Is there a neutral metalanguage?

Abstract

Logical pluralists are committed to the idea of a neutral metalanguage, which serves as a framework for debates in logic. Two versions of this neutrality can be found in the literature: an agreed upon collection of inferences, and a metalanguage that is neutral as such. I discuss both versions and show that they are not immune to Quinean criticism, which builds on the notion of meaning. In particular, I show that (i) the first version of neutrality is sub-optimal, and hard to reconcile with the theories of meaning for logical constants, and (ii) the second version collapses mathematically, if rival logics, as object languages, are treated with charity in the metalanguage. I substantiate (ii) by proving a collapse theorem that generalizes familiar results. Thus, the existence of a neutral metalanguage cannot be taken for granted, and meaning-invariant logical pluralism might turn out to be dubious.

Appendix

As the logics at issue are reflexive, transitive, and compact, and admit deduction theorems, we can use lattice-based semantics to give a general account of their semantics. I follow mainly Rasiowa’s notion of ${ implicative}\;{ algebras}$ (Rasiowa 1974) with slight differences. Recall that we are concerned here with a language $\mathcal {L}$ whose expressions are generated by some set of connectives over a countable set of atoms to be designated by At. Axioms and rules of inference for each logic (over $\mathcal {L}$) are defined in the usual way.

Definition 1

A ${ lattice}$ is a partial order $\langle A,\le \rangle$ such that for all $a,b\in A$ there exist $a\vee b,a\wedge b\in A$ with:

• 1. $a\wedge b\le a,b$ and if $c\le a,b$ then $a\wedge b\le c$.

• 2. $a,b\le a\vee b$ and if $a,b\le c$ then $a\vee b\le c$.

A lattice is ${ bounded}$ if there are $0,1\in A$ such that $0\le a\le 1$ for all $a\in A$.

For semantics, we have to define ${ implication}\,{ lattices}$:

Definition 2

An ${ implication}\,{ lattice}$ is a tuple $M=\langle A,\le ,Tr \rangle$ where:

• 1. $\langle A,\le \rangle$ is a bounded lattice.

• 2. For each n-ary logical constant, $*\in \mathcal {L}$, there is a corresponding n-ary operation $*':A^{n}\rightarrow A$.

• 3. In particular, $\wedge ,\vee ,\bot ,\top$ stand for conjunction, disjunction, bottom, and top respectively.

• 4. $Tr\subseteq A$ is a filter with respect to $\le$, $\top \in Tr$. We call Tr a ${ truth}\,{ filter}$.

Definition 3

Let L be some logic over $\mathcal {L}$, and $M=\langle A,\le ,Tr \rangle$ be an implication lattice. A functor $F:\mathcal {L}\rightarrow A$ is a ${ semantic}\ { functor}$ for L if the following conditions are met:

• 1. For each n-ary logical connective, $*\in \mathcal {L}$, and $\alpha _{1},\ldots ,\alpha _{n}\in \mathcal {L}$: $F(*(\alpha _{1},\ldots ,\alpha _{n}))=*'(F(\alpha _{1}),\ldots ,F(\alpha _{n}))$.

• 2. For each conditional, $\alpha \rightarrow \beta$: $F(\alpha \rightarrow \beta )\in Tr$ iff $F(\alpha )\le F(\beta )$. The interpretation of implications is thus constrained by the lattice partial order, though this restriction isn’t unique:—we might have various implications meeting this condition. Under this restriction each implication respects the consequence relation, so that deduction theorems can hold, as required.

The points of A are regarded as truth values. We say that M satisfies $\alpha$ ($M\vDash \alpha$) if $F(\alpha )\in Tr$.

Lemma 1

Let $F:\mathcal {L}\rightarrow A$ be a semantic functor and $\sigma :At\rightarrow At$ some permutation on the atomic propositions. We define by induction for any $\alpha \in \mathcal {L}\ \sigma (\alpha )$—the translation function generated by $\sigma$ in the obvious way. Then $F_{\sigma }=F\circ \sigma :\mathcal {L}\rightarrow A$ is a semantic functor as well.

The proof of Lemma 1 results trivially from the formality of logic. Hence a lattice is a model for some logic independently of some specific semantic functor. Note that not every lattice is a model for every logic. Soundness and completeness theorems may specify for each logic its class of models in terms of corresponding algebraic properties.

