Metamath Proof Explorer


Theorem 2itscplem2

Description: Lemma 2 for 2itscp . (Contributed by AV, 4-Mar-2023)

Ref Expression
Hypotheses 2itscp.a ⊢ φ → A ∈ ℝ
2itscp.b ⊢ φ → B ∈ ℝ
2itscp.x ⊢ φ → X ∈ ℝ
2itscp.y ⊢ φ → Y ∈ ℝ
2itscp.d ⊢ D = X − A
2itscp.e ⊢ E = B − Y
2itscp.c ⊢ C = D ⁢ B + E ⁢ A
Assertion 2itscplem2 ⊢ φ → C 2 = D 2 ⁢ B 2 + 2 ⁢ D ⁢ A ⁢ E ⁢ B + E 2 ⁢ A 2

Proof

Step Hyp Ref Expression
1 2itscp.a ⊢ φ → A ∈ ℝ
2 2itscp.b ⊢ φ → B ∈ ℝ
3 2itscp.x ⊢ φ → X ∈ ℝ
4 2itscp.y ⊢ φ → Y ∈ ℝ
5 2itscp.d ⊢ D = X − A
6 2itscp.e ⊢ E = B − Y
7 2itscp.c ⊢ C = D ⁢ B + E ⁢ A
8 7 oveq1i ⊢ C 2 = D ⁢ B + E ⁢ A 2
9 8 a1i ⊢ φ → C 2 = D ⁢ B + E ⁢ A 2
10 3 recnd ⊢ φ → X ∈ ℂ
11 1 recnd ⊢ φ → A ∈ ℂ
12 10 11 subcld ⊢ φ → X − A ∈ ℂ
13 5 12 eqeltrid ⊢ φ → D ∈ ℂ
14 2 recnd ⊢ φ → B ∈ ℂ
15 13 14 mulcld ⊢ φ → D ⁢ B ∈ ℂ
16 4 recnd ⊢ φ → Y ∈ ℂ
17 14 16 subcld ⊢ φ → B − Y ∈ ℂ
18 6 17 eqeltrid ⊢ φ → E ∈ ℂ
19 18 11 mulcld ⊢ φ → E ⁢ A ∈ ℂ
20 binom2 ⊢ D ⁢ B ∈ ℂ ∧ E ⁢ A ∈ ℂ → D ⁢ B + E ⁢ A 2 = D ⁢ B 2 + 2 ⁢ D ⁢ B ⁢ E ⁢ A + E ⁢ A 2
21 15 19 20 syl2anc ⊢ φ → D ⁢ B + E ⁢ A 2 = D ⁢ B 2 + 2 ⁢ D ⁢ B ⁢ E ⁢ A + E ⁢ A 2
22 13 14 sqmuld ⊢ φ → D ⁢ B 2 = D 2 ⁢ B 2
23 mul4r ⊢ D ∈ ℂ ∧ B ∈ ℂ ∧ E ∈ ℂ ∧ A ∈ ℂ → D ⁢ B ⁢ E ⁢ A = D ⁢ A ⁢ E ⁢ B
24 13 14 18 11 23 syl22anc ⊢ φ → D ⁢ B ⁢ E ⁢ A = D ⁢ A ⁢ E ⁢ B
25 24 oveq2d ⊢ φ → 2 ⁢ D ⁢ B ⁢ E ⁢ A = 2 ⁢ D ⁢ A ⁢ E ⁢ B
26 22 25 oveq12d ⊢ φ → D ⁢ B 2 + 2 ⁢ D ⁢ B ⁢ E ⁢ A = D 2 ⁢ B 2 + 2 ⁢ D ⁢ A ⁢ E ⁢ B
27 18 11 sqmuld ⊢ φ → E ⁢ A 2 = E 2 ⁢ A 2
28 26 27 oveq12d ⊢ φ → D ⁢ B 2 + 2 ⁢ D ⁢ B ⁢ E ⁢ A + E ⁢ A 2 = D 2 ⁢ B 2 + 2 ⁢ D ⁢ A ⁢ E ⁢ B + E 2 ⁢ A 2
29 9 21 28 3eqtrd ⊢ φ → C 2 = D 2 ⁢ B 2 + 2 ⁢ D ⁢ A ⁢ E ⁢ B + E 2 ⁢ A 2