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Manfred comments on Walkthrough of "Definability of Truth in Probabilistic Logic" - Less Wrong
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Comments (30)
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.