| Step | Hyp | Ref
| Expression |
| 1 | | cnegex2 8205 |
. . 3
⊢ (𝐴 ∈ ℂ →
∃𝑥 ∈ ℂ
(𝑥 + 𝐴) = 0) |
| 2 | 1 | 3ad2ant1 1020 |
. 2
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) →
∃𝑥 ∈ ℂ
(𝑥 + 𝐴) = 0) |
| 3 | | oveq2 5930 |
. . . 4
⊢ ((𝐴 + 𝐵) = (𝐴 + 𝐶) → (𝑥 + (𝐴 + 𝐵)) = (𝑥 + (𝐴 + 𝐶))) |
| 4 | | simprr 531 |
. . . . . . 7
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → (𝑥 + 𝐴) = 0) |
| 5 | 4 | oveq1d 5937 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → ((𝑥 + 𝐴) + 𝐵) = (0 + 𝐵)) |
| 6 | | simprl 529 |
. . . . . . 7
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → 𝑥 ∈ ℂ) |
| 7 | | simpl1 1002 |
. . . . . . 7
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → 𝐴 ∈ ℂ) |
| 8 | | simpl2 1003 |
. . . . . . 7
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → 𝐵 ∈ ℂ) |
| 9 | 6, 7, 8 | addassd 8049 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → ((𝑥 + 𝐴) + 𝐵) = (𝑥 + (𝐴 + 𝐵))) |
| 10 | | addlid 8165 |
. . . . . . 7
⊢ (𝐵 ∈ ℂ → (0 +
𝐵) = 𝐵) |
| 11 | 8, 10 | syl 14 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → (0 + 𝐵) = 𝐵) |
| 12 | 5, 9, 11 | 3eqtr3d 2237 |
. . . . 5
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → (𝑥 + (𝐴 + 𝐵)) = 𝐵) |
| 13 | 4 | oveq1d 5937 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → ((𝑥 + 𝐴) + 𝐶) = (0 + 𝐶)) |
| 14 | | simpl3 1004 |
. . . . . . 7
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → 𝐶 ∈ ℂ) |
| 15 | 6, 7, 14 | addassd 8049 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → ((𝑥 + 𝐴) + 𝐶) = (𝑥 + (𝐴 + 𝐶))) |
| 16 | | addlid 8165 |
. . . . . . 7
⊢ (𝐶 ∈ ℂ → (0 +
𝐶) = 𝐶) |
| 17 | 14, 16 | syl 14 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → (0 + 𝐶) = 𝐶) |
| 18 | 13, 15, 17 | 3eqtr3d 2237 |
. . . . 5
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → (𝑥 + (𝐴 + 𝐶)) = 𝐶) |
| 19 | 12, 18 | eqeq12d 2211 |
. . . 4
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → ((𝑥 + (𝐴 + 𝐵)) = (𝑥 + (𝐴 + 𝐶)) ↔ 𝐵 = 𝐶)) |
| 20 | 3, 19 | imbitrid 154 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶)) |
| 21 | | oveq2 5930 |
. . 3
⊢ (𝐵 = 𝐶 → (𝐴 + 𝐵) = (𝐴 + 𝐶)) |
| 22 | 20, 21 | impbid1 142 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝑥 + 𝐴) = 0)) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) ↔ 𝐵 = 𝐶)) |
| 23 | 2, 22 | rexlimddv 2619 |
1
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) ↔ 𝐵 = 𝐶)) |