Metamath Proof Explorer


Theorem fsneq

Description: Equality condition for two functions defined on a singleton. (Contributed by Glauco Siliprandi, 3-Mar-2021)

Ref Expression
Hypotheses fsneq.a φ A V
fsneq.b B = A
fsneq.f φ F Fn B
fsneq.g φ G Fn B
Assertion fsneq φ F = G F A = G A

Proof

Step Hyp Ref Expression
1 fsneq.a φ A V
2 fsneq.b B = A
3 fsneq.f φ F Fn B
4 fsneq.g φ G Fn B
5 eqfnfv F Fn B G Fn B F = G x B F x = G x
6 3 4 5 syl2anc φ F = G x B F x = G x
7 snidg A V A A
8 1 7 syl φ A A
9 2 eqcomi A = B
10 9 a1i φ A = B
11 8 10 eleqtrd φ A B
12 11 adantr φ x B F x = G x A B
13 simpr φ x B F x = G x x B F x = G x
14 fveq2 x = A F x = F A
15 fveq2 x = A G x = G A
16 14 15 eqeq12d x = A F x = G x F A = G A
17 16 rspcva A B x B F x = G x F A = G A
18 12 13 17 syl2anc φ x B F x = G x F A = G A
19 18 ex φ x B F x = G x F A = G A
20 simpl F A = G A x B F A = G A
21 2 eleq2i x B x A
22 21 biimpi x B x A
23 velsn x A x = A
24 22 23 sylib x B x = A
25 24 fveq2d x B F x = F A
26 25 adantl F A = G A x B F x = F A
27 24 fveq2d x B G x = G A
28 27 adantl F A = G A x B G x = G A
29 20 26 28 3eqtr4d F A = G A x B F x = G x
30 29 adantll φ F A = G A x B F x = G x
31 30 ralrimiva φ F A = G A x B F x = G x
32 31 ex φ F A = G A x B F x = G x
33 19 32 impbid φ x B F x = G x F A = G A
34 6 33 bitrd φ F = G F A = G A