Proof of Theorem add20
Step | Hyp | Ref
| Expression |
1 | | simpllr 529 |
. . . . . . . . 9
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 0 ≤ 𝐴) |
2 | | simplrl 530 |
. . . . . . . . . 10
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 𝐵 ∈ ℝ) |
3 | | simplll 528 |
. . . . . . . . . 10
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 𝐴 ∈ ℝ) |
4 | | addge02 8392 |
. . . . . . . . . 10
⊢ ((𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ) → (0 ≤
𝐴 ↔ 𝐵 ≤ (𝐴 + 𝐵))) |
5 | 2, 3, 4 | syl2anc 409 |
. . . . . . . . 9
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → (0 ≤ 𝐴 ↔ 𝐵 ≤ (𝐴 + 𝐵))) |
6 | 1, 5 | mpbid 146 |
. . . . . . . 8
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 𝐵 ≤ (𝐴 + 𝐵)) |
7 | | simpr 109 |
. . . . . . . 8
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → (𝐴 + 𝐵) = 0) |
8 | 6, 7 | breqtrd 4015 |
. . . . . . 7
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 𝐵 ≤ 0) |
9 | | simplrr 531 |
. . . . . . 7
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 0 ≤ 𝐵) |
10 | | 0red 7921 |
. . . . . . . 8
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 0 ∈
ℝ) |
11 | 2, 10 | letri3d 8035 |
. . . . . . 7
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → (𝐵 = 0 ↔ (𝐵 ≤ 0 ∧ 0 ≤ 𝐵))) |
12 | 8, 9, 11 | mpbir2and 939 |
. . . . . 6
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 𝐵 = 0) |
13 | 12 | oveq2d 5869 |
. . . . 5
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → (𝐴 + 𝐵) = (𝐴 + 0)) |
14 | 3 | recnd 7948 |
. . . . . 6
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 𝐴 ∈ ℂ) |
15 | 14 | addid1d 8068 |
. . . . 5
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → (𝐴 + 0) = 𝐴) |
16 | 13, 7, 15 | 3eqtr3rd 2212 |
. . . 4
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → 𝐴 = 0) |
17 | 16, 12 | jca 304 |
. . 3
⊢ ((((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) ∧ (𝐴 + 𝐵) = 0) → (𝐴 = 0 ∧ 𝐵 = 0)) |
18 | 17 | ex 114 |
. 2
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((𝐴 + 𝐵) = 0 → (𝐴 = 0 ∧ 𝐵 = 0))) |
19 | | oveq12 5862 |
. . 3
⊢ ((𝐴 = 0 ∧ 𝐵 = 0) → (𝐴 + 𝐵) = (0 + 0)) |
20 | | 00id 8060 |
. . 3
⊢ (0 + 0) =
0 |
21 | 19, 20 | eqtrdi 2219 |
. 2
⊢ ((𝐴 = 0 ∧ 𝐵 = 0) → (𝐴 + 𝐵) = 0) |
22 | 18, 21 | impbid1 141 |
1
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((𝐴 + 𝐵) = 0 ↔ (𝐴 = 0 ∧ 𝐵 = 0))) |