Metamath Proof Explorer


Theorem feqmptd

Description: Deduction form of dffn5 . (Contributed by Mario Carneiro, 8-Jan-2015)

Ref Expression
Hypothesis feqmptd.1 φF:AB
Assertion feqmptd φF=xAFx

Proof

Step Hyp Ref Expression
1 feqmptd.1 φF:AB
2 1 ffnd φFFnA
3 dffn5 FFnAF=xAFx
4 2 3 sylib φF=xAFx