Semidisquotation and the infinitary function of truth

Abstract

The infinitary function of the truth predicate consists in its ability to express infinite conjunctions and disjunctions. A transparency principle for truth states the equivalence between a sentence and its truth predication; it requires an introduction principle—which allows the inference from “snow is white” to “the sentence ‘snow is white’ is true”—and an elimination principle—which allows the inference from “the sentence ‘snow is white’ is true” to “snow is white”. It is commonly assumed that a theory of truth needs to satisfy a transparency principle to fulfil the infinitary function. Picollo and Schindler (Erkenntnis 83:899–928, 2017) argue against this idea. They prove that, given certain assumptions, an elimination principle is sufficient for the purpose. Then, they pose a challenge: to show why we should incorporate introduction principles to our theory of truth. In this essay I take on the challenge. I show that, given the authors’ assumptions, an introduction principle is also sufficient to perform the infinitary function.

Appendix

Recall that ${\mathcal {L}}$ contains a term $\ulcorner \sigma \urcorner$ denoting $\sigma$ (perhaps via some coding) for each expression $\sigma$ of ${\mathcal {L}}_T$. Suppose our base theory $\Sigma \subseteq {\mathcal {L}}$ is classical. Let $\phi (y)$ be a predicate of sentences of ${\mathcal {L}}$.

Proof of Proposition 10

Define $\Sigma '$ as $\Sigma +\{\phi (\ulcorner \psi \urcorner )\rightarrow \psi :\psi \in {\mathcal {L}}\}$ and define $\Gamma '$ as $\Gamma +\forall x(\phi (x)\rightarrow Tx)$. We show that $\Sigma '$ and $\Gamma '$ have the same T-free consequences.

Let $\chi \in {\mathcal {L}}$. Assume that $\Sigma '\vdash \chi$. By the argument from Sect. 2.1 and the fact that $\Sigma$ is classical, $\Gamma '\vdash \chi$. For the converse, we will use the following lemma:

Lemma 12

Any model ${\mathcal {M}}$ of $\Sigma '$ can be expanded to a model ${\mathcal {M}}^+$ of $\,\Gamma '$.

Proof

If $\Sigma '$ has no models, the Lemma holds trivially. So, let ${\mathcal {M}}=\langle D,I \rangle$ be a model where D is the domain and I the interpretation function, and suppose that ${\mathcal {M}}\models \Sigma '$. Define ${\mathcal {M}}^+=\langle D,I^+ \rangle$ to be the expansion of ${\mathcal {M}}$ s.t. $I^+(T)=\{I(\ulcorner \psi \urcorner ):{\mathcal {M}}\models \phi (\ulcorner \psi \urcorner )\}$. We prove that ${\mathcal {M}}^+\vDash \Gamma '$.

• (a) Let $\delta \in {\mathcal {L}}_T$. Assume ${\mathcal {M}}^+\models T\ulcorner \delta \urcorner$. Then, $I(\ulcorner \delta \urcorner )\in \{I(\ulcorner \psi \urcorner ): {\mathcal {M}}\models \phi (\ulcorner \psi \urcorner )\}$, and hence ${\mathcal {M}}\models \phi (\ulcorner \delta \urcorner )$. Besides, ${\mathcal {M}}\models \phi (\ulcorner \delta \urcorner )\rightarrow \delta$, so ${\mathcal {M}}\models \delta$. ${\mathcal {M}}^+$ expands ${\mathcal {M}}$, so ${\mathcal {M}}^+\models \delta$. Therefore, ${\mathcal {M}}^+\models T\ulcorner \delta \urcorner \rightarrow \delta$.

• (b) Let s be any map from the variables in ${\mathcal {L}}_T$ to D. Let $s'$ differ from s at most at variable x. Assume ${\mathcal {M}}^+,s'\models \phi (x)$. $\phi (y)$ is a predicate of sentences of ${\mathcal {L}}$ and ${\mathcal {M}}^+$ expands ${\mathcal {M}}$. Thus, $s'(x)\in \{I(\ulcorner \psi \urcorner ):{\mathcal {M}}\models \phi (\ulcorner \psi \urcorner )\}$. Hence, ${\mathcal {M}}^+,s'\models Tx$. Therefore, ${\mathcal {M}}^+,s'\models \phi (x)\rightarrow Tx$. Map $s'$ was arbitrary. Thus, ${\mathcal {M}}^+,s\models \forall x(\phi (x)\rightarrow Tx)$. Besides, $\forall x(\phi (x)\rightarrow Tx)$ is a sentence, so ${\mathcal {M}}^+\models \forall x(\phi (x)\rightarrow Tx)$.