Definition 4

A logic is said to be ${ sound}$ if $\varGamma \vdash \alpha$ implies $\varGamma \models \alpha$, i.e., for each lattice M and semantic functor, $F:L\rightarrow M$: $F(\varGamma )\subseteq Tr$ implies $F(\alpha )\in Tr$. A logic is called ${ complete}$ if $\varGamma \vdash \alpha$ entails $\varGamma \models \alpha$.

Let $L_{1},L_{2}$ be two ${ rival}\,{ logics}$, i.e., there are $\varGamma \subseteq \mathcal {L},\alpha \in \mathcal {L}$ s.t. $\varGamma \vdash _{L_{1}}\alpha ,\varGamma \nvdash _{L_{2}}\alpha$. We define the unifying metalanguage $\mathcal {L}'$ and the corresponding logic $L'$ as explained in Sect. 5. We define models for $\mathcal {L}'$ in the usual way.

Lemma 2

Let $C_{i}$ be the class of models for $L_{i}$ ($i=1,2$); then M is a model for $L'$ iff $M\in C_{1}\cap C_{2}$.

Proof

($\Leftarrow$) Assume $M\in C_{1}\cap C_{2}$. Then, there are two semantic functors for these logics: $F_{i}:\mathcal {L}\rightarrow A$. Define $F:\mathcal {L}'\rightarrow A$ as follows. Without loss of generality (following Lemma 1) we assume that $F_{1}\upharpoonleft At=F_{2}\upharpoonleft At$. By definition, for each logical constant $*$ except for implication: $F(*(\alpha _{1},\ldots ,\alpha _{n}))=*' (F(\alpha _{1}),\ldots ,F(\alpha _{n}))$ is well-defined. In the same way, for each implication $F(\alpha \rightarrow _{i}\beta )$ is well defined in terms of the operator $\rightarrow _{i}'$, since $M\in C_{i}$ (recall Definition 3). By induction, it is trivial to prove that all the rules of inferences and axioms for both logics are represented successfully in this way, for they are preserved by the original $F_{i}$’s, and each $\rightarrow _{i}$ represents the corresponding consequence relation. ($\Rightarrow$) Assume $F:\mathcal {L}\text {'}\rightarrow A$ is a semantic functor. For each i define: $F_{i}:\mathcal {L}\rightarrow A$ by $F_{i}(\alpha )=F(\tau _{i}(\alpha ))$ where $\tau _{i}$ is the translation function defined in Sect. 5. The rest of the proof is trivial. $\square$

Corollary 1

Soundness If $\varGamma \vdash \alpha$ then $\varGamma \models \alpha$.

Proof

By induction on the derivation of $\alpha$ from $\varGamma$. Each step is either a premise (true by definition), or an axiom or rule of inference representing $L_{1}$, preserved because $M\in C_{1}$, or an axiom or rule of inference representing $L_{2}$, preserved because $M\in C_{2}$, as Lemma 2 shows. $\square$

Definition 5

For technical reasons, we define a hierarchy of ${ incomplete}$ languages (namely, sets of propositions not necessarily closed under the connectives) such that: ${\mathcal {L}'} = \bigcup \nolimits _{{n \in \mathbb {N}}} {{\mathcal {L}}_{n} }$. Let $\mathcal {\mathcal {L}}_{1} =(\mathcal {L}\backslash \{\rightarrow \})\cup \{\rightarrow _{1}\}$, and define $\mathcal {L}_{2n+1}$ ($\mathcal {L}_{2n}$) by induction: $\alpha \in \mathcal {L}_{2n+1}\ (\mathcal {L}_{2n})$ iff:

$$\begin{aligned} \alpha \in \mathcal {L}_{2n}\;(\mathcal {L}_{2n-1})\;or\;\alpha =\beta \rightarrow _{1}\gamma \;(\alpha =\beta \rightarrow _{2}\gamma )\;where\;\beta , \gamma \in \mathcal {L}_{2n+1}\;(\mathcal {L}_{2n}) \end{aligned}$$

Let i(n) denote the index of the implication symbol (i.e., 1 or 2) added during this enrichment process at the level $\mathcal {L}_{n}$. That is, $i(n)={\left\{ \begin{array}{ll} 1 &{} n\;is\;{ odd}\\ 2 &{} n\;is\;{ even} \end{array}\right. }$

Before proving completeness, let me just explain what this hierarchy is good for. To prove completeness, we construct a saturated set which induces a lattice, and then show that this lattice is indeed a model for both logics, relying on the completeness theorems of both $L_{1},L_{2}$. In particular, we show by induction that up to the level of $\mathcal {L}_{n}$ we have a Tarski–Lindenbaum lattice for $L_{i(n)}$. Thus, for the entire language we have a model for $L'$. This will become clear while reading through the proof.

