An Expressivist Bilateral Meaning-is-Use Analysis of Classical Propositional Logic

Abstract

The connectives of classical propositional logic are given an analysis in terms of necessary and sufficient conditions of acceptance and rejection, i.e. the connectives are analyzed within an expressivist bilateral meaning-is-use framework. It is explained how such a framework differs from standard (bilateral) inferentialist frameworks and it is argued that it is better suited to address the particular issues raised by the expressivist thesis that the meaning of a sentence is determined by the mental state that it is conventionally used to express. Furthermore, it is shown that the classical requirements governing the connectives completely characterize classical logic, are conservative (indeed make the connectives redundant) and separable, are in bilateral harmony, are structurally preservative with respect to the classical coordination requirements and resolve the categoricity problem. These results are taken to show that one can give an expressivist bilateral meaning-is-use analysis of the connectives that confer on them a determinate coherent classical interpretation.

Technical Appendix I: Nelson Models

Most of the central results of the paper (proved in the next section) utilize the close relation between accept–reject states and Nelson models defined in this section. Nelson models are used for their technical power, not because they are taken to provide an intuitively satisfactory rendition of the basic notions of acceptance and rejection.

A Nelson model is a structure $N = (I, \sqsubseteq , V^{+}, V^{-})$ such that (i) $I$ is a set, (ii) $\sqsubseteq$ partially orders $I$, (iii) for each $p$ in the base language $V^{+}(p), V^{-}(p) \subseteq I$, and (iv) satisfies persistence ($\circ \in \{+,-\}$):

$$\begin{aligned} \text {If } u\sqsubseteq v \text { and } u\in V^{\circ }(p), \text { then } v\in V^{\circ }(p). \end{aligned}$$

On the basis of such a model one can define a positive and negative forcing relation $\models$, as follows:

$$\begin{aligned} \begin{array}{lll} u\models ^{+}p &{} \quad \hbox { if and only if }\quad &{} \quad u\in V^{+}(p).\\ u\models ^{-}p &{} \quad \hbox { if and only if }\quad &{} \quad u\in V^{-}(p).\\ u\models ^{+}\lnot A &{} \quad \hbox { if and only if }\quad &{} \quad u\models ^{-}A.\\ u \models ^{-}\lnot A &{} \quad \hbox { if and only if }\quad &{} \quad u \models ^{+}A.\\ u \models ^{+}A\supset B &{} \quad \hbox { if and only if }\quad &{} \quad \forall v: \hbox { if } u\sqsubseteq v \hbox { and } v\models ^{+}A, \hbox { then } v \models ^{+}B.\\ u \models ^{-}A\supset B &{} \quad \hbox { if and only if }\quad &{} \quad u \models ^{+}A \hbox { and } u \models ^{-}B.\\ \end{array} \end{aligned}$$

The accept–reject state $\Vdash$ induced by a Nelson model $N$ (with corresponding forcing relation $\models$) is defined (for any finite sets $\Gamma$ and $\Delta$ of sentences from the extended language and any $A$ from the extended language):

$$\begin{aligned} +\Gamma , - \Delta \Vdash + A&\text { if and only if } \forall v: \hbox { if } v\models ^{+}\Gamma \hbox { and } v\models ^{-}\Delta , \hbox { then } v \models ^{+}A.\\ +\Gamma , - \Delta \Vdash - A&\text { if and only if } \forall v: \hbox { if } v\models ^{+}\Gamma \hbox { and } v\models ^{-}\Delta , \hbox { then } v \models ^{-}A. \end{aligned}$$

Notation: $+\Gamma = \{+A\,|\, A\in \Gamma \}$ and $u\models ^{+}\Gamma$ is a short-hand for $\forall A\in \Gamma : u\models ^{+}A$.

I state here, without proofs (they employ standard techniques), some properties that will be employed later:

Lemma 1

• 1. The accept–reject state $\Vdash$ induced by $N$ satisfies, Reflexivity, Monotonicity and Cut, and the classical expressivist requirements for $\lnot$ and $\supset$.

• 2. Let $\Vdash$ be an accept–reject state induced by a Nelson model $N$. If there is no $u$ and $p$ such that $u\in V^{+}(p)$ and $u\in V^{-}(p)$, then $\Vdash$ satisfies Explosion.

• 3. Let $\Vdash$ be an accept–reject state induced by a Nelson model $N$. $\Vdash$ will satisfy Classical Reductio if $\models$ satisfies (for all $u, v\in I$ and atomic letters $p$):

  • (a) If there is no $v$ such that $u\sqsubseteq v$ and $v\in V^{-}(p)$, then $u\in V^{+}(p)$.

  • (b) If there is no $v$ such that $u\sqsubseteq v$ and $v\in V^{+}(p)$, then $u\in V^{-}(p)$.

