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AlexMennen comments on Robust Cooperation in the Prisoner's Dilemma - Less Wrong
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Oops, you're right. But how do you prove that every modal agent is defected against by at least one of FairBot and UnfairBot?
Proof: Let X be a modal agent, Phi(...) it's associated fully modalized formula, (K, R) a GL Kripke model and w minimal in K. Then, for all statement of the form ◻(...) we have w |- ◻(...) so Phi(...) reduce in w to a truth value which is independent of X opponent. As a result, we can't have both w |- X(FairBot) = C and w |- X(UnfairBot) = D and so we can't have both ◻(X(FairBot) = C) and ◻(X(UnfairBot) = D) and so we can't both have FairBot(X) = C and UnfairBot(X) = C. QED
I don't know what that means. Can you prove it without using Kripke semantics? (if that would complicate things enough to make it unpleasant to do so, don't worry about it; I believe you that you probably know what you're doing)
Proof without using Kripke semantic: Let X be a modal agent and Phi(...) it's associated fully modalized formula. Then if PA was inconsistent Phi(...) would reduce to a truth value independent of X opponent and so X would play the same move against both FairBot and UnfairBot (and this is provable in PA). But PA cannot prove it's own consistency so PA cannot both prove X(FairBot) = C and X(UnfairBot) = D and so we can't both have FairBot(X) = C and UnfairBot(X) = C. QED
Oh, I see. Thanks.