On the Computational Meaning of Axioms

Abstract

This paper investigates an anti-realist theory of meaning suitable for both logical and proper axioms. Unlike other anti-realist accounts such as Dummett–Prawitz verificationism, the standard framework of classical logic is not called into question. This account also admits semantic features beyond the inferential ones: computational aspects play an essential role in the determination of meaning. To deal with these computational aspects, a relaxation of syntax is necessary. This leads to a general kind of proof theory, where the objects of study are not typed objects like deductions, but rather untyped ones, in which formulas are replaced by geometrical configurations.

Appendices

Appendices

A Properties of System $\text{pLK}_{\text{R}}^{\mbox{ $\maltese$}}$

2.1 A.1 Proof of Proposition 5.3

By induction on the number of connectives in the sequent. Since the proofs for ∨ and for ∧ are similar, we only show the invertibility of the ∨. The base case for the induction is a sequent containing only one connective, i.e. a $\mbox{ $\maltese$}$ rule followed by a ∨-rule. In this case the $\mbox{ $\maltese$}^{At}$ rule itself gives a derivation of the premiss of the ∨-rule. Let us now assume that this is true for any sequent containing at most n connectives, and let us take a derivable sequent $\vdash \Gamma,A \vee B$, containing n + 1 connectives, and let π be one of its derivations. If π ends with a ∨-rule, the subderivation obtained dropping the last rule gives a derivation of the premiss. If π does not end with a ∨ rule, then it must end with a ∧ rule: Applying the induction hypothesis on the premisses we get two derivations π′₁ and π′₂ of $\vdash \Delta,A,B,C$ and $\vdash \Delta,A,B,D$ respectively; thus the desired derivation is:

2.2 A.2 Proof of Lemma 5.6

By induction on the number of connectives in $\vdash \Gamma$. The base case is obvious. Assume that the lemma is true when the number of connectives in $\Gamma$ is at most n. We show that the derivations of $\vdash \Gamma,F,G$ terminating with a rule where F is principal have the same set of leaves as the derivations terminating with a rule where G is principal. We only show how the proof is done in the case F = A ∧ B and G = C ∧ D, which is the more complicated case. By the invertibility of the ∧-rule we have the two derivations:

Using the induction hypothesis, we have that $\mathfrak{L}(\pi _{k}) = \mathfrak{L}(\rho _{k})$ for all k in {1, 2, 3, 4}. Using these equalities and the induction hypothesis once again, we obtain that:

• any derivation π of $\vdash \Gamma,A,C \wedge D$ satisfies $\mathfrak{L}(\pi ) = \mathfrak{L}(\pi _{1}) + \mathfrak{L}(\pi _{2})$;

• any derivation π of $\vdash \Gamma,B,C \wedge D$ satisfies $\mathfrak{L}(\pi ) = \mathfrak{L}(\pi _{3}) + \mathfrak{L}(\pi _{4})$;

• any derivation π of $\vdash \Gamma,A \wedge B,C$ satisfies $\mathfrak{L}(\pi ) = \mathfrak{L}(\pi _{1}) + \mathfrak{L}(\pi _{3})$;

• any derivation π of $\vdash \Gamma,A \wedge B,D$ satisfies $\mathfrak{L}(\pi ) = \mathfrak{L}(\pi _{2}) + \mathfrak{L}(\pi _{4})$;

We have therefore shown that if π is any derivation of $\vdash \Gamma,A \wedge B,C \wedge D$ terminating with a ∧-rule on A ∧ B and, if ρ is any derivation of the same sequent terminating with a ∧-rule on C ∧ D, they have the same set of leaves, namely:

$$\displaystyle{\mathfrak{L}(\pi ) =\sum _{i=1,\ldots,4}\mathfrak{L}(\pi _{i}) = \mathfrak{L}(\rho )}$$

The other cases are similar.

