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Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi$ and $\mathcal{M}_\psi\models\psi$, but $\text{PA}\vdash\neg(\phi\wedge\psi)$ and $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answerThis answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi$ and $\mathcal{M}_\psi\models\psi$, but $\text{PA}\vdash\neg(\phi\wedge\psi)$ and $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi$ and $\mathcal{M}_\psi\models\psi$, but $\text{PA}\vdash\neg(\phi\wedge\psi)$ and $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

Post Closed as "Not suitable for this site" by Eric Wofsey, David White, Ricardo Andrade, Benjamin Steinberg, Andrés E. Caicedo
fix body of question to be in accord with the title
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Benya
Benya
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Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi\wedge(\neg\psi)$$\mathcal{M}_\phi\models\phi$ and $\mathcal{M}_\psi\models(\neg\phi)\wedge\psi$$\mathcal{M}_\psi\models\psi$, but $\text{PA}\vdash\neg(\phi\wedge\psi)$ and $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi\wedge(\neg\psi)$ and $\mathcal{M}_\psi\models(\neg\phi)\wedge\psi$, but $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi$ and $\mathcal{M}_\psi\models\psi$, but $\text{PA}\vdash\neg(\phi\wedge\psi)$ and $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

fix bug in previous edit
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Benya
Benya
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Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi\wedge(\neg\psi)$ and $\mathcal{M}_\psi\models(\neg\phi)\wedge\psi$, but $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

(I I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist.) This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi\wedge(\neg\psi)$ and $\mathcal{M}_\psi\models(\neg\phi)\wedge\psi$, but $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

(I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist.)

Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi\wedge(\neg\psi)$ and $\mathcal{M}_\psi\models(\neg\phi)\wedge\psi$, but $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

Post Undeleted by Benya
Post Deleted by Benya
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Benya
Benya
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