The canonical model for $\Vdash$ is the Nelson model $N = (I, V^{+}, V^{-}, \sqsubseteq )$. Here $I$ is the set of suppositional contexts (so when $u\in I$, $u$ will be some set $\{+A_1, \ldots , +A_n, -B_1, \ldots , -B_m\}$), with the restriction that for some sentence $A$, either $u\not \Vdash +A$ or $u\not \Vdash -A$. Furthermore, $V^{+}$ and $V^{-}$ are defined:

$$\begin{aligned} V^{+}(p)&= \{u\in I\,|\, u \Vdash + p\}.\\ V^{-}(p)&= \{u\in I\,|\, u \Vdash - p\}. \end{aligned}$$

Finally, $\sqsubseteq$ is defined:

$$\begin{aligned} u\sqsubseteq v \text { if and only if }&\text {(i) } v\Vdash + A, \forall A\in u^{+}, \text { and}\\&\text {(ii) } v\Vdash - A, \forall A\in u^{-}. \end{aligned}$$

One can show using standard techniques that the canonical model for an accept–reject state induces that accept–reject state (so every accept–reject state is induced by some Nelson model).

1.1 Proof of Main Theorems

The proofs of Theorems 1 and 2 are straightforward.

Proof of Theorem 3

The proof idea is to use $\Vdash _b$ to construct a Nelson model and use such a model to construct an accept–reject state $\Vdash _e$ and then show that $\Vdash _e$ is a conservative extension of $\Vdash _b$ with the relevant properties.

Let $I$ be the set of suppositional contexts $u$ (restricted to the base language) such that for some $p$, either $u\not \Vdash _b +p$ or $u\not \Vdash _b - p$.

Define:

$$\begin{aligned} V^{+}(p)&= \{u\in I\,|\, u \Vdash _b + p\}.\\ V^{-}(p)&= \{u\in I\,|\, u \Vdash _b - p\}. \end{aligned}$$

Define:

$$\begin{aligned} u\sqsubseteq v \text { if and only if }&\text {(i) } v\Vdash _b + p, \forall p\in u^{+}, \text { and}\\&\text {(ii) } v\Vdash _b - p, \forall p\in u^{-}. \end{aligned}$$

$\sqsubseteq$ is a partial order. Through Reflexivity, $u\sqsubseteq u$ for all $u$. Assume that $u\sqsubseteq v$ and $v\sqsubseteq w$. So $\forall p\in u^{+}$: $v\Vdash _b + p$, and (ii) $\forall p\in u^{-}$: $v\Vdash _b - p$. Furthermore, $\forall p\in v^{+}$: $w\Vdash _b + p$, and (ii) $\forall p\in v^{-}$: $w\Vdash _b - p$. Assume that $p\in u^{+}$. Then $v\Vdash _b + p$. By Monotonicity, $v\cup w \Vdash _b + p$. By repeated applications of Cut, $w \Vdash _b + p$. Similarly, if $p\in u^{-}$, then $w \Vdash _b - p$. So $u \sqsubseteq w$.

$\sqsubseteq$ is persistent. Assume that $p\in u^{+}$ and $u\sqsubseteq v$. Then $v\Vdash _b + p$, so then $v\in V^{+}(p)$. Assume that $p\in u^{-}$ and $u\sqsubseteq v$. Then $v\Vdash _b - p$, so then $v\in V^{-}(p)$.

So we have a Nelson model.

Let $\Vdash _e$ be the accept–reject state induced by the forcing relation $\models$ in the above Nelson model.

One needs to show that $\Vdash _e$ is identical to $\Vdash _b$ when restricted to the base language. So let $\Gamma = \{p_1, \ldots , p_m\}$ and $\Delta = \{q_1, \ldots , q_m\}$ be sets of atomic sentences. $+p_1, \ldots , +p_n, -q_1, \ldots , -q_m \Vdash _e + r$ if and only if $\forall v:$ if $v \models ^{+}p_i$ (for each $i$) and $v\models ^{-}q_j$ (for each $j$), then $v \models ^{+}r$ if and only if $\forall v:$ if $v \in V^{+}(p_i)$ (for each $i$) and $v\in V^{+}(q_j)$ (for each $j$), then $v \in V^{+}(r)$ if and only if $\forall v:$ if $v\Vdash _b + p_i$ (for each $i$) and $v\Vdash _b - q_j$ (for each $j$), then $v \Vdash _b + r$ if and only if (given Monotonicity and Cut) $+p_1, \ldots , +p_n, -q_1, \ldots , -q_m \Vdash _b + r$.

So $\Vdash _e$ is a conservative extension of $\Vdash _b$ that (Lemma 1) satisfies the minimal structural requirements and the classical expressivist requirements for the connectives.

