Metamath Proof Explorer


Theorem ofnegsub

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

Ref Expression
Assertion ofnegsub ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F + f A × − 1 × f G = F − f G

Proof

Step Hyp Ref Expression
1 simp1 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → A ∈ V
2 simp2 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F : A ⟶ ℂ
3 2 ffnd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F Fn A
4 ax-1cn ⊢ 1 ∈ ℂ
5 4 negcli ⊢ − 1 ∈ ℂ
6 fnconstg ⊢ − 1 ∈ ℂ → A × − 1 Fn A
7 5 6 mp1i ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → A × − 1 Fn A
8 simp3 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → G : A ⟶ ℂ
9 8 ffnd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → G Fn A
10 inidm ⊢ A ∩ A = A
11 7 9 1 1 10 offn ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → A × − 1 × f G Fn A
12 3 9 1 1 10 offn ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F − f G Fn A
13 eqidd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F ⁡ x = F ⁡ x
14 5 a1i ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → − 1 ∈ ℂ
15 eqidd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → G ⁡ x = G ⁡ x
16 1 14 9 15 ofc1 ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → A × − 1 × f G ⁡ x = -1 ⁢ G ⁡ x
17 8 ffvelcdmda ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → G ⁡ x ∈ ℂ
18 17 mulm1d ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → -1 ⁢ G ⁡ x = − G ⁡ x
19 16 18 eqtrd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → A × − 1 × f G ⁡ x = − G ⁡ x
20 2 ffvelcdmda ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F ⁡ x ∈ ℂ
21 20 17 negsubd ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F ⁡ x + − G ⁡ x = F ⁡ x − G ⁡ x
22 3 9 1 1 10 13 15 ofval ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F − f G ⁡ x = F ⁡ x − G ⁡ x
23 21 22 eqtr4d ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ ∧ x ∈ A → F ⁡ x + − G ⁡ x = F − f G ⁡ x
24 1 3 11 12 13 19 23 offveq ⊢ A ∈ V ∧ F : A ⟶ ℂ ∧ G : A ⟶ ℂ → F + f A × − 1 × f G = F − f G