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2 questions
13
votes
1
answer
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How much determinacy do you need for second order arithmetic to be as strong as ZFC?
From Wikipedia (I couldn't find the original source):
$\text{ZFC} + \{\text{there are $n$ Woodin cardinals: $n$ is a natural number}\}$ is conservative over $\text{Z}_2$ with projective determinacy.
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10
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1
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How much of second-order arithmetic do you need for $\mathbf{\Sigma}^1_1$-determinacy to give you countable transitive models of $\mathsf{ZFC}$?
This is in some sense a follow-up to this question.
The answer there says that over $\mathsf{Z}_2$ (second-order arithmetic), (boldface) $\mathbf{\Sigma}^1_1$-determinacy is enough to entail the ...
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