2.3 A.3 Proof of Proposition 5.10

Suppose π is a derivation of $\vdash \Gamma$ in pLK _R , then by replacing every axiom rule by a $\mbox{ $\maltese$}$ rule we obtain a derivation π′ of $\vdash \Gamma$ in pLK$_{R}^{\mbox{ $\maltese$}}$. Every $\mbox{ $\maltese$}^{At}$ rule in π′ is correct, since the sequent was introduced by an axiom rule in pLK _R .

Conversely, if π′ is a derivation of $\vdash \Gamma$ in pLK$_{R}^{\mbox{ $\maltese$}}$ such that $\mathfrak{L}(\pi ')$ contains only correct sequent, then each sequent in $\mathfrak{L}(\pi ')$ can be derived from an axiom rule in pLK _R . Therefore we obtain a derivation π of $\vdash \Gamma$ in pLK _R by replacing every $\mbox{ $\maltese$}^{At}$ rule in π′ by an axiom rule.

2.4 A.4 Proof of Lemma 5.12

Notice that, by definition of satisfiability of a sequent (and associativity of ∨), $\delta \vDash \Delta,A,B$ if and only if $\delta \vDash \Delta,A \vee B$.

In the case of the ∧-rule, assume first that $\delta \vDash \Delta,A$ and $\delta \vDash \Delta,B$. Either there is satisfiable formula in $\Delta$, either both A and B are satisfiable, therefore $\delta \vDash \Delta,A \wedge B$. Conversely, assume that $\delta \nvDash \Delta,A$, then all formulas in $\Delta$ are unsatisfiable and A is not satisfiable, therefore A ∧ B is not satisfiable. We conclude that $\delta \nvDash \Delta,A \wedge B$. The lemma is then proved by simple induction.

2.5 A.5 Proof of Proposition 5.13

Suppose $\mathfrak{L}(\vdash \Gamma )$ contains only correct sequents, then for any valuation δ and sequents $\vdash \Delta$ in $\mathfrak{L}(\vdash \Gamma )$, $\delta \vDash \Delta$ from the definition of correct sequent. Then, by Lemma 5.12, $\delta \vDash \Gamma$.

Conversely, let us assume that $\mathfrak{L}(\vdash \Gamma )$ contains at least an incorrect sequent $\vdash P_{1},\ldots,P_{n},\neg Q_{1},\ldots \neg Q_{p}$, such that for all integer i ∈ [1, n] and j ∈ [1, p], P _i  ≢ Q _j . We can now take a valuation δ satisfying δ(P _i ) = 0 and δ(Q _j ) = 1. Then $\delta \nvDash P_{1},\ldots,P_{n},\neg Q_{1},\ldots \neg Q_{p}$ and by Lemma 5.12 this means that $\delta \nvDash \Gamma$.

B Properties of System LK$_{R}^{\mbox{ $\maltese$}}$

3.1 B.1 Proof of Lemma 5.19

Let π be such a derivation of the sequent $\vdash \Gamma$. Let $\mathfrak{L}(\pi )$ be the set of sequents introduced by $\mbox{ $\maltese$}$ rules in $\vdash \Gamma$, $\mathfrak{L}_{c}(\pi )$ the subset of $\mathfrak{L}(\pi )$ containing the sequents introduced by correct $\mbox{ $\maltese$}$ rules, and $\mathfrak{L}_{a}(\pi ) = \mathfrak{L}(\pi ) - \mathfrak{L}_{c}(\pi )$. By assumption, the sequents in $\mathfrak{L}_{a}(\pi )$ are introduced by admissible $\mbox{ $\maltese$}$ rules that are not correct. Hence there exists correct derivations π _i of $\vdash \Gamma _{i}$. Then, replacing the $\mbox{ $\maltese$}$ rules introducing the sequents $\vdash \Gamma _{i}$ in π by the derivations π _i , we obtain a derivation π′ of $\vdash \Gamma$ extending π and containing only correct $\mbox{ $\maltese$}$ rules.

