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May 22, 2015 at 17:00 comment added François G. Dorais @JoelDavidHamkins: You're right that $SH(\mathcal N)$ depends on how one clocks Turing machines. If $0 \notin SH(\mathcal N)$ then $SH(\mathcal N)$ isn't a semiring nor does it contain all parameter-free $\Sigma_1$-definable elements of $\mathcal N$ and the answer breaks down.
May 22, 2015 at 16:25 history edited François G. Dorais CC BY-SA 3.0
clarification
May 22, 2015 at 12:13 comment added Joel David Hamkins François, could you explain a bit more? Why doesn't $s=0$ fulfill that existential trivially?
May 22, 2015 at 2:51 history answered François G. Dorais CC BY-SA 3.0