Metamath Proof Explorer


Theorem eqfnfvd

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

Ref Expression
Hypotheses eqfnfvd.1
|- ( ph -> F Fn A )
eqfnfvd.2
|- ( ph -> G Fn A )
eqfnfvd.3
|- ( ( ph /\ x e. A ) -> ( F ` x ) = ( G ` x ) )
Assertion eqfnfvd
|- ( ph -> F = G )

Proof

Step Hyp Ref Expression
1 eqfnfvd.1
 |-  ( ph -> F Fn A )
2 eqfnfvd.2
 |-  ( ph -> G Fn A )
3 eqfnfvd.3
 |-  ( ( ph /\ x e. A ) -> ( F ` x ) = ( G ` x ) )
4 3 ralrimiva
 |-  ( ph -> A. x e. A ( F ` x ) = ( G ` x ) )
5 eqfnfv
 |-  ( ( F Fn A /\ G Fn A ) -> ( F = G <-> A. x e. A ( F ` x ) = ( G ` x ) ) )
6 1 2 5 syl2anc
 |-  ( ph -> ( F = G <-> A. x e. A ( F ` x ) = ( G ` x ) ) )
7 4 6 mpbird
 |-  ( ph -> F = G )