at the moment I try to figure out some details of Kunen's "Relative Interpretation" Definition (within the 2013 Edition of his "Set Theory", p. 99 to 100):
Definition If $\Lambda$ is some axioms in [the language] $\{\in\}$ for set theory, and $\mathcal{L}$ is a finite lexicon, a relative interpretation of $\mathcal{L}$ in $\Lambda$ consits of a non-empty class $A$ together with an assignment of a suitable semantic entity $s^\mathfrak{A}$ to each symbol of $\mathcal{L}$. Specifically,
If f is an $n$-ary function symbol [in $\mathcal{L}$] with $n > 0$, then $f^\mathfrak{A} : A^n \rightarrow A$ [...]
He mentions under this definition that in fact one does not assign to $f$ a function $f^\mathfrak{A}$ within $\{\in\}$ (since $A$ might be a proper class) but a formula $\phi(x_1, ... , x_n, y)$ such that \begin{align*} \Lambda \vdash \forall x_1, ..., x_n \in A . \exists ! y \in A . \phi(x_1, ... x_n, y) \end{align*}
... and so on for constants, etc. But how would one relativize \begin{align*} f(c) = c \end{align*}
as an example? Since $\phi_f(x,c) = \phi(c)$ does not make sense at all?
I took his example literally: let $\mathcal{L} = (0, +, \cdot)$ then I might define \begin{align*} \phi_0(y) & := \forall x\in A. x \notin y \\ \phi_{+}(x_1, x_2, y) & := y \in x_1 \vee y \in x_2 \wedge \neg (y \in x_1 \wedge y \in x_2)\\ \phi_{\cdot}(x_1, x_2, y) & := y \in x_1 \wedge y \in x_2 \end{align*}
but how to relativize $x_1 + x_2 = x_2 + x_1$ then?
Thanks!
PS: I'm aware of the more pedantic definition in Jech's Set Theory, but there it's only defined for $\mathcal{L} = \{\in\}$ and so it is in all the relevant questions on math.stack I have found so far.
In your example, $x_1+x_2=x_2+x_1$ gets relativized to be: $$\forall x_1\forall x_2\forall y\Bigl(\phi_+(x_1,x_2,y)\leftrightarrow\phi_+(x_2,x_1,y)\Bigr)$$
In general, replacing symbols by formulas only complicates things in terms of longer formulas, with more quantifiers and with longer proofs (well, depending on your inference rules, I guess).
It's a pain in the lower lower-back, but once you understand how it's done, it's not hard to see why this is true. And to understand how it's done, just look at the example above.