Metamath Proof Explorer


Theorem ofdivrec

Description: Function analogue of divrec , a division analogue of ofnegsub . (Contributed by Steve Rodriguez, 3-Nov-2015)

Ref Expression
Assertion ofdivrec ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → F × f A × 1 ÷ f G = F ÷ f G

Proof

Step Hyp Ref Expression
1 simp1 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → A ∈ V
2 simp2 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → F : A ⟶ ℂ
3 2 ffnd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → F Fn A
4 ax-1cn ⊢ 1 ∈ ℂ
5 fnconstg ⊢ 1 ∈ ℂ → A × 1 Fn A
6 4 5 mp1i ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → A × 1 Fn A
7 simp3 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → G : A ⟶ ℂ ∖ 0
8 7 ffnd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → G Fn A
9 inidm ⊢ A ∩ A = A
10 6 8 1 1 9 offn ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → A × 1 ÷ f G Fn A
11 3 8 1 1 9 offn ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → F ÷ f G Fn A
12 eqidd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → F ⁡ x = F ⁡ x
13 1cnd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → 1 ∈ ℂ
14 eqidd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → G ⁡ x = G ⁡ x
15 1 13 8 14 ofc1 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → A × 1 ÷ f G ⁡ x = 1 G ⁡ x
16 ffvelcdm ⊢ F : A ⟶ ℂ ∧ x ∈ A → F ⁡ x ∈ ℂ
17 2 16 sylan ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → F ⁡ x ∈ ℂ
18 ffvelcdm ⊢ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → G ⁡ x ∈ ℂ ∖ 0
19 eldifsn ⊢ G ⁡ x ∈ ℂ ∖ 0 ↔ G ⁡ x ∈ ℂ ∧ G ⁡ x ≠ 0
20 18 19 sylib ⊢ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → G ⁡ x ∈ ℂ ∧ G ⁡ x ≠ 0
21 7 20 sylan ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → G ⁡ x ∈ ℂ ∧ G ⁡ x ≠ 0
22 divrec ⊢ F ⁡ x ∈ ℂ ∧ G ⁡ x ∈ ℂ ∧ G ⁡ x ≠ 0 → F ⁡ x G ⁡ x = F ⁡ x ⁢ 1 G ⁡ x
23 22 eqcomd ⊢ F ⁡ x ∈ ℂ ∧ G ⁡ x ∈ ℂ ∧ G ⁡ x ≠ 0 → F ⁡ x ⁢ 1 G ⁡ x = F ⁡ x G ⁡ x
24 23 3expb ⊢ F ⁡ x ∈ ℂ ∧ G ⁡ x ∈ ℂ ∧ G ⁡ x ≠ 0 → F ⁡ x ⁢ 1 G ⁡ x = F ⁡ x G ⁡ x
25 17 21 24 syl2anc ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → F ⁡ x ⁢ 1 G ⁡ x = F ⁡ x G ⁡ x
26 3 8 1 1 9 12 14 ofval ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → F ÷ f G ⁡ x = F ⁡ x G ⁡ x
27 25 26 eqtr4d ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 ∧ x ∈ A → F ⁡ x ⁢ 1 G ⁡ x = F ÷ f G ⁡ x
28 1 3 10 11 12 15 27 offveq ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∖ 0 → F × f A × 1 ÷ f G = F ÷ f G