Theorem 1

Completeness: If $\varGamma \models \alpha$ then $\varGamma \vdash \alpha$.

Proof

Assume $\varGamma \nvdash \alpha$. We first construct a saturated set $\varGamma ^{+}\supseteq \varGamma$ s.t. $\varGamma ^{+}\nvdash \alpha$. Let $\varphi _{1}$ be an enumeration of $\mathcal {L}_{1}$ and $\varphi _{n+1}:\mathbb {N\rightarrow }\mathcal {L}_{n+1}\backslash \mathcal {L}_{n}$, an enumeration of $\mathcal {L}_{n+1}\backslash \mathcal {L}_{n}$. Then enumerate $\mathcal {L}'$ with $\varphi :\omega \times \omega \rightarrow \mathcal {L}'$ given by $\varphi (n,m)=\varphi _{n}(m)$. Now, we construct $\varGamma ^{+}$ by induction:

$$\begin{aligned} \varGamma _{1,0}= & {} \varGamma \cap \mathcal {L}_{1}\\ \varGamma _{n,m+1}= & {} {\left\{ \begin{array}{ll} \varGamma _{n,m}\cup \varphi (n,m)\} &{} if\;\varGamma _{n,m}\cup \{\varphi (n,m)\}\nvdash \alpha \\ \varGamma _{n,m} &{} { otherwise} \end{array}\right. }\\ \varGamma _{n}^{+}= & {} \bigcup _{m<\omega }\varGamma _{n,m}\\ \varGamma _{n+1,0}= & {} \varGamma _{n}^{+}\cup (\varGamma \cap \mathcal {L}_{n+1})\\ \varGamma ^{+}= & {} \bigcup _{n<\omega }\varGamma _{n,0} \end{aligned}$$

It is easy to verify that $\varGamma ^{+}$ is closed under the consequence relation and that $\varGamma ^{+}\nvdash \alpha$.

Now define $\delta \le \gamma$ if $\varGamma ^{+},\delta \vdash \gamma$, and $\delta \sim \gamma$ if $\delta \le \gamma$ and $\gamma \le \delta$. $\sim$ is then an equivalence relation. Define $A_{M}=\nicefrac {\mathcal {L}'}{\sim }$, so that $\langle A_{M},\le \rangle$ is a bounded lattice, since for all $\delta \in \mathcal {L}'$: $[\bot ]\le [\delta ]\le [\top ]$. Define $Tr=\nicefrac {\varGamma ^{+}}{\sim }$. Tr is by definition a truth filter, and so $M=\langle A_{M},\le ,Tr \rangle$ is an implication lattice. Define now: $F(\delta )=[\delta ]$. All we have to show is that $F:\mathcal {L}'\rightarrow A_{M}$ is a semantic functor for $L'$. If so, we get $Th(M)=\varGamma ^{+}$ by definition, and $M\nvDash \alpha$ as a result. In particular, we have to show that M is a Tarski–Lindenbaum lattice for both logics (see Font et al. 2003, pp. 22–24). If so, we get:

• (1) For each n-ary operator $*$ : $F(*(\delta _{1},\ldots ,\delta _{n}))=*'(F(\delta _{1}),\ldots ,F(\delta _{n}))$, where $*$ is the operator defined for semantic functors $F:\mathcal {L}'\rightarrow A_{M}$, where $M\in C_{1}\cap C_{2}$.

• (2) For each implication: $F(\delta \rightarrow _{i}\gamma )\in Tr$ iff $F(\delta )\le F(\gamma )$.

• (3) F satisfies all the axioms and the rules of inference.