3.2 B.2 Proof of Theorem 5.21

Suppose we have a derivation of π in LK _R  of a sequent $\vdash \Gamma$. Then, replacing every axiom rule by a $\mbox{ $\maltese$}$ yields a derivation π′ of $\vdash \Gamma$ in LK$_{R}^{\mbox{ $\maltese$}}$ . Moreover, the $\mbox{ $\maltese$}$ rules are all correct (since they were axiom rules in LK _R  ), hence admissible.

Conversely, suppose we have a derivation π′ of a sequent $\vdash \Gamma$ in LK$_{R}^{\mbox{ $\maltese$}}$ that contains only admissible rules. Then, by Lemma 5.19 we can find a derivation π″ extending π′ such that π″ contains only correct $\mbox{ $\maltese$}$ rules. Then, we can replace these $\mbox{ $\maltese$}$ rules by axiom rules to get a derivation π of $\vdash \Gamma$ in LK _R  .

3.3 B.3 Proof of Lemma 5.23

Suppose π contains at least one $\mbox{ $\maltese$}$ rule introducing a sequent $\vdash \Gamma$ containing a formula B that is not an atom, a negation of an atom or an existential formula. Then the principal connective in B is either a ∧, a ∨ or a $\forall$. Replacing the $\mbox{ $\maltese$}$ rule introducing $\vdash \Gamma$ by the rule introducing the principal connective and closing the derivation we obtain by $\mbox{ $\maltese$}$ rules then gives us a new derivation π ₁ of π. After a finite number of iterations of this process, we obtain the wanted extension.

3.4 B.4 Proof of Lemma 5.25

Suppose now that the sequent $\vdash \Gamma$, $\Gamma = A_{1},\ldots,A_{m}$, introduced by the non-admissible $\mbox{ $\maltese$}$ rule contains at least one quantifier. We fix an enumeration of the terms $t_{1},\ldots,t_{n},\ldots$ of the language and we will define an iterative process indexed by pairs (s, k _s ) where s is a finite sequence of integers (the first step will be indexed by the null sequence of length m which will be written (0) _m ) of length p _s and k _s is an integer in $[1,\ldots,p_{s}]$. The process we describe consists in extending the derivation by applying $\exists$ rules in a way that insures us that for all existential formula $\exists xA(x)$ and term t _i , there exists a step where the $\exists$ rule is used on the formula A[t _i ∕x]. To insure all terms appear at some point in the process we will use the enumeration but we need to keep track of the last term used for each existential formula. Moreover, applying a $\exists$ rule on a formula containing two existential connectives will produce new existential formulas on which we must apply the same procedure. The sequence will therefore keep track, for each existential formula, of the last term we used. Its length may vary, but due to our choice of existential rule it can only expand. The integer, on the other hand, will keep track of the last existential formula we decomposed, so that we can ensure that all formulas are taken into account.

First, let us write $A_{1},\ldots,A_{p_{(0)_{ m}}}$ the formulas in $\Gamma$ that contain quantifiers. By Lemma 5.23 we can suppose, without loss of generality, that the $\mbox{ $\maltese$}$ rules in π are simple. We will denote π by $\pi _{(0)_{m}}$, i.e. π will be the initial step of the process. The integer $k_{(0)_{m}}$ is defined to be 1, so we consider the derivation $\pi _{A_{1},t_{1}}$ obtained from π by replacing the $\mbox{ $\maltese$}$ rule introducing $\Gamma = \Delta,A_{1},\ldots,A_{p}$ by the derivation consisting of a $\mbox{ $\maltese$}$ rule introducing $\Delta,A'_{1}[t_{1}/x],A_{2},\ldots,A_{p_{(0)_{ m}}}$ followed by an existential rule introducing A ₁. It follows from Lemma 5.23 that this derivation can be extended to a simple derivation $\bar{\pi }_{A_{1},t_{1}}$. Then, by the non-admissibility of the $\mbox{ $\maltese$}$ rule, this derivation contains at least one non-admissible $\mbox{ $\maltese$}$ rule introducing a sequent $\vdash \Gamma '$. Amongst the formulas of $\Gamma$ are the all the formulas A _i for $1\leqslant i\leqslant p$, but $\Gamma '$ may contain more existential formulas. We thus denote by $A_{1},\ldots,A_{p}$ the existential formulas of $\Gamma '$. We write $(0)_{m}^{+} = (1,0,\ldots,0)$ the sequence of length p: we thus obtained an extension $\pi _{(0)_{m}^{+}} =\bar{\pi } _{A_{1},t_{1}}$ containing a non-admissible $\mbox{ $\maltese$}$ rule introducing a sequent $\Gamma _{(0)_{m}^{+}} =\bar{ \Gamma }$. Defining $k_{(0)_{m}^{+}} = 2$, we arrived at the next step, indexed by $((0)_{m}^{+},k_{(0)_{m}^{+}})$ and we can then iterate the process.

