(|A|)$$. Clause 3 follows from the previous two, as $$s_\mathbf{p}=l_\mathbf{p}\cdot w_\mathbf{p}$$. ■ We did not calculate the size of Socratic proof $$\mathbf{s}$$ of $$\vdash A$$ and we shall not do this. For the purpose of this article it is enough to observe that the size of $$\mathbf{s}$$, understood as the number of all occurrences of signs in $$\mathbf{s}$$, is clearly a function of $$|A|$$. As we have seen, there exists a function, $$\mathcal{F}$$, which assigns to $$\mathbf{s}$$ a proof of formula $$A$$ in $$\mathbb{A}$$ and which is such that the size of $$\mathbf{p}$$ is $$\mathcal{O}(|A|^2)$$. This means that: Theorem 7 Axiomatic system $$\mathbb{A}$$ simulates polynomially erotetic calculus $$\mathbb{E}^{\star\star}$$. 5.1 Back to the R-S system As we have pointed out in the Introduction and in Section 3, there is a close affinity between the right-sided version of the method of Socratic proofs and the R-S method. The affinity follows from the fact that the two methods are based on deriving a Conjunctive Normal Form. The Reader could see this on the examples presented in Sections 2 and 3. The consequence of the affinity of systems is that the erotetic one could be easily replaced with the R-S system in deriving the results presented in this article: the fundamental sequences would replace the base sequents and the rules of the R-S system would replace the rules of $$\mathbb{E^{**}}$$. This means that the algorithm presented in Section 5 works equally well for the case of the R-S system. It follows that axiomatic system $$\mathbb{A}$$ can be shown to simulate polynomially the R-S system. 6 Conclusions and further work The aim of this work was to automatize the process of proof-search in axiomatic system $$\mathbb{A}$$ for CPL. The results presented in this article allow us to transform any algorithm of proof-search by the method of Socratic proofs (in calculus $$\mathbb{E^{**}}$$) into an algorithm of proof-search in axiomatic system $$\mathbb{A}$$. Moreover, the estimations of the lengths and widths of proofs show that axiomatic system $$\mathbb{A}$$ simulates polynomially erotetic calculus $$\mathbb{E}^{\star\star}$$. It is a common perception that proving by means of axioms and rules is a difficult process, in which intuitions and high logical skills are necessarily involved. The results presented here neither prove nor disprove this common belief. We are convinced, however, that the detailed constructions and calculations presented here may shed a new light on the process of proof-search in axiomatic systems. We especially hope that the algorithm sketched in Section 5 will be implemented. Certainly, this is one of the perspectives for the future. The first author admits that the meticulous process of preparing the proofs in the axiomatic system of Tadeusz Batóg was the most difficult but also the most satisfying part of this work. The second author must admit that it would be a shame if we automatized it all. Funding The work of the second author was supported by the Polish National Science Center, grant no [2012/04/A/HS1/00715]. This work originated as a master’s thesis [16] (in Polish) defended by the first author under the supervision of the second author in the Department of Logic and Cognitive Science, Institute of Psychology, Adam Mickiewicz University. Part of the proofs have been changed and the issues concerning complexity have been added in this version. We wish to thank the Referee of the master’s thesis, Professor Andrzej Wiśniewski, and the anonymous Referees of the article for their detailed corrections, valuable comments and suggestions, which helped us to improve this article. Footnotes 1 We will use the term ‘axiomatic system’. 2 To get an idea of the progress in this field see e.g. [12]. Since 1992, the Workshop on Theorem Proving with Analytic Tableaux and Related Methods, and since 1997, the conference TABLEAUX Automated Reasoning with Analytic Tableaux and Related Methods have gathered together researchers in this discipline and published their proceedings. 3 Here by ‘tableau methods’ we mean analytic tableaux like those presented originally by Smullyan [32]. In [9] the authors propose an improved — among others, in terms of efficiency — version of analytic tableaux named the KE system, which was also popularized in [10] and [11]. 4 The acronym ‘RPQ’ is for ‘Right’, ‘Pure’, ‘Quantifier’, where ‘Pure’ is to indicate the fact that the proofs start with pure sentences of the language of $$\mathsf{FOL}$$, that is, sentences containing no parameters. 5 A terminological remark is in order here. What we mean by simulate here is that we can show that there is a function which to every Socratic proof assigns a proof in a sequent calculus, or a function which to every Socratic proof assigns a closed analytic tableau, etc. In terms of simulations used to examine the relative complexity of proof systems this means that the sequent calculus simulates erotetic calculus, or that the tableau system simulates erotetic calculus, where the question remains how fast it does it. 6 For an example of proof-search algorithm written in pseudocode see [26, Appendix]. Admittedly, the algorithm is written for the both-sided version of erotetic calculus $$\mathbb{E}^*$$, but the variant for $$\mathbb{E}^{**}$$ is easily obtained from it. 7 In the details concerning language and axiomatic system this work is based on [1]. 8 This is based on the system by Hilbert and Bernays presented in 1934 (see [17, p. 65]). There are differences, however, between the systems in [1] and [17]. In the latter the laws $$(A \rightarrow (A \rightarrow B)) \rightarrow (A \rightarrow B)$$ and $$(A \rightarrow B) \rightarrow ((B \rightarrow C) \rightarrow (A \rightarrow C))$$ are present instead of the Frege law $$(p \to (q \to r)) \to ((p \to q) \to (p \to r))$$, so each connective is defined by a group of three axioms and the system has 15 axioms. 9 The axiomatic system in [31] is based on axiom schemas and the solely rule modus ponens. In the quotation we have used ‘$$A$$’ instead of ‘$$\beta$$’, to avoid a mix-up of notation. 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