Metamath Proof Explorer


Theorem pm2.61ine

Description: Inference eliminating an inequality in an antecedent. (Contributed by NM, 16-Jan-2007) (Proof shortened by Andrew Salmon, 25-May-2011)

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

Proof

Step Hyp Ref Expression
1 pm2.61ine.1
 |-  ( A = B -> ph )
2 pm2.61ine.2
 |-  ( A =/= B -> ph )
3 nne
 |-  ( -. A =/= B <-> A = B )
4 3 1 sylbi
 |-  ( -. A =/= B -> ph )
5 2 4 pm2.61i
 |-  ph