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wantstojoinin comments on Logical Uncertainty as Probability - Less Wrong Discussion
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A natural number n can be even or odd: i.e. n%2=0 or n%2=1.
If X = {n is natural number} then you showed that we can use P(n%2=0|X) + P(n%2=1|X) = 1 and P(n%2=0|X) = P(n%2=1|X) together to get P(n%2=0|X) = 1/2.
The same logic works for the three statements n%3=0,n%3=1,n%3=2 to give us P(n%3=0|X) = P(n%3=1|X) = P(n%3=2|X) = 1/3.
But then the same logic also works for the two indistinguishable statements n%3=0,n%3=1 \/ n%3=2 to give us P(n%3=0|X) = P(n%3=1 \/ n%3=2) = 1/2.
But 1/2 = 1/3 is a contradiction, so we find that axiom 3 leads to inconsistencies.
n%3=0 is distinguishable from n%3=1∨n%3=2. If A="n%3=0", B="n%3=1", and C="n%3=2", then an isomorphism f that maps B∨C to A must satisfy f(B∨C) = f(B)∨f(C) = A, which is impossible.
I understand, what I wrote was wrong. What if we use n%3=0 and ~(n%3=0) though?