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Metamath Proof Explorer

Theorem eqfnfv

Description: Equality of functions is determined by their values. Special case of Exercise 4 of TakeutiZaring p. 28 (with domain equality omitted). (Contributed by NM, 3-Aug-1994) (Proof shortened by Andrew Salmon, 22-Oct-2011) (Proof shortened by Mario Carneiro, 31-Aug-2015)

Ref Expression
Assertion eqfnfv 
|- ( ( F Fn A /\ G Fn A ) -> ( F = G <-> A. x e. A ( F ` x ) = ( G ` x ) ) )

Proof

Step Hyp Ref Expression
1 dffn5 
 |-  ( F Fn A <-> F = ( x e. A |-> ( F ` x ) ) )
2 dffn5 
 |-  ( G Fn A <-> G = ( x e. A |-> ( G ` x ) ) )
3 eqeq12 
 |-  ( ( F = ( x e. A |-> ( F ` x ) ) /\ G = ( x e. A |-> ( G ` x ) ) ) -> ( F = G <-> ( x e. A |-> ( F ` x ) ) = ( x e. A |-> ( G ` x ) ) ) )
4 1 2 3 syl2anb 
 |-  ( ( F Fn A /\ G Fn A ) -> ( F = G <-> ( x e. A |-> ( F ` x ) ) = ( x e. A |-> ( G ` x ) ) ) )
5 fvex 
 |-  ( F ` x ) e. _V
6 5 rgenw 
 |-  A. x e. A ( F ` x ) e. _V
7 mpteqb 
 |-  ( A. x e. A ( F ` x ) e. _V -> ( ( x e. A |-> ( F ` x ) ) = ( x e. A |-> ( G ` x ) ) <-> A. x e. A ( F ` x ) = ( G ` x ) ) )
8 6 7 ax-mp 
 |-  ( ( x e. A |-> ( F ` x ) ) = ( x e. A |-> ( G ` x ) ) <-> A. x e. A ( F ` x ) = ( G ` x ) )
9 4 8 bitrdi 
 |-  ( ( F Fn A /\ G Fn A ) -> ( F = G <-> A. x e. A ( F ` x ) = ( G ` x ) ) )