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Manfred comments on Walkthrough of "Definability of Truth in Probabilistic Logic" - Less Wrong
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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.