Proof of Theorem addsubeq4
| Step | Hyp | Ref
| Expression |
| 1 | | eqcom 2083 |
. . 3
⊢ ((𝐶 − 𝐴) = (𝐵 − 𝐷) ↔ (𝐵 − 𝐷) = (𝐶 − 𝐴)) |
| 2 | | subcl 7307 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ) → (𝐶 − 𝐴) ∈ ℂ) |
| 3 | 2 | ancoms 264 |
. . . . 5
⊢ ((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐶 − 𝐴) ∈ ℂ) |
| 4 | | subadd 7311 |
. . . . . . 7
⊢ ((𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ (𝐶 − 𝐴) ∈ ℂ) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
| 5 | 4 | 3expa 1138 |
. . . . . 6
⊢ (((𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ) ∧ (𝐶 − 𝐴) ∈ ℂ) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
| 6 | 5 | ancoms 264 |
. . . . 5
⊢ (((𝐶 − 𝐴) ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
| 7 | 3, 6 | sylan 277 |
. . . 4
⊢ (((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
| 8 | 7 | an4s 552 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐵 − 𝐷) = (𝐶 − 𝐴) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
| 9 | 1, 8 | syl5bb 190 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 − 𝐴) = (𝐵 − 𝐷) ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
| 10 | | addcom 7245 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 + 𝐷) = (𝐷 + 𝐶)) |
| 11 | 10 | adantl 271 |
. . . . . 6
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐶 + 𝐷) = (𝐷 + 𝐶)) |
| 12 | 11 | oveq1d 5547 |
. . . . 5
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 + 𝐷) − 𝐴) = ((𝐷 + 𝐶) − 𝐴)) |
| 13 | | addsubass 7318 |
. . . . . . . 8
⊢ ((𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
| 14 | 13 | 3com12 1142 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
| 15 | 14 | 3expa 1138 |
. . . . . 6
⊢ (((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) ∧ 𝐴 ∈ ℂ) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
| 16 | 15 | ancoms 264 |
. . . . 5
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐷 + 𝐶) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
| 17 | 12, 16 | eqtrd 2113 |
. . . 4
⊢ ((𝐴 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 + 𝐷) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
| 18 | 17 | adantlr 460 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐶 + 𝐷) − 𝐴) = (𝐷 + (𝐶 − 𝐴))) |
| 19 | 18 | eqeq1d 2089 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) →
(((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐷 + (𝐶 − 𝐴)) = 𝐵)) |
| 20 | | addcl 7098 |
. . 3
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 + 𝐷) ∈ ℂ) |
| 21 | | subadd 7311 |
. . . . 5
⊢ (((𝐶 + 𝐷) ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
| 22 | 21 | 3expb 1139 |
. . . 4
⊢ (((𝐶 + 𝐷) ∈ ℂ ∧ (𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ)) → (((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
| 23 | 22 | ancoms 264 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 + 𝐷) ∈ ℂ) → (((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
| 24 | 20, 23 | sylan2 280 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) →
(((𝐶 + 𝐷) − 𝐴) = 𝐵 ↔ (𝐴 + 𝐵) = (𝐶 + 𝐷))) |
| 25 | 9, 19, 24 | 3bitr2rd 215 |
1
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 + 𝐵) = (𝐶 + 𝐷) ↔ (𝐶 − 𝐴) = (𝐵 − 𝐷))) |