4 deleted 9 characters in body

Let Con(PA) be the sentence of arithmetic which translates as "Peano Arithmetic is consistent." Then according to Godel's 2nd incompleteness theorem, assuming PA is consistent then PA can neither prove Con(PA) nor its negation. And in fact, if T contains PA and T is (omega-) consistent, then T can neither prove Con(T) nor its negation. In particular, if PA+Con(PA) is consistent then PA+Con(PA) can neither prove Con(PA+Con(PA)) nor its negation.

But consider the following piece of reasoning: if PA is consistent, then Con(PA) is true, so PA+Con(PA) is consistent, so Con(PA+Con(PA)) is true. My question is, why can't this reasoning by formalized in PA, so that within PA you can prove Con(PA) implies Con(PA+Con(PA))? If you could prove that, then since you can obviously prove Con(PA) within PA+Con(PA), you would be able to prove Con(PA+Con(PA)) within PA+Con(PA), which is a contradiction. So where am I going wrong?

We can even talk about this in terms of model theory. There are nonstandard models of PA in which Con(PA) does not hold: basically, you have infinitely large natural numbers, and you have infinitely long proofs of a contradiction in PA. This does not mean that PA is inconsistent, because there are no proofs of finite length of this contradiction. So are there also nonstandard models of PA+Con(PA) in which Con(PA+Con(PA)) does not hold? (That's a rhetorical question; clearly there must be, but what do they look like?)

Any help would be greatly appreciated.

3 fixed a typo

Let Con(PA) be the sentence of arithmetic which translates as "Peano Arithmetic is consistent." Then according to Godel's 2nd incompleteness theorem, assuming PA is consistent then PA can neither prove Con(PA) nor its negation. And in fact, if T contains PA and T is (omega-) consistent, then T can neither prove Con(T) nor its negation. In particular, if PA+Con(PA) is consistent then PA+Con(PA) can neither prove Con(PA+Con(PA)) nor its negation.

But consider the following piece of reasoning: if PA is consistent, then Con(PA) is true, so PA+Con(PA) is consistent, so Con(PA+Con(PA)) is true. My question is, why can't this reasoning by formalized in PA, so that within PA you can prove Con(PA) implies Con(PA+Con(PA))? If you could prove that, then since you can obviously prove Con(PA) within PA+Con(PA), you would be able to prove Con(PA+Con(PA)) within PA+Con(PA), which is a contradiction. So where am I going wrong?

We can even talk about this in terms of model theory: . There are nonstandard models of PA in which Con(PA) does not hold: basically, you have infinitely large natural numbers, and you have infinitely long proofs of a contradiction in PA. This does not mean that PA is inconsistent, because there are no proofs of finite length of this contradiction. So are there also nonstandard models of PA+Con(PA) in which Con(PA+Con(PA)) does not hold? (That's a rhetorical question; clearly there must be, but what do they look like?)

Any help would be greatly appreciated.

2 fixed typo

Let Con(PA) be the sentence of arithmetic which translates as "Peano Arithmetic is consistent." Then according to Godel's 2nd incompleteness theorem, assuming PA is consistent then PA can neither prove Con(PA) nor its negation. And in fact, if T contains PA and T is (omega-) consistent, then T can neither prove Con(T) nor its negation. In particular, if PA+Con(PA) is consistent then PA+Con(PA) can neither prove Con(PA+Con(PA)) nor its negation.

But consider the following piece of reasoning: if PA is consistent, then Con(PA) is true, so PA+Con(PA) is consistent, so Con(PA+Con(PA)) is true. My question is, why can't this reasoning by formalized in PA, so that within PA you can prove Con(PA) implies Con(PA+Con(PA))? If you could prove that, then since you can obviously prove Con(PA) within PA+Con(PA), you would be able to prove Con(PA+Con(PA)) within PA+Con(PA), which is a contradiction. So where am I going wrong?

We can even talk about this in terms of model theory. : There are nonstandard models of PA in which Con(PA) does not hold: basically, you have infinitely large natural numbers, and you have infinitely long proofs of a contradiction in PA. This does not mean that PA is inconsistent, because there are no proofs of finite length of this contradiction. So are there also nonstandard models of PA+Con(PA) in which Con(PA+Con(PA)) does not hold? (That's a rhetorical question; clearly there must be, but what do they look like?)

Any help would be greatly appreciated.