Metamath Proof Explorer


Theorem ofsubeq0

Description: Function analogue of subeq0 . (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Assertion ofsubeq0 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F − f G = A × 0 ↔ F = G

Proof

Step Hyp Ref Expression
1 simp2 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F : A ⟶ ℂ
2 1 ffnd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F Fn A
3 simp3 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → G : A ⟶ ℂ
4 3 ffnd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → G Fn A
5 simp1 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → A ∈ V
6 inidm ⊢ A ∩ A = A
7 eqidd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F ⁡ x = F ⁡ x
8 eqidd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → G ⁡ x = G ⁡ x
9 2 4 5 5 6 7 8 ofval ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F − f G ⁡ x = F ⁡ x − G ⁡ x
10 c0ex ⊢ 0 ∈ V
11 10 fvconst2 ⊢ x ∈ A → A × 0 ⁡ x = 0
12 11 adantl ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → A × 0 ⁡ x = 0
13 9 12 eqeq12d ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F − f G ⁡ x = A × 0 ⁡ x ↔ F ⁡ x − G ⁡ x = 0
14 1 ffvelcdmda ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F ⁡ x ∈ ℂ
15 3 ffvelcdmda ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → G ⁡ x ∈ ℂ
16 14 15 subeq0ad ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F ⁡ x − G ⁡ x = 0 ↔ F ⁡ x = G ⁡ x
17 13 16 bitrd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F − f G ⁡ x = A × 0 ⁡ x ↔ F ⁡ x = G ⁡ x
18 17 ralbidva ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → ∀ x ∈ A F − f G ⁡ x = A × 0 ⁡ x ↔ ∀ x ∈ A F ⁡ x = G ⁡ x
19 2 4 5 5 6 offn ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F − f G Fn A
20 10 fconst ⊢ A × 0 : A ⟶ 0
21 ffn ⊢ A × 0 : A ⟶ 0 → A × 0 Fn A
22 20 21 ax-mp ⊢ A × 0 Fn A
23 eqfnfv ⊢ F − f G Fn A ∧ A × 0 Fn A → F − f G = A × 0 ↔ ∀ x ∈ A F − f G ⁡ x = A × 0 ⁡ x
24 19 22 23 sylancl ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F − f G = A × 0 ↔ ∀ x ∈ A F − f G ⁡ x = A × 0 ⁡ x
25 eqfnfv ⊢ F Fn A ∧ G Fn A → F = G ↔ ∀ x ∈ A F ⁡ x = G ⁡ x
26 2 4 25 syl2anc ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F = G ↔ ∀ x ∈ A F ⁡ x = G ⁡ x
27 18 24 26 3bitr4d ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F − f G = A × 0 ↔ F = G