A Proof that Pr(H|E) > Pr(H) entails Pr(H|E) > Pr(H|~E)

A proof for the following conditional: Pr(H|E) > Pr(H) only if Pr(H|E) > Pr(H|~E).

1. P(H|E) > P(H)
2. P(H|E) > P(H) iff P(E|H) > P(E|~H) [Mackie's relevance criterion].
3. P(E|H) > P(E|~H) [MP, 1 & 2].
4. P(~E|H) = 1 - P(E|H) [subtraction].
5. P(~E|~H) = 1 - P(E|~H) [subtraction].
6. [1 - P(E|~H)] > [1 - P(E|H)] [from 3]. 
7. P(~E|~H) > P(~E|H) [from 4, 5, & 6].
8. P(H) > P(H|~E) iff P(~E|~H) > P(~E|H) [premise]. 
9. P(H) > P(H|~E) [MP, 7 & 8].
10. P(H|E) > P(H) > P(H|~E) [from (1), (9)].
11. P(H|E) > P(H|~E) [from 10].
12. Therefore, P(H|E) > P(H) only if P(H|E) > P(H|~E).