Finally, we need to show that if $\Vdash _b$ satisfies Explosion, then so does $\Vdash _e$ and if $\Vdash _b$ satisfies Classical Reductio, then so does $\Vdash _e$.

Explosion If $\Vdash _b$ satisfies Explosion, then there is no $u$ and no $p$ such that $u\models ^{+}p$ and $u\models ^{-}p$. By Lemma 1, $\Vdash _e$ satisfies Explosion.

Classical Reductio Assume that $\Vdash _b$ satisfies Classical Reductio. First establish the following two claims:

• 1. If there is no $v$ such that $u\sqsubseteq v$ and $v\in V^{-}(p)$, then $u\in V^{+}(p)$.

• 2. If there is no $v$ such that $u\sqsubseteq v$ and $v\in V^{+}(p)$, then $u\in V^{-}(p)$.

Assume that $u\not \models ^{+}p$. Then $u\not \in V^{+}(p)$ and so $u\not \Vdash +p$. By Classical Reductio, $u \cup \{ - p\}\not \Vdash +p$. But then $u \cup \{ - p\}\in I$ and as $u \cup \{ - p\}\in V^{-}(p)$, $u \cup \{ - p\} \models ^{-}p$. Furthermore, $u \sqsubseteq u \cup \{ - p\}$. So there is a $v$ such that $u\sqsubseteq v$ and $v\models ^{-}p$. The second case is analogous.

This establishes (1) and (2) above. From Lemma 1 it follows that $\Vdash _e$ satisfies Classical Reductio.

Proof of Theorem 4

Notation: where $\mathbf{S}= \{+A_1, \ldots , + A_n, -B_1, \ldots , -B_m\}$, let $S_\mathbf{S}= A_1 \wedge \cdots \wedge A_n \wedge B_1 \wedge \cdots \wedge B_m$.

Note that any sentence $A$ is classically equivalent to a (finite) set $\Gamma$ of sentences of either the form $p_1 \wedge \cdots \wedge p_n \wedge \lnot q_1\wedge \cdots \wedge \lnot q_m \supset r$ or the form $p_1 \wedge \cdots \wedge p_n \wedge \lnot q_1\wedge \cdots \wedge \lnot q_m \supset \lnot r$ (that is $A$ is a classical consequence of such a set $\Gamma$, and every element of $\Gamma$ is a consequence of $A$). Thus (from Theorem 2) where $\Gamma$ is such a set, a classical accept–reject state will satisfy:

$$\begin{aligned} \Vdash _e + A \text { if and only if }&\text {(i) } \mathbf{S}\Vdash _e + r, \text { for each } S_\mathbf{S}\supset r\in \Gamma , \text { and}\\&\text {(ii) } \mathbf{S}\Vdash _e - r, \text { for each } S_\mathbf{S}\supset \lnot r\in \Gamma . \end{aligned}$$

A classical accept–reject state $\Vdash _e$ defined on the extended language is uniquely determined by the set of categorically accepted sentences (Theorem 2). So as acceptance of a categorical sentence in turn is uniquely determined by the restriction of $\Vdash$ to the base language, it follows that as (Theorem 3) every classical accept–reject state $\Vdash _b$ defined on the base language has a classical conservative extension defined on the extended language, it has precisely one such extension.

Proof of Theorem 5

(1) Assume that $\Vdash$ satisfies the classical requirements for accepting $\lnot$.

Left-to-right, i.e. assume that $\Vdash$ satisfies the classical requirements for accepting $\lnot$.

Assume that $\not \Vdash -\lnot A$. Assume, for reductio, that $\Vdash + A$. By Reflexivity $+\lnot A\Vdash + \lnot A$, i.e. $+\lnot A\Vdash - A$. By Monotonicity $+\lnot A\Vdash + A$. So, by Explosion, $+\lnot A\Vdash - \lnot A$. By Classical Reductio, $\Vdash - \lnot A$. But then we have a contradiction, so $\not \Vdash +A$.

For the converse direction, Assume that $\not \Vdash +A$. Assume for reductio that $\Vdash - \lnot A$. By Reflexivity $- A\Vdash - A$, i.e. $-A \Vdash + \lnot A$. By Monotonicity $- A\Vdash - \lnot A$. By Explosion $-A\Vdash + A$, so by Classical Reductio $\Vdash + A$ and we have a contradiction. So $\not \Vdash -\lnot A$.

Right-to-left, i.e. assume that $\Vdash$ satisfies the classical requirements for rejecting $\lnot$. Assume that $\not \Vdash +\lnot A$. Assume, for reductio, that $\Vdash - A$. By Reflexivity $-\lnot A\Vdash - \lnot A$, i.e. $-\lnot A\Vdash + A$. By Monotonicity $-\lnot A\Vdash - A$. So, by Explosion, $-\lnot A\Vdash + \lnot A$. By Classical Reductio, $\Vdash + \lnot A$. But then we have a contradiction, so $\not \Vdash -A$.

