Quine Transform of a Language

This is a first application of this notion (and in fact where the name comes from!).

Let $\mathbf{L}=(\mathcal{L},\mathcal{A})$ be an interpreted language, such as  might be spoken/cognized by some agent $s$. Here $\mathcal{L}$ is the underlying (uninterpreted) syntax, and $\mathcal{A}$ is an extensional interpretation for $\mathcal{L}$-strings. So, $\mathcal{A}$ specifies, in the usual way, extensional meanings for $\mathbf{L}$'s syntactic components: connectives, quantifiers, names, predicates, etc. For example, if $t$ is a closed term, then its denotation in $\mathbf{L}$ is $t^{\mathcal{A}}$. If $\varphi$ is a sentence, then

$\varphi$ is true in $\mathbf{L}$ iff $\mathcal{A}\models \varphi$.

Let $\pi :A\to A$ be any permutation of $A$. Let ${\mathcal{A}}^{\pi }$ be the Quine transform of $\mathcal{A}$ under $\pi$.

Definition [Quine Transform of a Language]
The Quine transform of the language $\mathbf{L}$, written ${\mathbf{L}}^{\pi }$, is defined to be $(\mathcal{L},{\mathcal{A}}^{\pi })$.

The reason for being interested in this notion is that Quine argued (as I formulate it) that there cannot be a physical "fact of the matter" (by which Quine intends to include all "use-facts" or U-facts) discriminating between:

• agent $s$ cognizes/speaks $\mathbf{L}$.
• agent $s$ cognizes/speaks ${\mathbf{L}}^{\pi }$.

That is, according to Quine, it is indeterminate which language the agent $s$ speaks/cognizes. Quine's reasoning for this is a matter of dispute, of course. But note that the inteprertations $\mathcal{A}$ and ${\mathcal{A}}^{\pi }$ are not merely equivalent, in the technical sense, in making the same sentences $\varphi$ true; they are isomorphic.