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Oct 21, 2010 at 19:15 comment added user3462 Hi, thanks for the answer. I unfortunately did not mention the more general question that I asked in the title, but thankfully Stefan and Andres answered that. Thanks once again.
Oct 20, 2010 at 21:11 vote accept CommunityBot
Oct 20, 2010 at 20:27 comment added Bjørn Kjos-Hanssen In more detail: if $a\le_T b$ then $a=\Phi^b$ for some Turing functional $\Phi$. The set $\Phi^b$ is $\Sigma_1(b)$ definable (actually $\Delta_1(b)$ definable here since $\Phi^b$ is total).
Oct 20, 2010 at 19:24 history answered Bjørn Kjos-Hanssen CC BY-SA 2.5