A theory T is scapegoat if for every formula A with only one free variable there exist a closed term s such that T proves: (∃x(¬A(x)))⇒¬A(s) I think it is an expectable property for each theory since it says that if there is a x which A(x) does not hold for it the we can substitute a closed term instead of x . And we can expect for an aribtrary theory this statement can be proved in that theory. Please explain intuitively if I am wrong and what is my mistake. My second question is showing that PA is not scapegoat theory. I read "Scapegoat theory!" question but I cant understand it and its relation to Godel's incompleteness theorem.
Scapegoat theory and PA
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Theories with "built-in Skolem functions" have a stronger property: for any formula $\phi(v,w_1,\ldots,w_m)$ (together with the partition of variables), there is a function symbol $f$ such that $$T\vDash \forall v,w_1,\ldots,w_m [\phi(v,w_1,\ldots,w_m)\rightarrow \phi(f(w_1,\ldots,w_m),w_1,\ldots,w_m)]$$
It's easy to show that for any theory $T$ in a language $\mathcal{L}$, there is a Skolemization: a theory $T^*$ in a language $\mathcal{L}^*$ such that $T^*$ has built in Skolem functions and is conservative over $T$ (if $T^*\vDash\phi$ and $\phi$ is a formula in $\mathcal{L}$ then $T\vDash\phi$).
In that sense, scapegoat (and stronger) theories are common, and any sufficiently expressive language is one. In particular, the obstacle to PA having these properties isn't really related to Godel's theorem; it's because the underlying language has very terms.
Not every theory is Scapegoat, its actually very unreasonable. Take PA, there is the Godel sentence $G$ now $$G\equiv \neg \exists n (n \text{is a proof of} G)$$
We cannot prove $G$ in PA. If PA were a scapegoat theory there would be some fixed number $k$, say such that $$PA \vdash \exists n (n \text{is a proof of} G) \rightarrow (k \text{is a proof of} G)$$ or briefly, $$PA \vdash\neg G \rightarrow (k \text{is a proof of} G)$$ Now we can check to see if $k$ is a proof of $G$, it wont be cause we cant prove it. So we have $$PA \vdash\neg G \rightarrow \text{a falsehood}$$
therefore, $PA\vdash G$, a contradiction.