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Metamath Proof Explorer

Theorem pm2.24nel

Description: A contradiction concerning membership implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018)

Ref Expression
Assertion pm2.24nel 
|- ( A e. B -> ( A e/ B -> ph ) )

Proof

Step Hyp Ref Expression
1 df-nel 
 |-  ( A e/ B <-> -. A e. B )
2 pm2.24 
 |-  ( A e. B -> ( -. A e. B -> ph ) )
3 1 2 biimtrid 
 |-  ( A e. B -> ( A e/ B -> ph ) )