You could make f(1) large enough for Loebian cooperation to happen immediately. Then you could make f(2) large enough to simulate what happened in the first f(1)+1 steps, and so on. I still don’t completely understand how you’re going to use this thing, though.
Lemma 1: There’s a proof that M’(DefectBot)=D, and that there does not exist a shorter proof of M’(DefectBot)=C.
Proof: There’s a short proof that DefectBot(DefectBot)=D, and a proof of length (waiting time + short) that FB’(DefectBot)=DefectBot(FB’)=D. (It needs to wait until FB’ stops looking for only cooperative proofs.) Clearly there are no shorter proofs of contrary statements.
So there’s a proof that M’ skips to its default defection (since there’s nothing strictly better), and this proof has length less than (g(short) + g(wait + short) + short), where g is the function that bounds how long it takes to verify that something is the shortest proof of a certain type, in terms of the proof length. (I think g is just bounded by an exponential, but it doesn’t matter).
Thus, of course, Lemma 1 has a proof of length at most g(g(short) + g(wait + short) + short).
Lemma 2: There’s a proof that M’(FB’)=FB’(M’)=C, and that there does not exist a shorter proof of any alternative outcome.
Proof: Similar to Lemma 1. The initial proofs are a bit longer because we need the Löbian proof of length less than (waiting time) that FB’(FB’)=C, instead of the short proof that DefectBot(DefectBot)=D.
And so there’s a proof of length (g(wait+short) + g(wait) + short) that M’ goes to the sanity-check trying to prove that FB’(M’)=FB’(FB’). Now a proof of length L that FB’(M’)=C would imply a proof of length L + (g(wait+short) + g(wait) + 2*short) that M’(FB’)=C...
Oh, and here’s the impasse again, because the existence of a proof that M’(FB’)=C is no longer enough to prove that FB’(M’)=C, you also need that there’s no shorter proof that M’(FB’)=D. Unless there’s a clever way to finesse that (within something provable by the same system), it looks like that’s where this last-ditch fix fails.
So this is setting off all sorts of aesthetic flags in my head, but it appears to work:
Let FairBotPrime (FB’) be the agent that does this: first tries to prove Opp(FB’)=C for a long time (long enough for Löb to work), then simply tries to prove either Opp(FB’)=C or Opp(FB’)=D for the rest of the time. If it ever succeeds, it immediately halts and returns the same move as it just deduced. If it doesn’t deduce anything in time, then it defects by default.
Let MasqueradePrime (M’) be the agent that first tries on the DefectBot and FB’ masks: it looks through all proofs for one of the form “DefectBot(Opp)=X and Opp(DefectBot)=Y” and progresses to the next stage once it finds one (or once it runs out of allotted time), then does the same for the FB’ mask. Then, if either payout thus obtained exceeds the default (D,D) payout, it sanity-checks whichever mask gets the better result (let’s say it defaults to DefectBot if they tie): it tries to prove “Opp(M’)=Opp(chosen mask)”, and if so then it outputs the (pre-calculated) move that the chosen mask makes against the opponent. If something along this path fails, then it just outputs its default move (D).
I claim that M’ correctly defects against DefectBot and CooperateBot, and cooperates mutually with FB’ and M’. The key here is that at no point does it need to deduce that a proof search failed; it only needs to show that a certain proof is the first one that an agent finds. I’ll reply again with some lemmas.
You could make f(1) large enough for Loebian cooperation to happen immediately. Then you could make f(2) large enough to simulate what happened in the first f(1)+1 steps, and so on. I still don’t completely understand how you’re going to use this thing, though.
Lemma 1: There’s a proof that M’(DefectBot)=D, and that there does not exist a shorter proof of M’(DefectBot)=C.
Proof: There’s a short proof that DefectBot(DefectBot)=D, and a proof of length (waiting time + short) that FB’(DefectBot)=DefectBot(FB’)=D. (It needs to wait until FB’ stops looking for only cooperative proofs.) Clearly there are no shorter proofs of contrary statements.
So there’s a proof that M’ skips to its default defection (since there’s nothing strictly better), and this proof has length less than (g(short) + g(wait + short) + short), where g is the function that bounds how long it takes to verify that something is the shortest proof of a certain type, in terms of the proof length. (I think g is just bounded by an exponential, but it doesn’t matter).
Thus, of course, Lemma 1 has a proof of length at most g(g(short) + g(wait + short) + short).
Lemma 2: There’s a proof that M’(FB’)=FB’(M’)=C, and that there does not exist a shorter proof of any alternative outcome.
Proof: Similar to Lemma 1. The initial proofs are a bit longer because we need the Löbian proof of length less than (waiting time) that FB’(FB’)=C, instead of the short proof that DefectBot(DefectBot)=D.
And so there’s a proof of length (g(wait+short) + g(wait) + short) that M’ goes to the sanity-check trying to prove that FB’(M’)=FB’(FB’). Now a proof of length L that FB’(M’)=C would imply a proof of length L + (g(wait+short) + g(wait) + 2*short) that M’(FB’)=C...
Oh, and here’s the impasse again, because the existence of a proof that M’(FB’)=C is no longer enough to prove that FB’(M’)=C, you also need that there’s no shorter proof that M’(FB’)=D. Unless there’s a clever way to finesse that (within something provable by the same system), it looks like that’s where this last-ditch fix fails.
Drat.
So this is setting off all sorts of aesthetic flags in my head, but it appears to work:
Let FairBotPrime (FB’) be the agent that does this: first tries to prove Opp(FB’)=C for a long time (long enough for Löb to work), then simply tries to prove either Opp(FB’)=C or Opp(FB’)=D for the rest of the time. If it ever succeeds, it immediately halts and returns the same move as it just deduced. If it doesn’t deduce anything in time, then it defects by default.
Let MasqueradePrime (M’) be the agent that first tries on the DefectBot and FB’ masks: it looks through all proofs for one of the form “DefectBot(Opp)=X and Opp(DefectBot)=Y” and progresses to the next stage once it finds one (or once it runs out of allotted time), then does the same for the FB’ mask. Then, if either payout thus obtained exceeds the default (D,D) payout, it sanity-checks whichever mask gets the better result (let’s say it defaults to DefectBot if they tie): it tries to prove “Opp(M’)=Opp(chosen mask)”, and if so then it outputs the (pre-calculated) move that the chosen mask makes against the opponent. If something along this path fails, then it just outputs its default move (D).
I claim that M’ correctly defects against DefectBot and CooperateBot, and cooperates mutually with FB’ and M’. The key here is that at no point does it need to deduce that a proof search failed; it only needs to show that a certain proof is the first one that an agent finds. I’ll reply again with some lemmas.