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Mar 29, 2022 at 12:32 comment added Fedor Pakhomov @SamSanders You're welcome.
Mar 29, 2022 at 8:36 comment added Sam Sanders Thank you! This is a very nice answer.
Mar 29, 2022 at 8:36 vote accept Sam Sanders
Mar 29, 2022 at 8:37
Mar 28, 2022 at 21:31 history edited Fedor Pakhomov CC BY-SA 4.0
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Mar 28, 2022 at 21:29 comment added Fedor Pakhomov @NoahSchweber Yes, $\mathsf{RCA}_0^\star=\mathsf{EA}+\mathsf{I}\Delta^0_1+\Delta^0_1\textsf{-CA}$ and $\mathsf{WKL}_0^\star=\mathsf{RCA}_0^\star+\mathsf{WKL}$.
Mar 28, 2022 at 21:21 comment added Noah Schweber Is $\mathsf{WKL_0^\star}$ etc. just $\mathsf{WKL_0}$ but with induction weakened to $\Delta_0^{exp}$?
Mar 28, 2022 at 21:12 history answered Fedor Pakhomov CC BY-SA 4.0