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Let's take PA1( First order axioms of peano arithmetic ) for example. From godel's 2nd incompleteness theorem, PA1 can't prove its own consistency, more specifically it can't prove that the largest consistent subset of the theory is PA1 itself.

But PA1 has infinite axioms, so can PA1 prove atleast for a given finite set of axioms ( of PA1 ) that they are consistent, specifically that no contradictional proof exists which uses only those axioms ?

Or any formal theory of arithmetic for that matter, can it prove for a finite subset of its own axioms that they are consistent ?

Edit - What I know is that PA1 can prove that any finite subset of it is consistent, but I want to know if it can prove that a finite set of its axioms itself is consistent ? I believe PA1 or any other formal theory of arithmetic should be able to prove that for any given finite subset of its axioms its consistent.

jason
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note: As currently stated, this question is dangerously close to being a duplicate. While people deliberate whether it really counts as one, I'll try to at least give an answer that is not a duplicate of any existing one: instead of telling you how this is proven, I'll tell you a bit about the historical context of the result.

For any fixed finite set $\mathcal{F}$ of axioms of Peano arithmetic, one can obtain a proof of $\mathrm{Con}(\mathcal{F})$ inside Peano arithmetic. But the analogous proposition absolutely does not hold for every formal theory of arithmetic.

Mostowski studied this question in his 1952 article On models of axiomatic systems (Mostowski's article). His results immediately imply that Peano arithmetic is not finitely axiomatizable. Kreisel and Lévy put the result in the context of reflection principles in their 1968 article, and provide proof-theoretic arguments showing the same (and a lot more).

You should always keep in mind that in these results, the finite set $\mathcal{F}$ must be fixed externally: we have "for any finite set of axioms of PA, we can find a PA-proof...", and most definitely not "we can find a PA-proof that for any finite set of axioms of PA..."! If we had the latter, we could use the fact that any proof uses only finitely many axioms to conclude the consistency of PA inside of PA itself, which would spell doom for PA via Gödel's incompleteness theorem.

This sort of reflection is a special property enjoyed by PA, and not by every formal theory of arithmetic: there are finitely axiomatizable theories of formal arithmetic that are as strong as PA, and those will certainly not prove all finite sets of their own axioms consistent (just like there are finitely axiomatizable set theories that are as strong as ZFC).

Z. A. K.
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