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Nik Weaver
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One direction could go like this. Let ${\rm ZFC}^+$ be the theory in the language of set theory augmented by a constant symbol ${\bf M}$ with the axioms

$\bullet$ every axiom of ${\rm ZFC}$

$\bullet$ "${\bf M}$ is countable and transitive"

$\bullet$ the relativization of every axiom of ${\rm ZFC}$ to ${\bf M}$.

Then one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm ZFC}^+)$ in Peano arithmetic, and based on the page you linked, it looks like one can straightforwardly prove the consistency of any finite fragment of ${\rm Con}({\rm SEAR})$${\rm SEAR}$ in ${\rm ZFC}^+$. Thus one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm SEAR})$ in Peano arithmetic. (If ${\rm SEAR}$ were not consistent then some finite fragment ${\rm SEAR}_0$ would be inconsistent, and this fact would be verifiable in ${\rm ZFC}^+$, so that ${\rm ZFC}^+$ would prove both ${\rm Con}({\rm SEAR})$${\rm Con}({\rm SEAR}_0)$ and $\neg{\rm Con}({\rm SEAR})$$\neg{\rm Con}({\rm SEAR}_0)$ and hence be inconsistent.) For details see Chapter 7 of my book.

I imagine a similar argument would work in the reverse direction to prove ${\rm Con}({\rm SEAR}) \Rightarrow {\rm Con}({\rm ZFC})$ in Peano arithmetic, but I'm not familiar with ${\rm SEAR}$ so I can't say that for sure.

One direction could go like this. Let ${\rm ZFC}^+$ be the theory in the language of set theory augmented by a constant symbol ${\bf M}$ with the axioms

$\bullet$ every axiom of ${\rm ZFC}$

$\bullet$ "${\bf M}$ is countable and transitive"

$\bullet$ the relativization of every axiom of ${\rm ZFC}$ to ${\bf M}$.

Then one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm ZFC}^+)$ in Peano arithmetic, and based on the page you linked, it looks like one can straightforwardly prove ${\rm Con}({\rm SEAR})$ in ${\rm ZFC}^+$. Thus one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm SEAR})$ in Peano arithmetic. (If ${\rm SEAR}$ were not consistent then this fact would be verifiable in ${\rm ZFC}^+$, so that ${\rm ZFC}^+$ would prove both ${\rm Con}({\rm SEAR})$ and $\neg{\rm Con}({\rm SEAR})$ and hence be inconsistent.) For details see Chapter 7 of my book.

I imagine a similar argument would work in the reverse direction to prove ${\rm Con}({\rm SEAR}) \Rightarrow {\rm Con}({\rm ZFC})$ in Peano arithmetic, but I'm not familiar with ${\rm SEAR}$ so I can't say that for sure.

One direction could go like this. Let ${\rm ZFC}^+$ be the theory in the language of set theory augmented by a constant symbol ${\bf M}$ with the axioms

$\bullet$ every axiom of ${\rm ZFC}$

$\bullet$ "${\bf M}$ is countable and transitive"

$\bullet$ the relativization of every axiom of ${\rm ZFC}$ to ${\bf M}$.

Then one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm ZFC}^+)$ in Peano arithmetic, and based on the page you linked, it looks like one can straightforwardly prove the consistency of any finite fragment of ${\rm SEAR}$ in ${\rm ZFC}^+$. Thus one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm SEAR})$ in Peano arithmetic. (If ${\rm SEAR}$ were not consistent then some finite fragment ${\rm SEAR}_0$ would be inconsistent, and this fact would be verifiable in ${\rm ZFC}^+$, so that ${\rm ZFC}^+$ would prove both ${\rm Con}({\rm SEAR}_0)$ and $\neg{\rm Con}({\rm SEAR}_0)$ and hence be inconsistent.) For details see Chapter 7 of my book.

I imagine a similar argument would work in the reverse direction to prove ${\rm Con}({\rm SEAR}) \Rightarrow {\rm Con}({\rm ZFC})$ in Peano arithmetic, but I'm not familiar with ${\rm SEAR}$ so I can't say that for sure.

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Nik Weaver
  • 42.8k
  • 3
  • 112
  • 213

One direction could go like this. Let ${\rm ZFC}^+$ be the theory in the language of set theory augmented by a constant symbol ${\bf M}$ with the axioms

$\bullet$ every axiom of ${\rm ZFC}$

$\bullet$ "${\bf M}$ is countable and transitive"

$\bullet$ the relativization of every axiom of ${\rm ZFC}$ to ${\bf M}$.

Then one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm ZFC}^+)$ in Peano arithmetic, and based on the page you linked, it looks like one can straightforwardly prove ${\rm Con}({\rm SEAR})$ in ${\rm ZFC}^+$. Thus one can prove ${\rm Con}({\rm ZFC}) \Rightarrow {\rm Con}({\rm SEAR})$ in Peano arithmetic. (If ${\rm SEAR}$ were not consistent then this fact would be verifiable in ${\rm ZFC}^+$, so that ${\rm ZFC}^+$ would prove both ${\rm Con}({\rm SEAR})$ and $\neg{\rm Con}({\rm SEAR})$ and hence be inconsistent.) For details see Chapter 7 of my book.

I imagine a similar argument would work in the reverse direction to prove ${\rm Con}({\rm SEAR}) \Rightarrow {\rm Con}({\rm ZFC})$ in Peano arithmetic, but I'm not familiar with ${\rm SEAR}$ so I can't say that for sure.