$\square$

Assume that $\Gamma '\vdash \chi$. Suppose, for reductio, that $\Sigma '\nvdash \chi$. Then, there is a model ${\mathcal {M}}$ of $\Sigma '$ s.t. ${\mathcal {M}}\not \models \chi$. By our Lemma, there is an expansion ${\mathcal {M}}^+$ s.t. ${\mathcal {M}}^+\models \Sigma '\cup \Gamma '$. ${\mathcal {M}}^+\not \models \chi$, because ${\mathcal {M}}^+$ expands ${\mathcal {M}}$ and $\chi \in {\mathcal {L}}$. But ${\mathcal {M}}^+\models \chi$, because ${\mathcal {M}}^+\models \Gamma '$. A contradiction. Hence, $\Sigma '\vdash \chi$. $\square$

Proof of Proposition 11

Define $\Sigma '$ as $\Sigma +\{\phi (\ulcorner \psi \urcorner )\rightarrow \psi :\psi \in {\mathcal {L}}\}$ and define $\Gamma '$ as . We show that $\Sigma '$ and $\Gamma '$ have the same T-free consequences.

Let $\chi \in {\mathcal {L}}$. Assume that $\Sigma '\vdash \chi$. By the argument from Sect. 3.1 and the fact that $\Sigma$ is classical, $\Gamma '\vdash \chi$. For the converse, we will use the following lemma:

Lemma 13

Any model ${\mathcal {M}}$ of $\Sigma '$ can be expanded to a model ${\mathcal {M}}^+$ of $\,\Gamma '$.

Proof

Again, if $\Sigma '$ has no models, the Lemma holds trivially. So let ${\mathcal {M}}=\langle D,I \rangle$ be a model and suppose that ${\mathcal {M}}\models \Sigma '$. Define ${\mathcal {M}}^+=\langle D,I^+ \rangle$ to be the expansion of ${\mathcal {M}}$ s.t. $I^+(T)=\{I(\ulcorner \psi \urcorner ):\psi \in {\mathcal {L}}_T\} \setminus \{I(\ulcorner \lnot \psi \urcorner ):{\mathcal {M}}\models \phi (\ulcorner \psi \urcorner )\}$. We prove that ${\mathcal {M}}^+\vDash \Gamma '$.

• (a) Let $\delta \in {\mathcal {L}}_T$. Assume ${\mathcal {M}}^+\models \delta$. Then, $I(\ulcorner \delta \urcorner )\notin \{I(\ulcorner \lnot \psi \urcorner ):{\mathcal {M}}\models \phi (\ulcorner \psi \urcorner )\}$. (Assume the opposite. Then, there is a $\xi \in {\mathcal {L}}$ s.t. $\delta$ is identical to $\lnot \xi$ and ${\mathcal {M}}\models \phi (\ulcorner \xi \urcorner )$. ${\mathcal {M}}\models \phi (\ulcorner \xi \urcorner )\rightarrow \xi$, so ${\mathcal {M}}\models \xi$. ${\mathcal {M}}^+$ expands ${\mathcal {M}}$, so ${\mathcal {M}}^+\models \xi$. Thus, ${\mathcal {M}}^+\not \models \delta$. A contradiction.) Hence, ${\mathcal {M}}^+\models T\ulcorner \delta \urcorner$. Therefore, ${\mathcal {M}}^+\models \delta \rightarrow T\ulcorner \delta \urcorner$.

• (b) Let s be any map from the variables in ${\mathcal {L}}_T$ to D. Let $s'$ differ from s at most at variable x. Assume ${\mathcal {M}}^+,s'\models \phi (x)$. $\phi (y)$ is a predicate of sentences of ${\mathcal {L}}$ and ${\mathcal {M}}^+$ expands ${\mathcal {M}}$. Thus, . Hence, . Therefore, and . Map $s'$ was arbitrary. Thus, . Besides, is a sentence, so

$\square$

The proof continues as the analogous part of the proof of Proposition 10, but using Lemma 13 instead of Lemma 12. $\square$