Proof of Theorem bj-bary1
Step | Hyp | Ref
| Expression |
1 | | bj-bary1.s |
. . . . . . . . 9
⊢ (𝜑 → 𝑆 ∈ ℂ) |
2 | | bj-bary1.a |
. . . . . . . . 9
⊢ (𝜑 → 𝐴 ∈ ℂ) |
3 | 1, 2 | mulcld 10995 |
. . . . . . . 8
⊢ (𝜑 → (𝑆 · 𝐴) ∈ ℂ) |
4 | | bj-bary1.t |
. . . . . . . . 9
⊢ (𝜑 → 𝑇 ∈ ℂ) |
5 | | bj-bary1.b |
. . . . . . . . 9
⊢ (𝜑 → 𝐵 ∈ ℂ) |
6 | 4, 5 | mulcld 10995 |
. . . . . . . 8
⊢ (𝜑 → (𝑇 · 𝐵) ∈ ℂ) |
7 | 3, 6 | addcomd 11177 |
. . . . . . 7
⊢ (𝜑 → ((𝑆 · 𝐴) + (𝑇 · 𝐵)) = ((𝑇 · 𝐵) + (𝑆 · 𝐴))) |
8 | 7 | eqeq2d 2749 |
. . . . . 6
⊢ (𝜑 → (𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) ↔ 𝑋 = ((𝑇 · 𝐵) + (𝑆 · 𝐴)))) |
9 | 8 | biimpd 228 |
. . . . 5
⊢ (𝜑 → (𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) → 𝑋 = ((𝑇 · 𝐵) + (𝑆 · 𝐴)))) |
10 | 1, 4 | addcomd 11177 |
. . . . . . 7
⊢ (𝜑 → (𝑆 + 𝑇) = (𝑇 + 𝑆)) |
11 | 10 | eqeq1d 2740 |
. . . . . 6
⊢ (𝜑 → ((𝑆 + 𝑇) = 1 ↔ (𝑇 + 𝑆) = 1)) |
12 | 11 | biimpd 228 |
. . . . 5
⊢ (𝜑 → ((𝑆 + 𝑇) = 1 → (𝑇 + 𝑆) = 1)) |
13 | | bj-bary1.x |
. . . . . 6
⊢ (𝜑 → 𝑋 ∈ ℂ) |
14 | | bj-bary1.neq |
. . . . . . 7
⊢ (𝜑 → 𝐴 ≠ 𝐵) |
15 | 14 | necomd 2999 |
. . . . . 6
⊢ (𝜑 → 𝐵 ≠ 𝐴) |
16 | 5, 2, 13, 15, 4, 1 | bj-bary1lem1 35482 |
. . . . 5
⊢ (𝜑 → ((𝑋 = ((𝑇 · 𝐵) + (𝑆 · 𝐴)) ∧ (𝑇 + 𝑆) = 1) → 𝑆 = ((𝑋 − 𝐵) / (𝐴 − 𝐵)))) |
17 | 9, 12, 16 | syl2and 608 |
. . . 4
⊢ (𝜑 → ((𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) ∧ (𝑆 + 𝑇) = 1) → 𝑆 = ((𝑋 − 𝐵) / (𝐴 − 𝐵)))) |
18 | 13, 5, 2, 5, 14 | div2subd 11801 |
. . . . 5
⊢ (𝜑 → ((𝑋 − 𝐵) / (𝐴 − 𝐵)) = ((𝐵 − 𝑋) / (𝐵 − 𝐴))) |
19 | 18 | eqeq2d 2749 |
. . . 4
⊢ (𝜑 → (𝑆 = ((𝑋 − 𝐵) / (𝐴 − 𝐵)) ↔ 𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)))) |
20 | 17, 19 | sylibd 238 |
. . 3
⊢ (𝜑 → ((𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) ∧ (𝑆 + 𝑇) = 1) → 𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)))) |
21 | 2, 5, 13, 14, 1, 4 | bj-bary1lem1 35482 |
. . 3
⊢ (𝜑 → ((𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) ∧ (𝑆 + 𝑇) = 1) → 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴)))) |
22 | 20, 21 | jcad 513 |
. 2
⊢ (𝜑 → ((𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) ∧ (𝑆 + 𝑇) = 1) → (𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))))) |
23 | 2, 5, 13, 14 | bj-bary1lem 35481 |
. . . 4
⊢ (𝜑 → 𝑋 = ((((𝐵 − 𝑋) / (𝐵 − 𝐴)) · 𝐴) + (((𝑋 − 𝐴) / (𝐵 − 𝐴)) · 𝐵))) |
24 | | oveq1 7282 |
. . . . . 6
