Metamath Proof Explorer


Theorem feqmptd

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

Ref Expression
Hypothesis feqmptd.1 φ F : A B
Assertion feqmptd φ F = x A F x

Proof

Step Hyp Ref Expression
1 feqmptd.1 φ F : A B
2 1 ffnd φ F Fn A
3 dffn5 F Fn A F = x A F x
4 2 3 sylib φ F = x A F x