The Probability of a Contradiction is Always Zero: A Proof

The proposition in question is:

(X)Pr(X) = 0 if X is a logical contradiction (i.e., the probability of a contradiction is always zero)

Here's a proof (using the axioms and definitions of the probability calculus) for the above claim:

1. Φ→Φ  [asm]

2. Pr(Φ|Φ) = 1  [1, entailment]

3. Pr(Φ|Φ) + Pr(~Φ|Φ) = 1  [normality]

4. 1 - Pr(Φ|Φ) = Pr(~Φ|Φ)  [3]

5. Pr(~Φ|Φ) = 0  [2, 4]

6. Pr(Φ&~Φ) = Pr(~Φ|Φ)Pr(Φ)  [multiplication]

7. Pr(Φ&~Φ) = 0  [5, 6]