Identity and quantification

Abstract

It is a philosophical commonplace that quantification involves, invokes, or presupposes, the relation of identity. There seem to be two major sources for this belief: (1) the conviction that identity is implicated in the phenomenon of bound variable recurrence within the scope of a quantifier; (2) memories of Quine’s insistence that quantification requires absolute identity for the values of variables. With respect to (1), I show that the only extant argument for a dependence of variable recurrence on identity, due to John Hawthorne, fails. I further show that the function of variable recurrence is not subsumed under that of identity, so that a dependence of the former on the latter, if any, would have to be of a rather indirect nature. With respect to (2), I argue that the relevant passage in Quine fails to establish a connection between quantification and the identity relation, and indeed wasn’t intended by Quine to do so.

Appendices

Appendix 1

Here we establish the result that first-order logic with identity but without variable recurrence within the scope of a quantifier is strictly less expressive than full first-order logic with identity.

To see this, consider a first-order language $\mathcal {L}$ with identity whose signature is given by two unary predicate symbols F and G. The recurrence-free $\mathcal {L}$-formulas can be defined as follows: Whenever x and y are distinct variables, Fx, Gx, and $x=y$ are recurrence-free $\mathcal {L}$-formulas; whenever $\phi$ is a recurrence-free $\mathcal {L}$-formula, so is $\lnot \phi$; whenever $\phi$ and $\psi$ are recurrence-free $\mathcal {L}$-formulas such that no variable occurs free in both $\phi$ and $\psi$, $(\phi \wedge \psi )$ is a recurrence-free $\mathcal {L}$-formula; and whenever x is any variable and $\phi$ a recurrence-free $\mathcal {L}$-formula, $\exists x\phi$ is a recurrence-free $\mathcal {L}$-formula.

Now let ${\mathfrak {M}}_0$ be the model for $\mathcal {L}$ whose domain is given by the set $M = \{ 0,1\}$ and that interprets both F and G as the singleton set $\{ 0\}$. Let ${\mathfrak {M}}_1$ be the model for $\mathcal {L}$ whose domain is also the set M and that interprets F as $\{ 0\}$ (just like ${\mathfrak {M}}_0$), but interprets G as $\{ 1\}$. Clearly ${\mathfrak {M}}_0$ and ${\mathfrak {M}}_1$ can be distinguished by means of the non-recurrence-free $\mathcal {L}$-sentence $\exists x\, (Fx \wedge Gx)$, i.e. (2) in the main text, which is true in ${\mathfrak {M}}_0$ but false in ${\mathfrak {M}}_1$. However, as we will now prove, ${\mathfrak {M}}_0$ and ${\mathfrak {M}}_1$ make exactly the same recurrence-free $\mathcal {L}$-sentences true. It follows that the expressive power of first-order logic with identity is indeed diminished by disallowing variable recurrence. Hence the identity relation cannot subsume the semantic function of variable recurrence.

To obtain the desired result, we must prove a slightly more general fact:

Lemma

For every recurrence-free $\mathcal {L}$-formula $\phi$ there is a set $A_\phi$ included in the set $\mathsf {FV}(\phi )$ of variables occurring free in $\phi$ such that, for all variable assignments $\sigma$ in the set M, $\sigma$ satisfies $\phi$ in ${\mathfrak {M}}_0$ if and only if $\delta ^\phi _\sigma$ satisfies $\phi$ in ${\mathfrak {M}}_1$, where $\delta ^\phi _\sigma$ is the variable assignment that maps a variable x to $\sigma (x)$ if $x\not \in A_\phi$, but to $1-\sigma (x)$ if $x\in A_\phi$.

