Metamath Proof Explorer


Theorem bj-bary1lem1

Description: Lemma for bj-bary1 : computation of one of the two barycentric coordinates of a barycenter of two points in one dimension (complex line). (Contributed by BJ, 6-Jun-2019)

Ref Expression
Hypotheses bj-bary1.a ⊢ φ → A ∈ ℂ
bj-bary1.b ⊢ φ → B ∈ ℂ
bj-bary1.x ⊢ φ → X ∈ ℂ
bj-bary1.neq ⊢ φ → A ≠ B
bj-bary1.s ⊢ φ → S ∈ ℂ
bj-bary1.t ⊢ φ → T ∈ ℂ
Assertion bj-bary1lem1 ⊢ φ → X = S ⁢ A + T ⁢ B ∧ S + T = 1 → T = X − A B − A

Proof

Step Hyp Ref Expression
1 bj-bary1.a ⊢ φ → A ∈ ℂ
2 bj-bary1.b ⊢ φ → B ∈ ℂ
3 bj-bary1.x ⊢ φ → X ∈ ℂ
4 bj-bary1.neq ⊢ φ → A ≠ B
5 bj-bary1.s ⊢ φ → S ∈ ℂ
6 bj-bary1.t ⊢ φ → T ∈ ℂ
7 5 6 pncand ⊢ φ → S + T - T = S
8 oveq1 ⊢ S + T = 1 → S + T - T = 1 − T
9 pm5.31 ⊢ S + T - T = S ∧ S + T = 1 → S + T - T = 1 − T → S + T = 1 → S + T - T = 1 − T ∧ S + T - T = S
10 7 8 9 sylancl ⊢ φ → S + T = 1 → S + T - T = 1 − T ∧ S + T - T = S
11 eqtr2 ⊢ S + T - T = 1 − T ∧ S + T - T = S → 1 − T = S
12 11 eqcomd ⊢ S + T - T = 1 − T ∧ S + T - T = S → S = 1 − T
13 10 12 syl6 ⊢ φ → S + T = 1 → S = 1 − T
14 oveq1 ⊢ S = 1 − T → S ⁢ A = 1 − T ⁢ A
15 14 oveq1d ⊢ S = 1 − T → S ⁢ A + T ⁢ B = 1 − T ⁢ A + T ⁢ B
16 eqtr ⊢ X = S ⁢ A + T ⁢ B ∧ S ⁢ A + T ⁢ B = 1 − T ⁢ A + T ⁢ B → X = 1 − T ⁢ A + T ⁢ B
17 15 16 sylan2 ⊢ X = S ⁢ A + T ⁢ B ∧ S = 1 − T → X = 1 − T ⁢ A + T ⁢ B
18 1cnd ⊢ φ → 1 ∈ ℂ
19 18 6 1 subdird ⊢ φ → 1 − T ⁢ A = 1 ⁢ A − T ⁢ A
20 1 mullidd ⊢ φ → 1 ⁢ A = A
21 20 oveq1d ⊢ φ → 1 ⁢ A − T ⁢ A = A − T ⁢ A
22 19 21 eqtrd ⊢ φ → 1 − T ⁢ A = A − T ⁢ A
23 22 oveq1d ⊢ φ → 1 − T ⁢ A + T ⁢ B = A - T ⁢ A + T ⁢ B
24 17 23 sylan9eqr ⊢ φ ∧ X = S ⁢ A + T ⁢ B ∧ S = 1 − T → X = A - T ⁢ A + T ⁢ B
25 24 ex ⊢ φ → X = S ⁢ A + T ⁢ B ∧ S = 1 − T → X = A - T ⁢ A + T ⁢ B
26 13 25 sylan2d ⊢ φ → X = S ⁢ A + T ⁢ B ∧ S + T = 1 → X = A - T ⁢ A + T ⁢ B
27 6 1 mulcld ⊢ φ → T ⁢ A ∈ ℂ
28 6 2 mulcld ⊢ φ → T ⁢ B ∈ ℂ
29 1 27 28 subadd23d ⊢ φ → A - T ⁢ A + T ⁢ B = A + T ⁢ B - T ⁢ A
30 6 2 1 subdid ⊢ φ → T ⁢ B − A = T ⁢ B − T ⁢ A
31 30 eqcomd ⊢ φ → T ⁢ B − T ⁢ A = T ⁢ B − A
32 31 oveq2d ⊢ φ → A + T ⁢ B - T ⁢ A = A + T ⁢ B − A
33 29 32 eqtrd ⊢ φ → A - T ⁢ A + T ⁢ B = A + T ⁢ B − A
34 33 eqeq2d ⊢ φ → X = A - T ⁢ A + T ⁢ B ↔ X = A + T ⁢ B − A
35 26 34 sylibd ⊢ φ → X = S ⁢ A + T ⁢ B ∧ S + T = 1 → X = A + T ⁢ B − A
36 oveq1 ⊢ X = A + T ⁢ B − A → X − A = A + T ⁢ B − A - A
37 2 1 subcld ⊢ φ → B − A ∈ ℂ
38 6 37 mulcld ⊢ φ → T ⁢ B − A ∈ ℂ
39 1 38 pncan2d ⊢ φ → A + T ⁢ B − A - A = T ⁢ B − A
40 39 eqeq2d ⊢ φ → X − A = A + T ⁢ B − A - A ↔ X − A = T ⁢ B − A
41 36 40 imbitrid ⊢ φ → X = A + T ⁢ B − A → X − A = T ⁢ B − A
42 eqcom ⊢ X − A = T ⁢ B − A ↔ T ⁢ B − A = X − A
43 6 37 mulcomd ⊢ φ → T ⁢ B − A = B − A ⁢ T
44 43 eqeq1d ⊢ φ → T ⁢ B − A = X − A ↔ B − A ⁢ T = X − A
45 3 1 subcld ⊢ φ → X − A ∈ ℂ
46 4 necomd ⊢ φ → B ≠ A
47 2 1 46 subne0d ⊢ φ → B − A ≠ 0
48 37 6 45 47 rdiv ⊢ φ → B − A ⁢ T = X − A ↔ T = X − A B − A
49 48 biimpd ⊢ φ → B − A ⁢ T = X − A → T = X − A B − A
50 44 49 sylbid ⊢ φ → T ⁢ B − A = X − A → T = X − A B − A
51 42 50 biimtrid ⊢ φ → X − A = T ⁢ B − A → T = X − A B − A
52 35 41 51 3syld ⊢ φ → X = S ⁢ A + T ⁢ B ∧ S + T = 1 → T = X − A B − A