Proof of Theorem addsubeq4
Step | Hyp | Ref
| Expression |
1 | | eqcom 2042 |
. . 3
⊢ ((𝐶 − 𝐴) = (𝐵 − 𝐷) ↔ (𝐵 − 𝐷) = (𝐶 − 𝐴)) |
2 | | subcl 7210 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ) → (𝐶 − 𝐴) ∈ ℂ) |
3 | 2 | ancoms 255 |
. . . . 5
⊢ ((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐶 − 𝐴) ∈ ℂ) |
4 | | subadd 7214 |
. . . . . . 7
⊢ ((𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ (𝐶 − 𝐴) ∈ ℂ) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
5 | 4 | 3expa 1104 |
. . . . . 6
⊢ (((𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ) ∧ (𝐶 − 𝐴) ∈ ℂ) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
6 | 5 | ancoms 255 |
. . . . 5
⊢ (((𝐶 − 𝐴) ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
7 | 3, 6 | sylan 267 |
. . . 4
⊢ (((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
8 | 7 | an4s 522 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
9 | 1, 8 | syl5bb 181 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 − 𝐴) = (𝐵 − 𝐷) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
10 | | addcom 7150 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 + 𝐷) = (𝐷 + 𝐶)) |
11 | 10 | adantl 262 |
. . . . . 6
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐶 + 𝐷) = (𝐷 + 𝐶)) |
12 | 11 | oveq1d 5527 |
. . . . 5
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 + 𝐷) − 𝐴) = ((𝐷 + 𝐶) − 𝐴)) |
13 | | addsubass 7221 |
. . . . . . . 8
⊢ ((𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
14 | 13 | 3com12 1108 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
15 | 14 | 3expa 1104 |
. . . . . 6
⊢ (((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) ∧ 𝐴 ∈ ℂ) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
16 | 15 | ancoms 255 |
. . . . 5
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
17 | 12, 16 | eqtrd 2072 |
. . . 4
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 + 𝐷) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
18 | 17 | adantlr 446 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 + 𝐷) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
19 | 18 | eqeq1d 2048 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) →
(((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
20 | | addcl 7006 |
. . 3
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 + 𝐷) ∈ ℂ) |
21 | | subadd 7214 |
. . . . 5
⊢ (((𝐶 + 𝐷) ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
22 | 21 | 3expb 1105 |
. . . 4
⊢ (((𝐶 + 𝐷) ∈ ℂ ∧ (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ)) → (((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
23 | 22 | ancoms 255 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 + 𝐷) ∈ ℂ) → (((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
24 | 20, 23 | sylan2 270 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) →
(((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
25 | 9, 19, 24 | 3bitr2rd 206 |
1
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 + 𝐵) = (𝐶 + 𝐷) ↔ (𝐶 − 𝐴) = (𝐵 − 𝐷))) |