The lemma can be proved by induction on the recurrence-free formula $\phi$. If $\phi$ is atomic, it is of one of the forms Fx, Gx, and $x=y$. We let $A_{Fx}$ and $A_{x=y}$ be the empty set, so that $\delta ^{Fx}_\sigma = \delta ^{x=y}_\sigma = \sigma$ for every assignment $\sigma$. This has the desired result, since the interpretation of F is the same in ${\mathfrak {M}}_0$ and ${\mathfrak {M}}_1$, and likewise for $=$. For the remaining atomic case, we let $A_{Gx}$ be $\{ x\}$, so that $\delta ^{Gx}_\sigma (x) = 1-\sigma (x)$ for every assignment $\sigma$. Since i is in the interpretation of G in ${\mathfrak {M}}_0$ if and only if $1-i$ is in the interpretation of G in ${\mathfrak {M}}_1$, this too has the desired result.

If $\phi$ is compound, it is of one of the forms $\lnot \psi$, $\exists x\psi$, and $(\psi \wedge \theta )$. We let $A_{\lnot \psi }$ be $A_\psi$, so that $\delta ^{\lnot \psi }_\sigma$ is $\delta ^\psi _\sigma$. The induction hypothesis then immediately gives the desired result.

Let $A_{\exists x\psi }$ be $A_\psi {\setminus } \{ x\}$, so that, for $y\in \mathsf {FV}(\exists x\psi )$, $\delta ^{\exists x\psi }_\sigma (y)$ is $\delta ^{\psi }_\sigma (y)$. For assignments $\tau$ and elements $i\in M$, let us write $\tau \{ x:= i\}$ for the x-variant of $\tau$ that maps x to i. We note that the existence of an $i\in M$ such that $\delta ^\psi _{\sigma \{ x:= i\}}$ satisfies $\psi$ in ${\mathfrak {M}}_1$ is equivalent to the existence of a $j\in M$ such that $\delta ^\psi _\sigma \{ x:= j\}$ satisfies $\psi$ in ${\mathfrak {M}}_1$. This is because, if $x\not \in A_\psi$, $\delta ^\psi _{\sigma \{ x:= i\}}$ is $\delta ^\psi _\sigma \{ x:= i\}$ for each $i\in M$, while if $x\in A_\psi$, $\delta ^\psi _{\sigma \{ x:= i\}}$ is $\delta ^\psi _\sigma \{ x:= 1-i\}$. Further, for any $j\in M$, $\delta ^\psi _\sigma \{ x:= j\}$ is the same function as $\delta ^{\exists x\psi }_\sigma \{ x:= j\}$, since $\delta ^\psi _\sigma$ and $\delta ^{\exists x\psi }_\sigma$ differ at most in what they assign to x. With these observations in place, we see that $\sigma$ satisfies $\exists x\,\psi$ in ${\mathfrak {M}}_0$ if and only if for some $i\in M$, $\sigma \{ x:= i\}$ satisfies $\psi$ in ${\mathfrak {M}}_0$, if and only if (by induction hypothesis) for some $i\in M$, $\delta ^\psi _{\sigma \{ x:= i\}}$ satisfies $\psi$ in ${\mathfrak {M}}_1$, if and only if for some $j\in M$, $\delta ^\psi _\sigma \{ x:= j\}$ satisfies $\psi$ in ${\mathfrak {M}}_1$, if and only if for some $j\in M$, $\delta ^{\exists x\psi }_\sigma \{ x:= j\}$ satisfies $\psi$ in ${\mathfrak {M}}_1$, if and only if $\delta ^{\exists x\psi }_\sigma$ satisfies $\exists x\, \psi$ in ${\mathfrak {M}}_1$, as desired.

