Representational Impurities

Given an interpreted language $(L,I)$, of roughly the kind used in teaching predicate logic, there may be syntactically distinct strings which express the same proposition. For example,

If (John plays guitar and Paul drums) then Ringo whinges.
If [John plays guitar and Paul drums] then Ringo whinges.

These are syntactically distinct, because they are distinct strings of symbols. This has something to do with the use of brackets. In a sense, the brackets are a kind of "representational impurity". And, of course, brackets can be eliminated, at the cost of introducing Polish notation (for connectives and predicates). In a sense, Polish notation "quotients out" a certain kind of representational impurity.

Also, the following syntactically distinct string express the same proposition,

There is someone $x$ such that, for all $y$, $x$ does not like $y$.
There is someone $y$ such that, for all $x$, $y$ does not like $x$.

These are again syntactically distinct, because they are distinct strings of symbols. But they are equivalent in the sense that they are the result of applying a certain kind of substitution based on a permutation $\pi: Val(L) \rightarrow Var(L)$ of variables of the language $L$. This is sometimes called relabelling of variables, and explains why, for example,

$\int_0^1 x^2 dy = \int_0^1 y^2 dy$

The equivalences can be proved in predicate logic. So, the choice of variables is another kind of representational impurity. In fact variables can be eliminated, at the cost of introducing combinatory logic (what Quine calls predicate-functor logic). This is another example of "quotienting out" certain representational impurity.

So, brackets and choice of variables are examples of representational impurities.
And the "quotienting out" of these representational impurities in syntactic strings, by either going Polish, or going to predicate-functors, is not controversial. They are quite clear-cut cases of convention. (The "quotiented-out" languages are of course, very inconvenient! The resulting formulas are almost unintelligible to ordinary cognition.)

What is the underlying idea? First, one has an interpreted language $(L,I)$, and the interpretation function $I$ for $L$ generates certain semantic equivalence relations on the strings of $L$. In particular, a synonymy relation, which I'll write $\phi \equiv_{(L,I)} \theta$. To quotient out, one identifies:

i. a new interpreted language $(L^{\prime}, I^{\prime})$,
ii. a translation $^{\circ}: L \rightarrow L^{\prime}$, such that,

(a) if $\phi \equiv_{(L,I)} \theta$, then $\phi^{\circ} = \theta^{\circ}$
(b) $I \models \phi$ iff $I^{\prime} \models \phi^{\circ}$

So, equivalent but distinct $L$-strings are mapped to a single string in $L^{\prime}$.
For example, $\exists x \phi(x)$ and $\exists y \phi(y)$ in the ordinary quantificational language $L$ with variables will be mapped to the same syntactic string in the predicate-functor language $L^{\prime}$ without variables.

One might wonder if this process of eliminating representational impurities reaches a limit. This would mean that all synonymies have been eliminated. I suspect that there isn't a limit in the syntactic sense. And I also suspect that the standard criticisms of the correspondence theory as "mirroring" (a criticism one finds in the writings of many idealists and pragmatists) is really an expression of this conclusion: that syntactic representations are always "impure". These "impurities" are solely contributions of the representer, and are not there in unadorned reality.

Fortunately, such criticisms have little effect after Tarski (1935), "Der Wahrheitsbegriff". Tarski showed how truth is to be defined without assuming a mirroring relation. As we quotient out impurities, the translation and the (Tarskian) truth definition guarantee that required equivalences hold. That is, if $\phi$ and $\theta$ are syntactically distinct but equivalent in $(L,I)$, then each element of the equivalence class $[\phi]$ of equivalent $L$-strings has the same truth conditions as its image $\phi^{\circ}$ in $L^{\prime}$.