Metamath Proof Explorer


Theorem addinvcom

Description: A number commutes with its additive inverse. Compare remulinvcom . (Contributed by SN, 5-May-2024)

Ref Expression
Hypotheses addinvcom.a ⊢ φ → A ∈ ℂ
addinvcom.b ⊢ φ → B ∈ ℂ
addinvcom.1 ⊢ φ → A + B = 0
Assertion addinvcom ⊢ φ → B + A = 0

Proof

Step Hyp Ref Expression
1 addinvcom.a ⊢ φ → A ∈ ℂ
2 addinvcom.b ⊢ φ → B ∈ ℂ
3 addinvcom.1 ⊢ φ → A + B = 0
4 ssidd ⊢ φ → ℂ ⊆ ℂ
5 simpl ⊢ A + x = 0 ∧ x + A = 0 → A + x = 0
6 5 rgenw ⊢ ∀ x ∈ ℂ A + x = 0 ∧ x + A = 0 → A + x = 0
7 6 a1i ⊢ φ → ∀ x ∈ ℂ A + x = 0 ∧ x + A = 0 → A + x = 0
8 sn-negex12 ⊢ A ∈ ℂ → ∃ x ∈ ℂ A + x = 0 ∧ x + A = 0
9 1 8 syl ⊢ φ → ∃ x ∈ ℂ A + x = 0 ∧ x + A = 0
10 0cn ⊢ 0 ∈ ℂ
11 sn-subeu ⊢ A ∈ ℂ ∧ 0 ∈ ℂ → ∃! x ∈ ℂ A + x = 0
12 1 10 11 sylancl ⊢ φ → ∃! x ∈ ℂ A + x = 0
13 riotass2 ⊢ ℂ ⊆ ℂ ∧ ∀ x ∈ ℂ A + x = 0 ∧ x + A = 0 → A + x = 0 ∧ ∃ x ∈ ℂ A + x = 0 ∧ x + A = 0 ∧ ∃! x ∈ ℂ A + x = 0 → ι x ∈ ℂ | A + x = 0 ∧ x + A = 0 = ι x ∈ ℂ | A + x = 0
14 4 7 9 12 13 syl22anc ⊢ φ → ι x ∈ ℂ | A + x = 0 ∧ x + A = 0 = ι x ∈ ℂ | A + x = 0
15 oveq2 ⊢ x = B → A + x = A + B
16 15 eqeq1d ⊢ x = B → A + x = 0 ↔ A + B = 0
17 16 riota2 ⊢ B ∈ ℂ ∧ ∃! x ∈ ℂ A + x = 0 → A + B = 0 ↔ ι x ∈ ℂ | A + x = 0 = B
18 2 12 17 syl2anc ⊢ φ → A + B = 0 ↔ ι x ∈ ℂ | A + x = 0 = B
19 3 18 mpbid ⊢ φ → ι x ∈ ℂ | A + x = 0 = B
20 14 19 eqtrd ⊢ φ → ι x ∈ ℂ | A + x = 0 ∧ x + A = 0 = B
21 reurmo ⊢ ∃! x ∈ ℂ A + x = 0 → ∃* x ∈ ℂ A + x = 0
22 5 rmoimi ⊢ ∃* x ∈ ℂ A + x = 0 → ∃* x ∈ ℂ A + x = 0 ∧ x + A = 0
23 12 21 22 3syl ⊢ φ → ∃* x ∈ ℂ A + x = 0 ∧ x + A = 0
24 reu5 ⊢ ∃! x ∈ ℂ A + x = 0 ∧ x + A = 0 ↔ ∃ x ∈ ℂ A + x = 0 ∧ x + A = 0 ∧ ∃* x ∈ ℂ A + x = 0 ∧ x + A = 0
25 9 23 24 sylanbrc ⊢ φ → ∃! x ∈ ℂ A + x = 0 ∧ x + A = 0
26 oveq1 ⊢ x = B → x + A = B + A
27 26 eqeq1d ⊢ x = B → x + A = 0 ↔ B + A = 0
28 16 27 anbi12d ⊢ x = B → A + x = 0 ∧ x + A = 0 ↔ A + B = 0 ∧ B + A = 0
29 28 riota2 ⊢ B ∈ ℂ ∧ ∃! x ∈ ℂ A + x = 0 ∧ x + A = 0 → A + B = 0 ∧ B + A = 0 ↔ ι x ∈ ℂ | A + x = 0 ∧ x + A = 0 = B
30 2 25 29 syl2anc ⊢ φ → A + B = 0 ∧ B + A = 0 ↔ ι x ∈ ℂ | A + x = 0 ∧ x + A = 0 = B
31 20 30 mpbird ⊢ φ → A + B = 0 ∧ B + A = 0
32 31 simprd ⊢ φ → B + A = 0