Metamath Proof Explorer


Theorem fneqeql

Description: Two functions are equal iff their equalizer is the whole domain. (Contributed by Stefan O'Rear, 7-Mar-2015)

Ref Expression
Assertion fneqeql F Fn A G Fn A F = G dom F G = A

Proof

Step Hyp Ref Expression
1 eqfnfv F Fn A G Fn A F = G x A F x = G x
2 eqcom x A | F x = G x = A A = x A | F x = G x
3 rabid2 A = x A | F x = G x x A F x = G x
4 2 3 bitri x A | F x = G x = A x A F x = G x
5 1 4 bitr4di F Fn A G Fn A F = G x A | F x = G x = A
6 fndmin F Fn A G Fn A dom F G = x A | F x = G x
7 6 eqeq1d F Fn A G Fn A dom F G = A x A | F x = G x = A
8 5 7 bitr4d F Fn A G Fn A F = G dom F G = A