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 inss1 F G F
3 dmss F G F dom F G dom F
4 2 3 ax-mp dom F G dom F
5 fndm F Fn A dom F = A
6 5 adantr F Fn A G Fn A dom F = A
7 4 6 sseqtrid F Fn A G Fn A dom F G A
8 7 biantrurd F Fn A G Fn A A dom F G dom F G A A dom F G
9 eqss dom F G = A dom F G A A dom F G
10 8 9 syl6rbbr 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