Yeah, the details of actually proving this are looking like they contain more subtleties than I expected, but I tentatively agree with your analysis. Here’s what may be another proof,. Not only is PA not finitely axiomatizable, but any consistent extension of PA isn’t (I think this is true, the same proof that works for PA should go through here, but I haven’t checked the details). So PA+ConMinus(PA) still isn’t finitely axiomatizable. So now, pick any of the axioms created in the axiom schema of induction that are needed in PA + ConMinus(PA), Q+ConMinus(PA) also can’t prove any of those (since it is strictly weaker than PA+ConMinus(PA),) but all of those statements are theorems of PA (since they are in fact axioms). Does this work?
Overall, this is requiring a lot more subtlety than I initially thought was involved which may make Qiaochu Yuan’s example a better one.
Not only is PA not finitely axiomatizable, but any consistent extension of PA isn’t (I think this is true, the same proof that works for PA should go through here, but I haven’t checked the details)
Well if we had this, we would know immediately that Q + Con(PA) is not an extension of PA (which is what we originally wanted), because it certainly is finitely axiomatizable. I know there are several proofs that PA is not finitely axiomatizable, but I have not read any of them, so can’t comment on the strengthened statement, though it sounds true.
Yeah, the details of actually proving this are looking like they contain more subtleties than I expected, but I tentatively agree with your analysis. Here’s what may be another proof,. Not only is PA not finitely axiomatizable, but any consistent extension of PA isn’t (I think this is true, the same proof that works for PA should go through here, but I haven’t checked the details). So PA+ConMinus(PA) still isn’t finitely axiomatizable. So now, pick any of the axioms created in the axiom schema of induction that are needed in PA + ConMinus(PA), Q+ConMinus(PA) also can’t prove any of those (since it is strictly weaker than PA+ConMinus(PA),) but all of those statements are theorems of PA (since they are in fact axioms). Does this work?
Overall, this is requiring a lot more subtlety than I initially thought was involved which may make Qiaochu Yuan’s example a better one.
Well if we had this, we would know immediately that Q + Con(PA) is not an extension of PA (which is what we originally wanted), because it certainly is finitely axiomatizable. I know there are several proofs that PA is not finitely axiomatizable, but I have not read any of them, so can’t comment on the strengthened statement, though it sounds true.