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