## New answers tagged theories-of-arithmetic

2
votes

### Is diamond consistent with 2nd order PA?

In the case of $\text{GBC}^-$, the answer is no, $\Diamond$ is not provable from the assumption that all reals are constructible. It is consistent with $\text{GBC}^-$ and even $\text{KM}^-$ that $\...

15
votes

### Con(PA) via non-well-foundedness?

This is a completely standard perspective in work on models-of-PA, a view that informs dozens of arguments. That is simply the nature of nonstandard models, that things they think are well founded are ...

6
votes

### In what sense does the sentence $\operatorname{con}(\mathsf{PA})$ "say" that $\mathsf{PA}$ is consistent?

"So, in light of these objections, what viewpoints do logicians have about the relationship between the assertion that $PA$
is consistent, and the arithmetic sentence $con(PA)
$?"
Logicians ...

8
votes

Accepted

### In what sense does the sentence $\operatorname{con}(\mathsf{PA})$ "say" that $\mathsf{PA}$ is consistent?

The PA sentence “$\newcommand{\Con}{\text{Con}}\Con(\newcommand{\PA}{\text{PA}}\PA)$” says that PA is consistent in exactly the same ways that the PA sentences representing, say, the fundamental ...

9
votes

### In what sense does the sentence $\operatorname{con}(\mathsf{PA})$ "say" that $\mathsf{PA}$ is consistent?

There is really nothing peculiar about Con(PA) in this regard. Let's take a simpler statement, such as
$$(\exists x \exists y \exists z : xxx + yyy - zzz = 114) \vee (\exists x \exists y \exists z : ...

3
votes

I am going to answer the question in the title - In what sense does the sentence con(PA) "say" that PA is consistent? - rather than the question in the body. This is not what the OP wants, ...

5
votes

Let's go back to the continuum hypothesis. Some, probably most, people think that it really is objectively true or false, despite having different truth values in different models, because they ...

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