# Does Turing determinacy imply full determinacy?

The axiom of Turing determinacy is a weakening of the full axiom of determinacy, $AD$, in which only games with payoff sets which are $\equiv_T$-invariant are demanded to be determined.

In "Turing determinacy and the continuum hypothesis" (published in 1989), Ramez Sami writes:

"The main question so far unsettled in this particular domain can be roughly put this way: is it true that for any "reasonable" pointclass $\Gamma$ we have: Turing-Det$(\Gamma)\implies$Det$(\Gamma)$? In particular is it the case that: [over $ZF+DC$, presumably] Turing $AD$ implies $AD$?"

My question is, what is the status of this question currently? Do we know whether Turing $AD$ is strictly weaker than $AD$? The only recent work I know of around Turing determinacy is from the reverse mathematical side (http://www.math.cornell.edu/~shore/papers/pdf/TDet21.pdf); I'm not at all familiar with the set theory on the subject.

(I vaguely recall that Turing determinacy implies that every Suslin set is determined, but I can't remember where I supposedly learned this "fact.")

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This is open. In $L(\mathbb R)$ the answer is yes. Hugh has several proofs of this, and it remains one of the few unpublished results in the area. The latest version of the statement (that I know of) is the claim in your parenthetical remark at the end. This gives determinacy in $L(\mathbb R)$ using, for example, a reflection argument.
(I mentioned this a while ago somewhere on this site. Maybe that's where you heard of it? This can be used to prove that $\omega$-board determinacy is equiconsistent with determinacy. I seem to recall that's how the topic came up.)