cousin_it comments on A model of UDT without proof limits - Less Wrong

13 Post author: cousin_it 20 March 2012 07:41PM

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Comment author: cousin_it 20 March 2012 08:12:35PM *  3 points [-]

If A's proof system is sound, that directly implies that A won't generate "spurious" proofs.

Nope. In the example problem from the post, A's proof system will happily prove statements like "A()==1 implies U()==100". After all, A() finishes in finite time and returns 2, so any proof system worth its salt will eventually prove that A()==1 is false by simulating A, and a false statement implies anything. The challenge is to not let the agent stumble upon spurious proofs before it returns a value.

Comment author: tgb 20 March 2012 10:27:40PM 3 points [-]

To make this post more obvious to people like me and JGWeissman, it would be a good idea to include this in the original post! I came out of this post thinking "Why did I just read that?" and couldn't figure it out until this response.

Comment author: cousin_it 20 March 2012 11:55:39PM 1 point [-]

OK, thanks! Included this reasoning in the post.

Comment author: Dmytry 21 March 2012 04:06:20AM 0 points [-]

ahh, good, i did re-read stuff and i was like - did i skip some major piece of it? am i going insane lately?

Now it makes a lot more sense.

Comment author: cousin_it 21 March 2012 12:33:31PM *  0 points [-]

Sorry, the omission was due to illusion of transparency on my part. I'd spent so much time thinking about avoiding spurious counterfactuals that I couldn't imagine other people not seeing that as the most important problem :-)

Comment author: twanvl 22 March 2012 12:07:32PM 0 points [-]

any proof system worth its salt will eventually prove that A()==1 is false by simulating A,

Not if A includes the proof search! No sound proof system will be able to say anything interesting about programs that use the system itself.

Comment author: cousin_it 22 March 2012 12:26:24PM *  1 point [-]

If the program doesn't terminate, the proof system may be unable to prove that. But if the program terminates in finitely many steps, any good enough proof system allows for a proof that just simulates all the steps one by one. Of course the program itself won't reach that particular proof because the proof is too long.