For the converse direction, Assume that $\not \Vdash -A$. Assume for reductio that $\Vdash + \lnot A$. By Reflexivity $+ A\Vdash + A$, i.e. $+A \Vdash - \lnot A$. By Monotonicity $+ A\Vdash + \lnot A$. By Explosion $+A\Vdash - A$. But then by Classical Reductio $\Vdash - A$ and we have a contradiction. So $\not \Vdash +\lnot A$.

(2) Left-to-right, i.e. assume that $\Vdash$ satisfies the classical requirements for accepting $\supset$.

Assume that $\Vdash - A\supset B$. By Reflexivity and Explosion $- A, +A \Vdash + B$ and so $-A\Vdash + A\supset B$. By Monotonicity $-A\Vdash - A\supset B$ and so by Explosion $-A\Vdash + A$. By Classical Reductio, $\Vdash + A$. By Reflexivity and Explosion $+ B, +A \Vdash + B$ and so $+B\Vdash + A\supset B$. By Monotonicity $+B\Vdash - A\supset B$ and so by Explosion $+B\Vdash - B$. By Classical Reductio, $\Vdash - B$.

Assume that $\Vdash + A$ and $\Vdash - B$. By Reflexivity $+A\supset B\Vdash + A\supset B$, i.e. $+A\supset B, +A\Vdash + B$. By Monotonicity $+A\supset B, +A\Vdash - B$ and so by Explosion $+A\supset B, +A\Vdash - A$. By Reflexivity $+A\supset B, +A\Vdash + A$ so by Explosion $+A\supset B, +A \Vdash - A\supset B$. By Monotonicity $+A\supset B\Vdash + A$ and so by Cut $+A\supset B\Vdash - A\supset B$. By Classical Reductio, $\Vdash - A\supset B$.

Right-to-left, i.e. assume that $\Vdash$ satisfies the classical requirements for rejecting $\supset$.

Assume that $\Vdash + A\supset B$. By Reflexivity $+A, -B\Vdash + A$ and $+A, -B\Vdash - B$. So $+A, -B\Vdash - A\supset B$. By Monotonicity $+A, -B\Vdash + A\supset B$. By Explosion $+A, -B\Vdash + B$. By Classical Reductio $+A\Vdash + B$.

Assume that $+A\Vdash + B$. By Monotonicity $+A, -A\supset B\Vdash + B$. By Reflexivity $+A, -A\supset B\Vdash - A\supset B$ and so $+A,-A\supset B\Vdash - B$. By Explosion $+A, -A\supset B \Vdash + A\supset B$. By Classical Reductio, $+A\Vdash + A\supset B$. By Reflexivity, $-A\supset B \Vdash - A\supset B$. Thus $-A\supset B\Vdash + A$. By Monotonicity, $-A\supset B, +A\Vdash + A\supset B$. So, by Cut, $-A\supset B\Vdash + A\supset B$. By Classical Reductio, $\Vdash + A\supset B$.

Proof of Theorem 6

Let $\Vdash$ be an accept–reject state satisfying the minimal structural requirements and the expressive classical requirements on the connectives as well as the restricted requirements Explosion(R) and Classical Reductio(R). Let $N$ be the canonical model for $\Vdash$.

For Explosion assume that $u\in I$ and, for reductio, that $u\models ^{+}p$ and $u\models ^{-}p$. It follows that $u\Vdash + p$ and $u\Vdash - p$. By Explosion(R) $u\Vdash + q$ and $u\Vdash - q$ for all $q$. But then, this is easy to show, $u\Vdash + A$ and $u\Vdash - A$ for all $A$. But this contradicts the assumption that $u\in I$. So there is no $u\in I$ and no $p$ such that $u\models ^{+}p$ and $u\models ^{-}p$. So by Lemma 1, $\Vdash$ satisfies Explosion.

For Classical Reductio, establish the following two claims:

• 1. If there is no $v$ such that $u\sqsubseteq v$ and $v\in V^{-}(p)$, then $u\in V^{+}(p)$.

• 2. If there is no $v$ such that $u\sqsubseteq v$ and $v\in V^{+}(p)$, then $u\in V^{-}(p)$.

Assume that $u\not \in V^{+}(p)$. It follows that $u\not \Vdash +p$. By Classical Reductio(R) it follows that $u, -p\not \Vdash +p$. So $u\cup \{-p\}\in I$. Note that $u\sqsubseteq u\cup \{-p\}$ and $u\cup \{-p\}\in V^{-}(p)$. This establishes (1), the proof of (2) is analogous. From Lemma 1 it follows that $\Vdash$ satisfies Classical Reductio.