The proof rests on the completeness of both $L_{1},L_{2}$. We first define for all $n\in \mathbb {N}$ a projection translation function $\eta _{n}:\mathcal {L}_{n}\rightarrow \mathcal {L}\cup \{p_{n}^{k}\}_{k\in \mathbb {N}}$, where $\{p_{n}^{k}\}_{k\in \mathbb {N}}$ are countably many new atoms. $\eta _{1}$ is the identity function. $\eta _{n+1}$ is defined by induction:

$$\begin{aligned} \eta _{n+1}(\delta )={\left\{ \begin{array}{ll} \delta &{} if\ \delta \in At\\ \eta _{n+1}(\beta )*\eta _{n+1}(\gamma ) &{} if\ \delta =\beta *\gamma ,\ where*\ne \rightarrow _{1},\rightarrow _{2}\\ p_{n+1}^{k} &{} if\ \delta =\beta \rightarrow _{i(n)}\gamma ,\ where\ p_{n+1}^{k}\,wasn't\,added\,yet\\ \eta _{n+1}(\beta )\rightarrow \eta _{n+1}(\gamma ) &{} if\ \delta =\beta \rightarrow _{i(n+1)}\gamma \end{array}\right. } \end{aligned}$$

All $\eta _{n}$’s are by definition bijective. I now argue that $\eta _{n}(\varGamma _{n}^{+})$ is closed under the consequence relation of $L_{i(n)}$. Assume $\eta _{n}(\varGamma _{n}^{+})\vdash _{L_{i(n)}} \delta$; we are about to show by induction on the length of this proof that $\delta \in \eta _{n}(\varGamma _{n}^{+})$. For length=1, if $\delta$ is not an axiom then immediately $\delta \in \eta _{n}(\varGamma _{n}^{+})$. If it is an axiom of $L_{i(n)}$, then $\eta _{n}^{-1}(\delta )$ is an axiom of $L'$ whose principal operator is not $\rightarrow _{i(n+1)}$. Thus, $\eta _{n}^{-1}(\delta )\in \mathcal {L}_{n}$. By the construction, $\eta _{n}^{-1}(\delta )\in \varGamma _{n}^{+}$, so $\delta \in \eta _{n}(\varGamma _{n}^{+})$. For length>1, suppose that $\delta$ is derived from $\delta _{1},\ldots ,\delta _{k}$ in $L_{i(n)}$. By the induction hypothesis, $\delta _{1},\ldots ,\delta _{k}\in \eta _{n} (\varGamma _{n}^{+})$. Thus, $\eta _{n}^{-1}(\delta _{1}),\ldots , \eta _{n}^{-1}(\delta _{k})\in \varGamma _{n}^{+}$. By the conjunction introduction-rule of $L'$: $\varGamma _{n}^{+}\vdash {\bigwedge \nolimits _{1\le j\le k}}\eta ^{-1}(\delta _{j})$. The above mentioned derivation in $L_{i(n)}$ is by the construction represented in $L'$, with $\varGamma _{n}^{+}\vdash {\bigwedge \nolimits {1\le j\le k}}\eta ^{-1}(\delta _{j})\rightarrow _{i(n)}\eta ^{-1}(\delta )$. By MP: $\varGamma _{n}^{+}\vdash \eta ^{-1}(\delta )$. Hence, by the construction: $\eta _{n}^{-1}(\delta )\in \varGamma _{n}^{+}$. Therefore, $\delta \in \eta _{n}^{-1}(\varGamma _{n}^{+})$.

Define now: $F_{n}:\mathcal {L}\cup \{p_{n}^{k}\}_{k\in \mathbb {N}}\rightarrow A_{M}$ by $F_{n}(\delta )=F(\eta _{n}^{-1}(\delta ))$. It is easy to verify that due to the soundness and completeness of $L_{i(n)}$, together with the representational properties of $L'$, $F_{n}$ is a well defined semantic functor for $L_{i(n)}$, and $F_{n}(\mathcal {L}\cup \{p_{n}^{k}\}_{k\in \mathbb {N}})\subseteq A_{M}$ forms a Tarski–Lindenbaum lattice for $L_{i(n)}$. In conclusion: $M\in C_{i(n)}$, because $A_{M}\supseteq F_{n}(\mathcal {L}\cup \{p_{n}^{k}\}_{k\in \mathbb {N}})$.

