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.