Finally, let $A_{(\psi \wedge \theta )}$ be $A_\psi \cup A_\theta$. We note that $\delta ^\psi _\sigma$ and $\delta ^{(\psi \wedge \theta )}_\sigma$ agree on $\mathsf {FV}(\psi )$, and $\delta ^\theta _\sigma$ and $\delta ^{(\psi \wedge \theta )}_\sigma$ agree on $\mathsf {FV}(\theta )$. To see the former, observe that for every $x\in \mathsf {FV}(\psi )$, $x\in A_\psi \cup A_\theta$ if and only if $x\in A_\psi$, since $(\psi \wedge \theta )$ is recurrence-free, so that $\psi$ and $\theta$ do not share any free variables, and x cannot be in $A_\theta \subseteq \mathsf {FV}(\theta )$. The corresponding claim for $\theta$ follows analogously. We then have that $\sigma$ satisfies $(\psi \wedge \theta )$ in ${\mathfrak {M}}_0$ if and only if $\sigma$ satisfies both $\psi$ and $\theta$ in ${\mathfrak {M}}_0$, if and only if (by induction hypothesis) $\delta ^\psi _\sigma$ satisfies $\psi$ in ${\mathfrak {M}}_1$ and $\delta ^\theta _\sigma$ satisfies $\theta$ in ${\mathfrak {M}}_1$. By the fact just observed, and since satisfaction depends only upon what is assigned to the variables actually occurring free in a formula, this is the case if and only if $\delta ^{(\psi \wedge \theta )}_\sigma$ satisfies $\psi$ in ${\mathfrak {M}}_1$ and $\delta ^{(\psi \wedge \theta )}_\sigma$ satisfies $\theta$ in ${\mathfrak {M}}_1$, i.e. if and only if $\delta ^{(\psi \wedge \theta )}_\sigma$ satisfies $(\psi \wedge \theta )$ in ${\mathfrak {M}}_1$, as desired. This completes the proof of our lemma.

The result concerning recurrence-free sentences now follows from the lemma by observing that a sentence (closed formula) is true in a model if and only if it is satisfied by at least one assignment. Thus a recurrence-free sentence $\phi$ is true in ${\mathfrak {M}}_0$ if and only if there is an assignment $\sigma$ that satisfies $\phi$ in ${\mathfrak {M}}_0$; by the lemma, this is the case if and only if there is an assignment (to wit, $\delta ^\phi _\sigma$, which in the case of sentences just is $\sigma$) that satisfies $\phi$ in ${\mathfrak {M}}_1$, if and only if $\phi$ is true in ${\mathfrak {M}}_1$.⁸

In an analogous fashion, one can prove the corresponding result about first-order languages in predicate functor notation: PFL-languages with an identity predicate but without the functor $\mathsf {Ref}$ are, in general, strictly less expressive than the same languages with $\mathsf {Ref}$.

Appendix 2

Here we recall the syntax and semantics of predicate functor logic.

A language $\mathcal {L}$ of predicate functor logic with identity is characterized by its set of non-logical primitive predicates, each of which has an arity $n\ge 0$. The primitive symbols of such a language $\mathcal {L}$ are, in addition to its non-logical primitive predicates, the functors $\mathsf {Der}$, $\mathsf {Ref}$, $\mathsf {Inv}$, $\mathsf {inv}$, and $\mathsf {Neg}$, all of arity 1, the functor $\times$ of arity 2, and the identity predicate I of arity 2 (as well as left and right parentheses for grouping).

The predicates of $\mathcal {L}$ and their arities are defined inductively as follows. Every n-ary non-logical primitive predicate P of $\mathcal {L}$ is an $\mathcal {L}$-predicate of arity n. The identity predicate I is an $\mathcal {L}$-predicate of arity 2. If $\phi$ is an n-ary $\mathcal {L}$-predicate, $\mathsf {Der}\phi$ is an $(n-1)$-ary $\mathcal {L}$-predicate (except when $n=0$, in which case $\mathsf {Der}\phi$ is also 0-ary), $\mathsf {Ref}\phi$ is an $(n-1)$-ary $\mathcal {L}$-predicate (except when $n=0$, in which case $\mathsf {Ref}\phi$ is also 0-ary), and $\mathsf {Inv}\phi$, $\mathsf {inv}\phi$, as well as $\mathsf {Neg}\phi$, are n-ary $\mathcal {L}$-predicates. If $\phi$ is an n-ary, and $\psi$ an m-ary, $\mathcal {L}$-predicate, $(\phi \times \psi )$ is an $(n+m)$-ary $\mathcal {L}$-predicate. The 0-ary $\mathcal {L}$-predicates are also called $\mathcal {L}$-sentences.

