Identity and Aboutness

Abstract

This paper develops a theory of propositional identity which distinguishes necessarily equivalent propositions that differ in subject-matter. Rather than forming a Boolean lattice as in extensional and intensional semantic theories, the space of propositions forms a non-interlaced bilattice. After motivating a departure from tradition by way of a number of plausible principles for subject-matter, I will provide a Finean state semantics for a novel theory of propositions, presenting arguments against the convexity and nonvacuity constraints which Fine (Journal of Philosophical Logic, 4545, 199–226 13, 14, 15) introduces. I will then move to compare the resulting logic of propositional identity (PI¹) with Correia’s (The Review of Symbolic Logic, 9, 103–122 9) logic of generalised identity (GI), as well as the first degree fragment of Angell’s (2) logic of analytic containment (AC). The paper concludes by extending PI¹ to include axioms and rules for a subject-matter operator, providing a much broader theory of subject-matter than the principles with which I will begin.

Appendix

I will begin by considering propositional languages of the form ${\mathscr{L}}=\angle {\mathbb {L},\vec {\mathcal {Q}}}\rangle$ where each $\mathcal {Q}_{i}^{n}\in \vec {\mathcal {Q}}$ is an n-ary sentential operator for some $n\in \mathbb {N}$, and $\vec {\mathcal {Q}}$ includes the extensional connectives ‘¬’, ‘∧’, and ‘∨’ along with the propositional identity operator ‘≡’. The well-formed sentences in $\texttt {wfs}(\mathbb {L})$ consist of the sentence letters in $\mathbb {L}$ along with any $\mathcal {Q}_{i}^{n}(\vec {O})$, where $\mathcal {Q}_{i}^{n}\in \vec {\mathcal {Q}}$ and $\vec {O}$ is a sequence of n well-formed sentences. Assuming a background classical logic in which $\ulcorner {A\rightarrow B}\urcorner$ abbreviates $\ulcorner {\neg A\vee B}\urcorner$, we may consider the following:

$$ \begin{array}{llrl} \textbf{Ref}\quad \quad A\equiv A.&\textbf{Trans}\quad (A\equiv B)\rightarrow[(B\equiv C)\rightarrow(A\equiv C)].\\ \textbf{Sym}\quad (A\equiv B)\rightarrow(B\equiv A).&\textbf{Imps}\quad (A\equiv B)\rightarrow(A\rightarrow B).\\ \textbf{LL}\quad (A\equiv B)\rightarrow(C\rightarrow C_{(B/A)}).&\textbf{Func}\quad (A\equiv B)\rightarrow[\mathcal{Q}(\vec{O})\equiv\mathcal{Q}(\vec{O}_{(B/A)})]. \end{array} $$

As above, ${\mathscr{L}}$ is transparent just in case all instances of Func hold, where $\ulcorner {\mathcal {Q}(\vec {O}_{(B/A)})}\urcorner$ is the result of replacing one or more instances of $\ulcorner {A}\urcorner$ which occur as members of the sequence $\ulcorner {\vec {O}}\urcorner$ with $\ulcorner {B}\urcorner$, where similarly $\ulcorner {\mathcal {Q}(\vec {O}_{[B/A]})}\urcorner$ is the result of replacing all instances of $\ulcorner {A}\urcorner$ which occur as members of the sequence $\ulcorner {\vec {O}}\urcorner$ with $\ulcorner {B}\urcorner$. Additionally, I will take $\ulcorner {C_{(B/A)}}\urcorner$ to be the result of replacing one or more instances of $\ulcorner {A}\urcorner$ as it occurs anywhere in $\ulcorner {C}\urcorner$ with $\ulcorner {B}\urcorner$, as well as taking $\ulcorner {C_{[B/A]}}\urcorner$ to be the result of replacing all instances of $\ulcorner {A}\urcorner$ as it occurs anywhere in $\ulcorner {C}\urcorner$ with $\ulcorner {B}\urcorner$. We may now prove the following propositions.

P1 If ${\mathscr{L}}$ is transparent, then Ref and Imps entail both Sym and Trans.

Proof

Assuming that ${\mathscr{L}}$ is transparent, $(A\equiv B)\rightarrow [(A\equiv A)\equiv (B\equiv A)]$ follows, where $[(A\equiv A)\equiv (B\equiv A)]\rightarrow [(A\equiv A)\rightarrow (B\equiv A)]$ holds by Imps, and so $(A\equiv B)\rightarrow (B\equiv A)$ follows from Ref by propositional logic. Again by transparency, $(A\equiv B)\rightarrow ([(A\equiv C)\rightarrow (A\equiv C)]\equiv [(B\equiv C)\rightarrow (A\equiv C)])$, and so $(A\equiv B)\rightarrow ([(A\equiv C)\rightarrow (A\equiv C)]\rightarrow [(B\equiv C)\rightarrow (A\equiv C)])$ by Imps. Since $(A\equiv C)\rightarrow (A\equiv C)$ holds by propositional logic, we may conclude as desired that $(A\equiv B)\rightarrow [(B\equiv C)\rightarrow (A\equiv C)]$. □

P1 If ${\mathscr{L}}$ is transparent, then Ref and Imps entail $(A\equiv B)\rightarrow (C\equiv C_{[B/A]})$.

