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.
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.