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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!

user52534
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  • You probably meant recursively axiomatizable. Also, $\Rightarrow$ is usually used to denote implication in metalanguage, not the formal language (as opposed to $\rightarrow$). Which part of the hypothesis of $(\star\star)$ does $G$ not satisfy??? – tomasz Apr 04 '13 at 18:18
  • sorry about that, I think the parentheses will clarify. $G$ does not necessarily contain all of $PA$. Are you saying that $(\star\star)$ is true without the condition that $T\supset PA$? – user52534 Apr 04 '13 at 18:29

1 Answers1

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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.

Lawrence Wong
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