A model ${\mathfrak {M}}$ for $\mathcal {L}$ is a pair $(M, \mathcal {I})$, where M, the domain of ${\mathfrak {M}}$, is a non-empty set and $\mathcal {I}$ is a function assigning to each n-ary primitive predicate P of $\mathcal {L}$ an n-ary function $P^{\mathfrak {M}}$ from $M^n$ to the set $\{ 0,1\}$ of truth values. $M^0$ is the singleton set containing the empty tuple $\langle \rangle$ as a member, so a function f from $M^0$ to $\{0,1\}$ maps $\langle \rangle$ to $f(\langle \rangle )\in \{ 0,1\}$, and we will simply identify such f with $f(\langle \rangle )$.

For each n-ary $\mathcal {L}$-predicate $\phi$, we define the interpretation $\phi ^{\mathfrak {M}}: M^n \rightarrow \{0,1\}$ of $\phi$ in ${\mathfrak {M}}$ recursively as follows: Where P is an n-ary primitive non-logical predicate of $\mathcal {L}$, $P^{\mathfrak {M}}$ is already given by ${\mathfrak {M}}$. The interpretation $I^{\mathfrak {M}}$ of the identity predicate I in ${\mathfrak {M}}$ is true identity, i.e. the binary function on M that maps each pair (a, a) to 1 and every other pair to 0. Where $\phi$ is an $(n+1)$-ary $\mathcal {L}$-predicate, $(\mathsf {Der}\phi )^{\mathfrak {M}}$ is the function that maps any n-tuple $(a_1,\ldots ,a_n)$ to $\max \{ \phi ^{\mathfrak {M}}(a_1,\ldots ,a_n,a)\, |\, a\in M \}$; if $\phi$ is 0-ary, $(\mathsf {Der}\phi )^{\mathfrak {M}}$ is just $\phi ^{\mathfrak {M}}$. Where $\phi$ is an $(n+1)$-ary $\mathcal {L}$-predicate, $(\mathsf {Ref}\phi )^{\mathfrak {M}}$ is the function that maps any n-tuple $(a_1,\ldots ,a_n)$ to $\phi ^{\mathfrak {M}}(a_1,\ldots ,a_n,a_n)$; if $\phi$ is 0-ary, $(\mathsf {Ref}\phi )^{\mathfrak {M}}$ is just $\phi ^{\mathfrak {M}}$. If $\phi$ is an n-ary $\mathcal {L}$-predicate, $(\mathsf {Inv}\phi )^{\mathfrak {M}}$ is the function that maps any n-tuple $(a_1,\ldots ,a_n)$ to $\phi ^{\mathfrak {M}}(a_n,a_1,\ldots ,a_{n-1})$, $(\mathsf {inv}\phi )^{\mathfrak {M}}$ maps $(a_1,\ldots ,a_n)$ to $\phi ^{\mathfrak {M}}(a_1,\ldots ,a_{n-2},a_n,a_{n-1})$, and $(\mathsf {Neg}\phi )^{\mathfrak {M}}$ maps any n-tuple $(a_1,\ldots ,a_n)$ to $1 - \phi ^{\mathfrak {M}}(a_1,\ldots ,a_n)$. Finally, if $\phi$ is an n-ary and $\psi$ an m-ary $\mathcal {L}$-predicate, $(\phi \times \psi )^{\mathfrak {M}}$ maps any $(n+m)$-tuple $(a_1,\ldots ,a_n,b_1,\ldots ,b_m)$ to $\min \{ \phi ^{\mathfrak {M}}(a_1,\ldots ,a_n), \psi ^{\mathfrak {M}}(b_1,\ldots ,b_m)\}$.

If $\phi$ is a sentence of $\mathcal {L}$, we say that $\phi$ is true in ${\mathfrak {M}}$ just in case $\phi ^{\mathfrak {M}}= 1$; otherwise the sentence $\phi$ is false in ${\mathfrak {M}}$.