A question about consistent axiomatizable extensions of PA Given $T\supset PA$ to be consistent and  axiomatizable, I've been told that when $G\subset T$   is finite, and $\phi$  is a universal sentence, then:
($\star$)   $PA\vdash ((Pr_G(\underline\phi)\wedge con_G)\implies \phi) $
I can see that $PA\vdash ((Pr_T(\underline\phi)\wedge con_T)\implies \phi) $ is true for universal $\phi$ which follows easily from the fact that:
($\star\star$) Given $T\supset PA$ to be consistent and  axiomatizable, then  $PA\vdash(\phi\implies Pr_T(\underline \phi ))$  for every existential sentence $\phi$
But I don't see why ($\star$) is true (if it even is true), since $G$
 doesn't satisfy the hypothesis of ($\star\star$).
Any help is appreciated 
Thanks!
 A: A good reference on this topic is Richard Kaye's book Models of Peano Arithmetic.
Take any finite (or even recursive) $G\subseteq T$.  Without loss of generality, suppose $G\supseteq\mathrm{PA}^-$.  (See Chapter 2 in Kaye's book for the definition of $\mathrm{PA}^-$.)  Let $\phi$ be a universal sentence and $M\models\mathrm{PA}$ such that $M\models\mathrm{Pr}_G(\phi)+\mathrm{Con}_G$.  By the Arithmetized Completeness Theorem (which is a formalized version of the Completeness Theorem in $\mathrm{PA}$, see Section 13.2 of Kaye's book), the model $M$ knows there is a model $K\models G$.
As $G\supseteq\mathrm{PA}^-$, we can consider $K$ as an extension of $M$.  To see this, observe that the $M$-version of the language of arithmetic contains a closed term $\underline n$ for every $n\in M$, and $\mathrm{PA}^-$ is strong enough to show that $\underline m\not=\underline n$ for all distinct $m,n\in M$.
Now, since $M\models\mathrm{Pr}_G(\phi)$ and $M$ thinks $K\models G$, we know (and $M$ knows) $K\models\phi$.  Thus $M\models\phi$ because $\phi$ is universal, and universal formulas are preserved downwards.
