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 x A F x = G x

Proof

Step Hyp Ref Expression
1 dffn5 F Fn A F = x A F x
2 dffn5 G Fn A G = x A G x
3 eqeq12 F = x A F x G = x A G x F = G x A F x = x A G x
4 1 2 3 syl2anb F Fn A G Fn A F = G x A F x = x A G x
5 fvex F x V
6 5 rgenw x A F x V
7 mpteqb x A F x V x A F x = x A G x x A F x = G x
8 6 7 ax-mp x A F x = x A G x x A F x = G x
9 4 8 bitrdi F Fn A G Fn A F = G x A F x = G x