More generally, suppose we are at step (s, k _s ) with $s = (s(0),\ldots,s(p))$: we have a simple derivation π _s with a non-admissible $\mbox{ $\maltese$}$ rule introducing a sequent $\Gamma _{s} = \Delta _{s},A_{1},\ldots,A_{p_{s}}$ ($\Delta _{s}$ contains only atoms and negations of atoms). We obtain a derivation $\pi _{A_{k_{s}},t_{s(k_{ s})+1}}$ by replacing the $\mbox{ $\maltese$}$ rule introducing $\Gamma _{s}$ with the derivation:

This derivation $\pi _{A_{k_{s}},t_{s(k_{ s})+1}}$ can then be extended by Lemma 5.23 to a simple derivation $\bar{\pi }_{A_{k_{s}},t_{s(k_{ s})+1}}$ which contains a non-admissible $\mbox{ $\maltese$}$ rule. The sequent $\Gamma '$ introduced by this rule contains all the formulas $A_{1},\ldots,A_{p_{s}}$ and may contain additional existential formulas $A_{p_{s}+1},\ldots,A_{n}$. Let s ⁺ to be the sequence of length n defined by $(s(0),\ldots,s(k_{s} - 1),s(k_{s}) + 1,s(k_{s} + 1),\ldots,s(p_{s}),0,\ldots,0)$, and:

$$\displaystyle{ k_{s^{+}} = \left \{\begin{array}{ll} k_{s} + 1&\mbox{ if $k_{s} + 1\leqslant n$}\\ 1 &\text{ otherwise} \end{array} \right. }$$

Let us write $n = p_{s^{+}}$. We thus obtained the next step in the process, indexed by $(s^{+},k_{s}^{+})$: a simple derivation $\pi _{s^{+}} =\bar{\pi } _{A_{k_{s}},t_{s(k_{ s})+1}}$ with a non-admissible $\mbox{ $\maltese$}$ rule introducing a sequent $\Gamma _{s^{+}} = \Gamma ' = \Delta _{s^{+}},A_{1},\ldots,A_{p_{ s^{+}}}$.

We claim that for all pairs (i, j) of natural numbers (different from 0), there is a step s in the process such that s(i) = j. We will write len(s) the length of a sequence s. Notice the formulas $A_{p_{s}+1},\ldots,A_{p_{ s^{+}}}$ are instantiations of an existential formulas A _i for $1\leqslant i\leqslant p_{s}$ and therefore the number of existential connectives in a B _j is strictly less than the number of existential connectives in the corresponding A _i . We will show that the value of k _s returns to 1 in a finite number of steps using this remark. We will denote the number of existential connectives in a formula A by ♮(A). Suppose that we are at a given step s such that k _s  = 1 and write $o_{s} = (\max _{k_{s}<i\leqslant p_{s}}\natural (A_{i}),p_{s} - k_{s})$. This pair somehow measures the number of steps one has to make before k _s returns to 1. Then, after p _s − k _s steps in the process – let us write the resulting step as s ¹, we have $k_{s^{1}} = p_{s}$ and $p_{s^{1}} - p_{s}$ new formulas, each one such that $\natural (A) <\max _{k_{s}<i\leqslant p_{s}}\natural (A_{i})$. Therefore, the pair $o_{s^{1}} = (\max _{k_{ s^{1}}<i\leqslant p_{s^{1}}}\natural (A_{i}),p_{s^{1}} - p_{s})$. Since $\max _{k_{ s^{1}}<i\leqslant p_{s^{1}}}\natural (A_{i}) <\max _{k_{s}<i\leqslant p_{s}}\natural (A_{i})$, we have $o_{s^{1}} < o_{s}$ in the lexicographical order and this is enough to show the claim.

