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Metamath Proof Explorer

Theorem pm2.61dane

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 30-Nov-2011)

Ref Expression
Hypotheses pm2.61dane.1 
|- ( ( ph /\ A = B ) -> ps )
pm2.61dane.2 
|- ( ( ph /\ A =/= B ) -> ps )
Assertion pm2.61dane 
|- ( ph -> ps )

Proof

Step Hyp Ref Expression
1 pm2.61dane.1 
 |-  ( ( ph /\ A = B ) -> ps )
2 pm2.61dane.2 
 |-  ( ( ph /\ A =/= B ) -> ps )
3 1 ex 
 |-  ( ph -> ( A = B -> ps ) )
4 2 ex 
 |-  ( ph -> ( A =/= B -> ps ) )
5 3 4 pm2.61dne 
 |-  ( ph -> ps )