An informational interpretation of weak relevant logic and relevant property theory

Abstract

This paper extends the theory of situated inference from Mares (Relevant logic: a philosophical interpretation. Cambridge University Press, Cambrdge, 2004) to treat two weak relevant logics, B and DJ. These logics are interesting because they can be used as bases for consistent naïve theories, such as naïve set theory. The concepts of a situation and of information that are employed by the theory of situated inference are used to justify various aspects of these logics and to give an interpretation of the notion of set that is represented in the naïve set theories that are based on them.

Appendices

Appendix I: The logics B and DJ

The logic B.

Axioms:

• 1. $A\rightarrow B$

• 2. $(A\wedge B)\rightarrow A$; $(A\wedge B)\rightarrow B$;

• 3. $A\rightarrow (A\vee B)$; $B\rightarrow (A\vee B)$;

• 4. $((A\rightarrow B)\wedge (A\rightarrow C))\rightarrow (A\rightarrow (B\wedge C))$;

• 5. $((A\rightarrow C)\wedge (B\rightarrow C))\rightarrow ((A\vee B)\rightarrow C)$;

• 6. $(A\wedge (B\vee C))\rightarrow ((A\wedge B)\vee (A\wedge C))$;

• 7. $A\leftrightarrow \lnot \lnot A$.

Rules:

• $\vdash A\rightarrow B, \ \vdash A \ \Longrightarrow \ \vdash B$;

• $\vdash A, \ \vdash B \ \Longrightarrow \ \vdash A\wedge B$;

• $\vdash A\rightarrow B, \ \vdash B\rightarrow C \ \Longrightarrow \ \vdash A\rightarrow C$;

• $\vdash A\rightarrow \lnot B \ \Longrightarrow \ \vdash B\rightarrow \lnot A$.

To obtain DJ add the conjunctive syllogism axiom, $((A\rightarrow B)\wedge (B\rightarrow C))\rightarrow (A\rightarrow C)$, and the contraposition axiom, $(A\rightarrow \lnot B)\rightarrow (B\rightarrow \lnot A)$.

Appendix II: Natural deduction systems for B and DJ

For the logic B:

• $\rightarrow$ E From $A\rightarrow B_{\alpha }$ and $A_{\beta }$ to infer $B_{\alpha \circ \beta }$.

• $\rightarrow$ I From a subproof that proves $B_{\alpha \circ a}$ from the hypothesis $A_{a}$ to discharge the hypothesis, end the subproof and infer $A\rightarrow B_{\alpha }$.

• $\wedge$ E From $A\wedge B_{\alpha }$ to infer $A_{\alpha }$ or $B_{\alpha }$.

• $\wedge$ I From $A_{\alpha }$ and $B_{\alpha }$ to infer $A\wedge B_{\alpha }$.

• $\vee$ E From $A\vee B_{\beta }$, $A\rightarrow C_{\alpha }$, and $B\rightarrow C_{\alpha }$ to infer $C_{\alpha \circ \beta }$.

• $\lnot \hbox {E}_1$ From $A\rightarrow B_o$ and $\lnot B_{\alpha }$ to infer $\lnot A_{\alpha }$.

• $\lnot \hbox {E}_2$ From $\lnot \lnot A_{\alpha }$ to infer $A_{\alpha }$.

• $\lnot$ I From $A_{\alpha }$ to infer $\lnot \lnot A_{\alpha }$.

• Dist From $A\wedge (B\vee C)_{\alpha }$ to infer $(A\wedge B)\vee (A\wedge C)_{\alpha }$.

Left Push From $A_{\alpha }$ to infer $A_{o\circ \alpha }$.

Left Pop From $A_{o\circ \alpha }$ to infer $A_{\alpha }$

• To obtain DJ add the rules: (DJ) From $A_{\alpha \circ (\alpha \circ \beta )}$ to infer $A_{\alpha \circ \beta }$; (Contrap) From $A\rightarrow B_{\alpha }$ and $\lnot B_{\beta }$ to infer $\lnot A_{\alpha \circ \beta }$.

Appendix III: Semantic proof sketch of triviality for full DJ-frames

Lemma 4

In all Routley–Meyer models, for all $X\subseteq S$ that are closed upwards under $\le$, $0\circ X = X$.