The last result is true for all $n\in \mathbb {N}$. Moreover, $F_{n}(\mathcal {L}\cup \{p_{n}^{k}\}_{k\in \mathbb {N}})=F(\mathcal {L}_{n})$, and hence $F(\mathcal {L}_{n})\in C_{i(n)}$. By definition $F(\mathcal {L}_{n})\subseteq F(\mathcal {L}_{n+1})$ (because $\mathcal {L}_{n}\subseteq \mathcal {L}_{n+1}$), and $F(\mathcal {L}_{n+1})\in C_{i(n+1)}$. To sum up, for all $n\in \mathbb {N}\cup \{0\}$:

$$\begin{aligned} F(\mathcal {L}_{2n+1})\in & {} C_{1}\\ F(\mathcal {L}_{2n})\in & {} C_{2} \\ F(\mathcal {L}_{n})\subseteq & {} F(\mathcal {L}_{n+1}) \end{aligned}$$

These three results give us:

$$\begin{aligned} \underset{n\in \mathbb {N}}{\bigcup F(\mathcal {L}_{2n}})= & {} F\left( \underset{n\in \mathbb {N}}{\bigcup \mathcal {L}_{2n}}\right) =\underset{n\in \mathbb {N}}{F\left( \bigcup \mathcal {L}_{n}\right) }=A_{M} =\underset{n\in \mathbb {N}}{F\left( \bigcup \mathcal {L}_{n}\right) }\\= & {} \underset{n\in \mathbb {N}}{F\left( \bigcup \mathcal {L}_{2n+1}\right) } =\underset{n\in \mathbb {N}}{\bigcup F(\mathcal {L}_{2n+1}}) \end{aligned}$$

But ${\bigcup F(\mathcal {L}_{2n+1}})_{n\in \mathbb {N}},{\bigcup F(\mathcal {L}_{2n}})_{n\in \mathbb {N}}$ form implication lattices that belong to $C_{1},C_{2}$, respectively. So $M\in C_{1}\cap C_{2}$, and F forms a Tarski–Lindenbaum lattice for both logics. Thus, M is a model for $L'$ such that $M\nvDash \alpha$. $\square$

Theorem 2

(collapse): Let $L_{1},L_{2}$ be two rival logics, and L’ the unifying system. Assume without loss of generality that for some $\alpha ,\beta \in \mathcal {L}$: $\alpha \vdash _{L_{1}}\beta ,\alpha \nvdash _{L_{2}}\beta$. The representational roles of the implications collapse: $\vdash \tau _{1}(\alpha )\rightarrow _{1}\tau _{1}(\beta )$ entails $\vdash \tau _{2}(\alpha )\rightarrow _{2}\tau _{2}(\beta )$.

Proof

We first assume that $\alpha ,\beta$ do not involve implications, so that: $\tau _{i}(\alpha )=\alpha ,\tau _{i}(\beta )=\beta$, and show that $\alpha \vdash _{L_{1}}\beta$ entails $\vdash \alpha \rightarrow _{2}\beta$ (this by itself shows that the representational roles of the implications collapse). Assume then that $\alpha \vdash _{L_{1}}\beta$ where $\alpha ,\beta$ do not involve implications. Therefore, $\vdash \alpha \rightarrow _{1}\beta$ as $L'$ represents $L_{1}$ (this direction is straight forward, as stated in Sect. 5). Hence, for all $M=\langle A_{M}, \le ,Tr \rangle$ such that $M\in C_{1}\cap C_{2}$ and semantic functor: $F:\mathcal {L}':\rightarrow A_{M}$: $F(\alpha \rightarrow _{1}\beta )\in Tr$. By definition, for all these models: $F(\alpha )\le F(\beta )$. But then: $F(\alpha \rightarrow _{2}\beta )\in Tr$. That is, for all $M\in C_{1}\cap C_{2}$ and $F:\mathcal {L}'\rightarrow A_{M}$: $M\models \alpha \rightarrow _{2}\beta$. By the completeness of $L'$ we get: $\vdash \alpha \rightarrow _{2}\beta$ even though $\alpha \nvdash _{L_{2}}\beta$. The proof for $\alpha ,\beta$ that do involve implications is done by induction on the number of implications in $\alpha$ and $\beta$. Since I’ve already shown that the representational roles collapse, it is left to the reader. $\square$