Proof of Theorem div2subap
Step | Hyp | Ref
| Expression |
1 | | subcl 8111 |
. . 3
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 − 𝐵) ∈ ℂ) |
2 | | subcl 8111 |
. . . . 5
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 − 𝐷) ∈ ℂ) |
3 | 2 | 3adant3 1012 |
. . . 4
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 − 𝐷) ∈ ℂ) |
4 | | apneg 8523 |
. . . . . . . 8
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 # 𝐷 ↔ -𝐶 # -𝐷)) |
5 | 4 | biimp3a 1340 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -𝐶 # -𝐷) |
6 | | simp1 992 |
. . . . . . . . 9
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 𝐶 ∈ ℂ) |
7 | 6 | negcld 8210 |
. . . . . . . 8
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -𝐶 ∈ ℂ) |
8 | | simp2 993 |
. . . . . . . . 9
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 𝐷 ∈ ℂ) |
9 | 8 | negcld 8210 |
. . . . . . . 8
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -𝐷 ∈ ℂ) |
10 | | apadd2 8521 |
. . . . . . . 8
⊢ ((-𝐶 ∈ ℂ ∧ -𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (-𝐶 # -𝐷 ↔ (𝐶 + -𝐶) # (𝐶 + -𝐷))) |
11 | 7, 9, 6, 10 | syl3anc 1233 |
. . . . . . 7
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (-𝐶 # -𝐷 ↔ (𝐶 + -𝐶) # (𝐶 + -𝐷))) |
12 | 5, 11 | mpbid 146 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 + -𝐶) # (𝐶 + -𝐷)) |
13 | 6 | negidd 8213 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 + -𝐶) = 0) |
14 | 6, 8 | negsubd 8229 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 + -𝐷) = (𝐶 − 𝐷)) |
15 | 12, 13, 14 | 3brtr3d 4018 |
. . . . 5
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 0 # (𝐶 − 𝐷)) |
16 | | 0cnd 7906 |
. . . . . 6
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → 0 ∈ ℂ) |
17 | | apsym 8518 |
. . . . . 6
⊢ ((0
∈ ℂ ∧ (𝐶
− 𝐷) ∈ ℂ)
→ (0 # (𝐶 −
𝐷) ↔ (𝐶 − 𝐷) # 0)) |
18 | 16, 3, 17 | syl2anc 409 |
. . . . 5
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (0 # (𝐶 − 𝐷) ↔ (𝐶 − 𝐷) # 0)) |
19 | 15, 18 | mpbid 146 |
. . . 4
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → (𝐶 − 𝐷) # 0) |
20 | 3, 19 | jca 304 |
. . 3
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → ((𝐶 − 𝐷) ∈ ℂ ∧ (𝐶 − 𝐷) # 0)) |
21 | | div2negap 8645 |
. . . 4
⊢ (((𝐴 − 𝐵) ∈ ℂ ∧ (𝐶 − 𝐷) ∈ ℂ ∧ (𝐶 − 𝐷) # 0) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐴 − 𝐵) / (𝐶 − 𝐷))) |
22 | 21 | 3expb 1199 |
. . 3
⊢ (((𝐴 − 𝐵) ∈ ℂ ∧ ((𝐶 − 𝐷) ∈ ℂ ∧ (𝐶 − 𝐷) # 0)) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐴 − 𝐵) / (𝐶 − 𝐷))) |
23 | 1, 20, 22 | syl2an 287 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷)) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐴 − 𝐵) / (𝐶 − 𝐷))) |
24 | | negsubdi2 8171 |
. . 3
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → -(𝐴 − 𝐵) = (𝐵 − 𝐴)) |
25 | | negsubdi2 8171 |
. . . 4
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → -(𝐶 − 𝐷) = (𝐷 − 𝐶)) |
26 | 25 | 3adant3 1012 |
. . 3
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷) → -(𝐶 − 𝐷) = (𝐷 − 𝐶)) |
27 | 24, 26 | oveqan12d 5870 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷)) → (-(𝐴 − 𝐵) / -(𝐶 − 𝐷)) = ((𝐵 − 𝐴) / (𝐷 − 𝐶))) |
28 | 23, 27 | eqtr3d 2205 |
1
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 # 𝐷)) → ((𝐴 − 𝐵) / (𝐶 − 𝐷)) = ((𝐵 − 𝐴) / (𝐷 − 𝐶))) |