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 $\phi$ is a sentence, then

$\phi$ is true in $\mathbf{L}$ iff $\mathcal{A} \models \phi$.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 $\mathbf{L}^{\pi}$.

*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 $\phi$ true; they are

*isomorphic.*

## No comments:

## Post a Comment