# 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 syl6bb
` |-  ( ( F Fn A /\ G Fn A ) -> ( F = G <-> A. x e. A ( F ` x ) = ( G ` x ) ) )`