Proof

Suppose first that $x\in 0\circ X$. Then there is a $y\in 0$ and a $z\in X$ such that Ryzx. Since $y\in 0$, by the definition of $\le$, $z\le x$. Since X is an up-set, $x\in X$. Generalizing, $0\circ X\subseteq X$.

Now suppose that $x\in X$. Since $\le$ is a partial order, $x\le x$. Thus, by the definition of $\le$, there is some $y\in 0$ such that Ryxx. So $x\in 0\circ X$. Generalizing, $X\subseteq 0\circ X$ and hence $0\circ X=X$. $\square$

Lemma 5

For all X, Y up-sets of members of S, $X\subseteq Y$ iff $0\subseteq X\Rightarrow Y$.

Proof

First, suppose that $X\subseteq Y$. Now suppose that $o\in O$, Roxy, and $x\in X$. By the definition of $\le$, $x\le y$. Since X is an up-set and $x\in X$, $y\in X$. Since $X\subseteq Y$, $y\in Y$. Thus, by the definition of $\Rightarrow$, $o\in X\Rightarrow Y$. Generalising, $0\subseteq X\Rightarrow Y$.

Second, suppose that $0\subseteq X\Rightarrow Y$. Suppose that $x\in X$. By semantic postulate 1, $\le$ is reflexive, so $x\le x$. By the definition of $\le$, there is some $o\in 0$ such that Roxx. But, by the definition of $\Rightarrow$, if Roxx and $x\in X$, then $x\in Y$. Hence $X\subset Y$. $\square$

Lemma 6

$(X\circ Y)\subseteq Z$ iff $X\subseteq (Y\Rightarrow Z)$.

Proof

First assume $(X\circ Y)\subseteq Z$. Now suppose that $x\in X$, Rxyz, and $y\in Y$. By the definition of $\circ$, $z\in X\circ Y$. By the assumption, $z\in Z$. Hence $x\in Y\Rightarrow Z$.

Now assume that $X\subseteq Y\Rightarrow Z$. Also assume that $z\in X\circ Y$. By the definition of $\circ$, there are x and y such that Rxyz, $x\in X$, and $y\in Y$. By the assumption and the definition of $\Rightarrow$, $z\in Z$. $\square$

The semantic version of naïve comprehension. For each propositional function $\psi$ there is some $i\in D$ such that

$$\begin{aligned} \varphi _{\in i} (f) = \psi (f). \end{aligned}$$

As I showed in Sect. 9, from naïve comprehension the following can be derived for full frames:

$$\begin{aligned} \varphi _{\in } (f[k/1,k/2]) = ((\varphi _{\in }(f[k/1,k/2]) \circ \varphi _{\in } (f[k/1,k/2]))\Rightarrow \pi ). \end{aligned}$$

Theorem 7

In every full-frame for DJ that satisfies naïve comprehension, $0\subseteq \pi$ for every proposition $\pi$.

Proof

$\varphi _{\in } (f[k/1,k/2]) = ((\varphi _{\in }(f[k/1,k/2]) \circ \varphi _{\in } (f[k/1,k/2]))\Rightarrow \pi )$, so $\varphi _{\in } (f[k/1,k/2]) \ \subseteq \ ((\varphi _{\in }(f[k/1,k/2]) \circ \varphi _{\in } (f[k/1,k/2]))\Rightarrow \pi )$. By lemma 6, $(\varphi _{\in } (f[k/1,k/2])\circ ((\varphi _{\in }(f[k/1,k/2]) \circ (\varphi _{\in } (f[k/1,k/2])) \subseteq \pi$. By proposition 3, $X\circ Y\subseteq X\circ (X\circ Y)$, so $((\varphi _{\in }(f[k/1,k/2]) \circ (\varphi _{\in } (f[k/1,k/2])) \subseteq \pi$. Thus, by lemma 5, $0\subseteq (((\varphi _{\in }(f[k/1,k/2]) \circ (\varphi _{\in } (f[k/1,k/2])) \Rightarrow \pi$. But this implies that $0\subseteq \varphi _{\in } (f[k/1,k/2])$. Thus, $0\circ 0 \subseteq (\varphi _{\in } (f[k/1,k/2])\circ \varphi _{\in } (f[k/1,k/2]))$. By lemma 4, $0 \subseteq (\varphi _{\in } (f[k/1,k/2])\circ \varphi _{\in } (f[k/1,k/2]))$ hence $0\subseteq \pi$. $\square$