Proof of Theorem div2subap
| Step | Hyp | Ref
| Expression |
| 1 | | subcl 8242 |
. . 3
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 − 𝐵) ∈ ℂ) |
| 2 | | subcl 8242 |
. . . . 5
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 − 𝐷) ∈ ℂ) |
| 3 | 2 | 3adant3 1019 |
. . . 4
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 − 𝐷) ∈ ℂ) |
| 4 | | apneg 8655 |
. . . . . . . 8
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 # 𝐷 ↔ -𝐶 # -𝐷)) |
| 5 | 4 | biimp3a 1356 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -𝐶 # -𝐷) |
| 6 | | simp1 999 |
. . . . . . . . 9
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 𝐶 ∈ ℂ) |
| 7 | 6 | negcld 8341 |
. . . . . . . 8
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -𝐶 ∈ ℂ) |
| 8 | | simp2 1000 |
. . . . . . . . 9
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 𝐷 ∈ ℂ) |
| 9 | 8 | negcld 8341 |
. . . . . . . 8
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -𝐷 ∈ ℂ) |
| 10 | | apadd2 8653 |
. . . . . . . 8
⊢ ((-𝐶 ∈ ℂ ∧ -𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (-𝐶 # -𝐷 ↔ (𝐶 + -𝐶) # (𝐶 + -𝐷))) |
| 11 | 7, 9, 6, 10 | syl3anc 1249 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (-𝐶 # -𝐷 ↔ (𝐶 + -𝐶) # (𝐶 + -𝐷))) |
| 12 | 5, 11 | mpbid 147 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 + -𝐶) # (𝐶 + -𝐷)) |
| 13 | 6 | negidd 8344 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 + -𝐶) = 0) |
| 14 | 6, 8 | negsubd 8360 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 + -𝐷) = (𝐶 − 𝐷)) |
| 15 | 12, 13, 14 | 3brtr3d 4065 |
. . . . 5
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 0 # (𝐶 − 𝐷)) |
| 16 | | 0cnd 8036 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 0 ∈ ℂ) |
| 17 | | apsym 8650 |
. . . . . 6
⊢ ((0
∈ ℂ ∧ (𝐶
− 𝐷) ∈ ℂ)
→ (0 # (𝐶 −
𝐷) ↔ (𝐶 − 𝐷) # 0)) |
| 18 | 16, 3, 17 | syl2anc 411 |
. . . . 5
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (0 # (𝐶 − 𝐷) ↔ (𝐶 − 𝐷) # 0)) |
| 19 | 15, 18 | mpbid 147 |
. . . 4
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 − 𝐷) # 0) |
| 20 | 3, 19 | jca 306 |
. . 3
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → ((𝐶 − 𝐷) ∈ ℂ ∧ (𝐶 − 𝐷) # 0)) |
| 21 | | div2negap 8779 |
. . . 4
⊢ (((𝐴 − 𝐵) ∈ ℂ ∧ (𝐶 − 𝐷) ∈ ℂ ∧ (𝐶 − 𝐷) # 0) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐴 − 𝐵) / (𝐶 − 𝐷))) |
| 22 | 21 | 3expb 1206 |
. . 3
⊢ (((𝐴 − 𝐵) ∈ ℂ ∧ ((𝐶 − 𝐷) ∈ ℂ ∧ (𝐶 − 𝐷) # 0)) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐴 − 𝐵) / (𝐶 − 𝐷))) |
| 23 | 1, 20, 22 | syl2an 289 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷)) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐴 − 𝐵) / (𝐶 − 𝐷))) |
| 24 | | negsubdi2 8302 |
. . 3
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → -(𝐴 − 𝐵) = (𝐵 − 𝐴)) |
| 25 | | negsubdi2 8302 |
. . . 4
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → -(𝐶 − 𝐷) = (𝐷 − 𝐶)) |
| 26 | 25 | 3adant3 1019 |
. . 3
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -(𝐶 − 𝐷) = (𝐷 − 𝐶)) |
| 27 | 24, 26 | oveqan12d 5944 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷)) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐵 − 𝐴) / (𝐷 − 𝐶))) |
| 28 | 23, 27 | eqtr3d 2231 |
1
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷)) → ((𝐴 − 𝐵) / (𝐶 − 𝐷)) = ((𝐵 − 𝐴) / (𝐷 − 𝐶))) |