3.5 B.5 Proof of Theorem 5.26

If the sequent $\vdash \Gamma$ introduced by the non-admissible $\mbox{ $\maltese$}$ rule does not contain any quantifiers, then the proof reduces to the proof of Proposition 5.13. Indeed, the derivation of $\vdash \Gamma$ in $pLK_{R}^{\mbox{ $\maltese$}}$ is incorrect (if it were correct, it would contradict the assumption since any correct derivation in $pLK_{R}^{\mbox{ $\maltese$}}$ is a correct derivation in LK$_{R}^{\mbox{ $\maltese$}}$ ), hence we can find a model $\mathcal{M}$ such that $\mathcal{M}\not\vDash \Gamma$.

If $\vdash \Gamma$ contains existential formulas, we use Lemma 5.25 to obtain a sequence (π _i ) _i ∈ N of extensions. From this sequence of extensions, one can obtain a sequence of sequents $\vdash \Gamma _{i}$ where for each i, there exists N such that $\vdash \Gamma _{i+1}$ is the premise of a rule whose conclusion is $\vdash \Gamma _{i}$ in all derivations π _j with $j\geqslant N$. Moreover, this sequence can be chosen so as to contain all instances of the subformulas of $\Gamma$. We now define a model whose base set is the set of terms. The interpretations of function symbols and constants are straightforward. The only thing left to define is the interpretation of predicates: if P is a n-ary predicate symbol, then $(t_{1},\ldots,t_{n})$ is in the interpretation of P if and only if $\forall i\geqslant 0$, $Pt_{1}\ldots t_{n}\notin \Gamma _{i}$.

We can now check that $\mathcal{M}\not\vDash \Gamma$. We chose A a formula in $\Gamma$ and prove by induction on the size of the formula A that $\mathcal{M}\not\vDash A$:

• if A is an atomic formula, then $\mathcal{M}\not\vDash A$ by definition of the model;

• if A = B ∧ C, then there exists a sequent $\vdash \Gamma _{i}$ such that either $B \in \Gamma _{i}$ or $C \in \Gamma _{i}$. We suppose $B \in \Gamma _{i}$ without loss of generality. Then, by the induction hypothesis, we have $\mathcal{M}\not\vDash B$, hence $\mathcal{M}\not\vDash A$;

• if A = B ∨ C, then there exists a sequent $\vdash \Gamma _{i}$ containing both B and C. By induction, these two formulas are not satisfied in the model $\mathcal{M}$, hence $\mathcal{M}\not\vDash A$;

• if $A = \forall xB(x)$, then there is a sequent $\vdash \Gamma _{i}$ containing B[y∕x]. By induction, $\mathcal{M}\nvdash B[y/x]$, hence $\mathcal{M}\not\vDash A$;

• if $A = \exists xB(x)$, then for every term t there exists a sequent $\vdash \Gamma _{i}$ such that $B[t/x] \in \Gamma _{i}$. By the induction hypothesis, $\mathcal{M}\not\vDash B[t/x]$. This being true for all term t, we conclude that $\mathcal{M}\not\vDash A$.

This concludes the proof: since all formula $A \in \Gamma$ is such that $\mathcal{M}\not\vDash A$, we have that $\mathcal{M}\not\vDash \Gamma$.