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benkuhn comments on Walkthrough of "Definability of Truth in Probabilistic Logic" - Less Wrong
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Comments (30)
I'm having a hard time showing that P maps equivalent statements to the same value, or that it's bounded. In particular, I'm having difficulty showing P(p && q) = P(q && p); it seems like none of the axioms lets you switch the order of two atoms. What's the trick I'm missing?
How about if axiom 1 read "P(x) = P(x && y) + P(¬y && x)"?
Then we could say P(q && p) = P(q && p && ¬p) + P(p && q && p) = P(p && q && p) by axioms 1' and 3, and similar for P(p && q).
So P(q && p) = P(p && q && p) = P(p && q).
I think your proof secretly turns ¬¬p into p in the first step, which isn't legit. However, your modification does give equality on logically equivalent sentences, in the following way:
Suppose p <=> ~q. Then P(p) = P(p && q) + P(~q && p) = P(~q && p) by 3 P(~q) = P(~q && p) + P(~p && ~q) = P(~q && p) by 3
Hence if p is equivalent to q and at least one starts with a ~ sign, then P is equal on them. Now suppose p <=> q with neither one starting with a ~ sign: we have
P(p) = P(~~p) = P(~~q) = P(q)
and we're done.
It does, drat.
Ok. So to fix/complete my proof we need P(¬¬p && q && p) = P(p && q && p). Hm. Ok. So to prove they're equal we just add on another term using axiom 1 and then rule out the contradiction in such away that we show they're both equal to the same thing.
Or, once you have equality for logically equivalent sentences, note that (p && q) <=> (q && p) and hence we have directly that P of the two sides are equal.
Yes, the axioms seem incomplete, or perhaps it was simply meant to be implied that "P(p) = P(q & p) + P(¬q & p)" also. Otherwise as far as I can tell there's no axiom that lets you relate any expression containing "P(p" to an expression containing "P(q", unless the arguments of P(·) are each a tautology or contradiction (which is unhelpful).
Well, a tautology can be made up of non-tautological things; we could conceivably have some sentence phi(p, q) that's a tautology if p <=> q, such that P(p) = f(P(phi(p,q))) = f(P(phi(q,p))) = P(p). I think this is what ygert is trying to do. I don't have much hope for this approach, though.
I'd think that the way you'd prove it is with the fact that (p && q)<=>(q && p) is a tautology. I don't have an exact proof at the moment; let me work on it.
After working on the problem, I am convinced we also need an order-swapped version of axiom 1. If we had that, we could prove that any pair of equivalent statements have that same value: the general case of the problem benkuhn posed.
(If A and B are equivalent, then P(A&~B)=P(B&~A)=0 as contradictions, and so by axiom 1:
P(A)=P(A&B)+P(A&~B)=P(A&B)+0=P(A&B)
P(B)=P(B&A)+P(B&~A)=P(B&A)+0=P(B&A)
So close. If only we could swap the orders, we'd have P(A)=P(B).)
I tried applying the proof of the theorem to the problem, as P(q) = P(p) for equivalent statements p,q is clear from the claim P(q) = mu({M: M|= q}) so our desired result should be part of the proof of Theorem 1. However it seems that our desired result is implicitly assumed in the statement "Axiom 1 implies". Setting P(A|B) = P(B && A)/P(B) (note the reversal of order, without being allowed to replace equivalent statements inside P the mess is even greater if we pick the natural order. Also we want P(B) =/= 0) and writing psi for phi j we can write the claim as:
P(Ti && phi)/P(Ti) = P(Ti && psi && phi)/P(Ti && psi) * P(Ti && psi)/P(Ti) + P(Ti && ~psi && phi)/P(Ti && ~psi) * P(Ti && ~psi)/P(Ti)
(Here I ignored some annoying brackets, actually there's more of a mess as its not clear if P((A && B) && C) = P(A && (B && C))). Multiplying both sides by P(Ti) and simplifying now gives:
P(Ti && phi) = P(Ti && psi && phi) + P(Ti && ~psi && phi)
which does not follow from Axiom 1 as the added statements are in the middle of our statement. I am convinced that our claim (P(p) = P(q) for equivalent statements p,q) is required for the proof in the paper and should be added as an extra axiom.
On a sidenote:unde the assumption above (equivalent statements get equal probabilities) axiom 3 is a consequence of axioms 1 and 2. This can be seen as follows: Let q be a contradiction, so ~q is a tautology. By axioms 1 and 2 we have P(q) = P(q && q) + P(q && ~q). But ~q is a tautology, so p && ~q and p are equivalent for every p. In particular we have P(q && ~q) = P(q). But q and (q && q) are also equivalent, so the above can be written as P(q) = P(q) + P(q) so P(q) = 0.
I, too, am now doubtful that axioms 1-3 are sufficient. I've updated the post accordingly.
Yeah, I couldn't find anything either.
As Manfred and I showed above, replacing axiom 1 with "P(x) = P(x && y) + P(¬y && x)" gives a sufficient condition, though.