⊢ (𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) → (𝑆 · 𝐴) = (((𝐵 − 𝑋) / (𝐵 − 𝐴)) · 𝐴)) |
25 | | oveq1 7282 |
. . . . . 6
⊢ (𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴)) → (𝑇 · 𝐵) = (((𝑋 − 𝐴) / (𝐵 − 𝐴)) · 𝐵)) |
26 | 24, 25 | oveqan12d 7294 |
. . . . 5
⊢ ((𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))) → ((𝑆 · 𝐴) + (𝑇 · 𝐵)) = ((((𝐵 − 𝑋) / (𝐵 − 𝐴)) · 𝐴) + (((𝑋 − 𝐴) / (𝐵 − 𝐴)) · 𝐵))) |
27 | 26 | a1i 11 |
. . . 4
⊢ (𝜑 → ((𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))) → ((𝑆 · 𝐴) + (𝑇 · 𝐵)) = ((((𝐵 − 𝑋) / (𝐵 − 𝐴)) · 𝐴) + (((𝑋 − 𝐴) / (𝐵 − 𝐴)) · 𝐵)))) |
28 | | eqtr3 2764 |
. . . 4
⊢ ((𝑋 = ((((𝐵 − 𝑋) / (𝐵 − 𝐴)) · 𝐴) + (((𝑋 − 𝐴) / (𝐵 − 𝐴)) · 𝐵)) ∧ ((𝑆 · 𝐴) + (𝑇 · 𝐵)) = ((((𝐵 − 𝑋) / (𝐵 − 𝐴)) · 𝐴) + (((𝑋 − 𝐴) / (𝐵 − 𝐴)) · 𝐵))) → 𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵))) |
29 | 23, 27, 28 | syl6an 681 |
. . 3
⊢ (𝜑 → ((𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))) → 𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)))) |
30 | | oveq12 7284 |
. . . 4
⊢ ((𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))) → (𝑆 + 𝑇) = (((𝐵 − 𝑋) / (𝐵 − 𝐴)) + ((𝑋 − 𝐴) / (𝐵 − 𝐴)))) |
31 | 5, 13 | subcld 11332 |
. . . . . . 7
⊢ (𝜑 → (𝐵 − 𝑋) ∈ ℂ) |
32 | 13, 2 | subcld 11332 |
. . . . . . 7
⊢ (𝜑 → (𝑋 − 𝐴) ∈ ℂ) |
33 | 5, 2 | subcld 11332 |
. . . . . . 7
⊢ (𝜑 → (𝐵 − 𝐴) ∈ ℂ) |
34 | 5, 2, 15 | subne0d 11341 |
. . . . . . 7
⊢ (𝜑 → (𝐵 − 𝐴) ≠ 0) |
35 | 31, 32, 33, 34 | divdird 11789 |
. . . . . 6
⊢ (𝜑 → (((𝐵 − 𝑋) + (𝑋 − 𝐴)) / (𝐵 − 𝐴)) = (((𝐵 − 𝑋) / (𝐵 − 𝐴)) + ((𝑋 − 𝐴) / (𝐵 − 𝐴)))) |
36 | 5, 13, 2 | npncand 11356 |
. . . . . . 7
⊢ (𝜑 → ((𝐵 − 𝑋) + (𝑋 − 𝐴)) = (𝐵 − 𝐴)) |
37 | 33, 34, 36 | diveq1bd 11799 |
. . . . . 6
⊢ (𝜑 → (((𝐵 − 𝑋) + (𝑋 − 𝐴)) / (𝐵 − 𝐴)) = 1) |
38 | 35, 37 | eqtr3d 2780 |
. . . . 5
⊢ (𝜑 → (((𝐵 − 𝑋) / (𝐵 − 𝐴)) + ((𝑋 − 𝐴) / (𝐵 − 𝐴))) = 1) |
39 | 38 | eqeq2d 2749 |
. . . 4
⊢ (𝜑 → ((𝑆 + 𝑇) = (((𝐵 − 𝑋) / (𝐵 − 𝐴)) + ((𝑋 − 𝐴) / (𝐵 − 𝐴))) ↔ (𝑆 + 𝑇) = 1)) |
40 | 30, 39 | syl5ib 243 |
. . 3
⊢ (𝜑 → ((𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))) → (𝑆 + 𝑇) = 1)) |
41 | 29, 40 | jcad 513 |
. 2
⊢ (𝜑 → ((𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))) → (𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) ∧ (𝑆 + 𝑇) = 1))) |
42 | 22, 41 | impbid 211 |
1
⊢ (𝜑 → ((𝑋 = ((𝑆 · 𝐴) + (𝑇 · 𝐵)) ∧ (𝑆 + 𝑇) = 1) ↔ (𝑆 = ((𝐵 − 𝑋) / (𝐵 − 𝐴)) ∧ 𝑇 = ((𝑋 − 𝐴) / (𝐵 − 𝐴))))) |