Metamath Proof Explorer


Theorem eqfnfvd

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

Ref Expression
Hypotheses eqfnfvd.1 φFFnA
eqfnfvd.2 φGFnA
eqfnfvd.3 φxAFx=Gx
Assertion eqfnfvd φF=G

Proof

Step Hyp Ref Expression
1 eqfnfvd.1 φFFnA
2 eqfnfvd.2 φGFnA
3 eqfnfvd.3 φxAFx=Gx
4 3 ralrimiva φxAFx=Gx
5 eqfnfv FFnAGFnAF=GxAFx=Gx
6 1 2 5 syl2anc φF=GxAFx=Gx
7 4 6 mpbird φF=G