Proof

Assuming that ${\mathscr{L}}$ is transparent, the proof proceeds by induction on the complexity of $C\in \texttt {wfs}(\mathbb {L})$. Of course, if $C\in \mathbb {L}$, then $(A\equiv B)\rightarrow (C\equiv C_{[B/A]})$ holds by propositional logic if A occurs in C, and C ≡ C_[B/A] holds by Ref otherwise, where $(A\equiv B)\rightarrow (C\equiv C_{[B/A]})$ follows by propositional logic.

Assume for induction that $(A\equiv B)\rightarrow (C\equiv C_{[B/A]})$ holds whenever comp(C) ≤ n, and let comp(C) = n + 1. Assuming that ${C}={\mathcal {Q}^{n}(\vec {D})}$, we may observe that for all 1 ≤ i ≤ n that $(A\equiv B)\rightarrow (D_{i}\equiv D_{i[B/A]})$ follows by hypothesis, where $(D_{i}\equiv D_{i[B/A]})\rightarrow [\mathcal {Q}^{n}(\vec {E})\equiv \mathcal {Q}^{n}(\vec {E}_{[D_{i[B/A]}/D_{i}]})]$ by the transparency of ${\mathscr{L}}$ for any $\vec {E}$. Assuming A ≡ B, it follows for all 1 ≤ m ≤ n that $\mathcal {Q}^{n}(\vec {D}_{[D_{1[B/A]}/D_{1}]\ldots [D_{m[B/A]}/D_{m}]})\equiv \mathcal {Q}^{n}(\vec {D}_{[D_{1[B/A]}/D_{1}]\ldots [D_{m+1[B/A]}/D_{m+1}]})$. Given that Ref and Imps, it follows by P1 that Trans holds, and so by n − 1 applications of Trans, $\mathcal {Q}^{n}(\vec {D})\equiv \mathcal {Q}^{n}(\vec {D}_{[D_{1[B/A]}/D_{1}]\ldots [D_{n[B/A]}/D_{n}]})$. We may then observe that ${\mathcal {Q}^{n}(\vec {D}_{[D_{1[B/A]}/D_{1}]\ldots [D_{n[B/A]}/D_{n}]}}={\mathcal {Q}^{n}(\vec {D})_{[B/A]}}$, and so it follows by discharging our assumption that $(A\equiv B)\rightarrow (\mathcal {Q}^{n}(\vec {D})\equiv \mathcal {Q}^{n}(\vec {D})_{[B/A]})$. Since ${C}={\mathcal {Q}^{n}(\vec {D})}$, we may conclude that $(A\equiv B)\rightarrow (C\equiv C_{[B/A]})$. □

P1 Assuming Ref and Imps, then ${\mathscr{L}}$ is transparent just in case LL holds.

Proof

Assume Ref and Imps. Letting LL hold in ${\mathscr{L}}$ where $\mathcal {Q}$ is an operator in ${\mathscr{L}}$, it follows that $(A\equiv B)\rightarrow ([\mathcal {Q}(\vec {O})\equiv \mathcal {Q}(\vec {O})]\rightarrow [\mathcal {Q}(\vec {O})\equiv \mathcal {Q}(\vec {O})_{(B/A)}])$. However, given Ref, $\mathcal {Q}(\vec {O})\equiv \mathcal {Q}(\vec {O})$, and so $(A\equiv B)\rightarrow [\mathcal {Q}(\vec {O})\equiv \mathcal {Q}(\vec {O})_{(B/A)}]$. In particular, $(A\equiv B)\rightarrow [\mathcal {Q}(\vec {O})\equiv \mathcal {Q}(\vec {O}_{(B/A)})]$ as in Func. Generalising on $A,B,\vec {O},$ and $\mathcal {Q}$, we may conclude that ${\mathscr{L}}$ is transparent.

Assume instead that ${\mathscr{L}}$ is transparent. Letting $p\in \mathbb {L}$ where p does not occur in B or C, it follows by L1 that $(p\equiv B)\rightarrow (C_{(p/A)}\equiv C_{(p/A)[B/p]})$. However, C_(p/A)[B/p] = C_(B/A), and so $(p\equiv B)\rightarrow (C_{(p/A)}\equiv C_{(B/A)})$. Thus it follows that, $(A\equiv B)\rightarrow (C\equiv C_{(B/A)})$ and so LL follows by Imps. □