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