Metamath Proof Explorer


Theorem fneqeql2

Description: Two functions are equal iff their equalizer contains the whole domain. (Contributed by Stefan O'Rear, 9-Mar-2015)

Ref Expression
Assertion fneqeql2 F Fn A G Fn A F = G A dom F G

Proof

Step Hyp Ref Expression
1 fneqeql F Fn A G Fn A F = G dom F G = A
2 eqss dom F G = A dom F G A A dom F G
3 inss1 F G F
4 dmss F G F dom F G dom F
5 3 4 ax-mp dom F G dom F
6 fndm F Fn A dom F = A
7 6 adantr F Fn A G Fn A dom F = A
8 5 7 sseqtrid F Fn A G Fn A dom F G A
9 8 biantrurd F Fn A G Fn A A dom F G dom F G A A dom F G
10 2 9 bitr4id F Fn A G Fn A dom F G = A A dom F G
11 1 10 bitrd F Fn A G Fn A F = G A dom F G