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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.

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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.

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  • $\begingroup$ thanks! I thought of this, but what disturbs me: somehow, here, $=$ gets shifted to $\leftrightarrow$. This holds in a general way if one relativize via the given procedure? $\endgroup$ – aphorisme Jul 23 '14 at 18:41
  • $\begingroup$ Just think about this in general, $=$ is an equivalence relation. If $\varphi(x,y,z)$ is the formula $x+y=z$ then $x_1+x_2=x_2+x_1$ is translated to $\leftrightarrow$ quite immediately. Within the same language. $\endgroup$ – Asaf Karagila Jul 23 '14 at 18:50
  • $\begingroup$ ah... I see! Thanks again! $\endgroup$ – aphorisme Jul 23 '14 at 19:16
  • $\begingroup$ Again a question: The point here might be, that $=$ in $\mathcal{L}$ is somehow something different to the $=$ in $\{\in\}$, because it does compare different "things", or? $\endgroup$ – aphorisme Jul 23 '14 at 19:26
  • $\begingroup$ No, $=$ is part of the underlying logic. $\endgroup$ – Asaf Karagila Jul 23 '14 at 19:29

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