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May 22, 2013 at 16:37 answer added François G. Dorais timeline score: 5
May 22, 2013 at 4:33 vote accept Scott Aaronson
May 22, 2013 at 3:17 comment added François G. Dorais Ah, I guess your promise problems are what computability theorists call mass problems?
May 22, 2013 at 3:16 answer added Joel David Hamkins timeline score: 25
May 22, 2013 at 3:15 comment added Scott Aaronson François: Yes, I know that CG is not a set; rather, it's what you call a separation problem and what complexity theorists would call a promise problem (the promise being that M halts). But the notion of Turing-reducibility can be generalized to promise problems in a fairly straightforward way, and once you do that $CG\le_T PROVELOOP$ becomes meaningful. Your last observation, implying that in fact $CG\lt_T PROVELOOP$, is quite interesting and not something I knew -- thanks for that!
May 22, 2013 at 3:05 comment added François G. Dorais Note that GC is not a well-defined set and so $GC \leq_T PROVELOOP$ is not meaningful. There are many solutions to the separation problem, some of which are $<_T HALT$ while others are way up in the stratosphere. What you have shown is that PROVELOOP computes a separating set and is therefore not computable since the separation problem has no computable solution. In fact, it is known that if $X$ is a solution to the separation problem, then there is another solution $Y$ such that $0 <_T Y <_T X$. So there is a solution to the separation problem which is $<_T PROVELOOP$.
May 22, 2013 at 2:53 answer added Noah Schweber timeline score: 4
May 22, 2013 at 2:16 history edited Scott Aaronson CC BY-SA 3.0
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May 22, 2013 at 2:11 history edited Scott Aaronson CC BY-SA 3.0
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May 22, 2013 at 2:03 history asked Scott Aaronson CC BY-SA 3.0