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Karl comments on Robust Cooperation in the Prisoner's Dilemma - Less Wrong
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UnfairBot defect against PrudentBot.
Proof: For UnfairBot to cooperate with PrudentBot, PA would have to prove that PrudentBot defect against UnfairBot which would require PA to prove that "PA does not prove that UnfairBot cooperate with PrudentBot or PA+1 does not prove that UnfairBot defect against DefectBot" but that would require PA to prove it's own consistency which it cannot do. QED
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.