Metamath Proof Explorer


Theorem eqfnfvd

Description: Deduction for equality of functions. (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Hypotheses eqfnfvd.1 φ F Fn A
eqfnfvd.2 φ G Fn A
eqfnfvd.3 φ x A F x = G x
Assertion eqfnfvd φ F = G

Proof

Step Hyp Ref Expression
1 eqfnfvd.1 φ F Fn A
2 eqfnfvd.2 φ G Fn A
3 eqfnfvd.3 φ x A F x = G x
4 3 ralrimiva φ x A F x = G x
5 eqfnfv F Fn A G Fn A F = G x A F x = G x
6 1 2 5 syl2anc φ F = G x A F x = G x
7 4 6 mpbird φ F = G