Metamath Proof Explorer


Theorem 2itscplem1

Description: Lemma 1 for 2itscp . (Contributed by AV, 4-Mar-2023)

Ref Expression
Hypotheses 2itscp.a φ A
2itscp.b φ B
2itscp.x φ X
2itscp.y φ Y
2itscp.d D = X A
2itscp.e E = B Y
Assertion 2itscplem1 φ E 2 B 2 + D 2 A 2 - 2 D A E B = D A E B 2

Proof

Step Hyp Ref Expression
1 2itscp.a φ A
2 2itscp.b φ B
3 2itscp.x φ X
4 2itscp.y φ Y
5 2itscp.d D = X A
6 2itscp.e E = B Y
7 2 recnd φ B
8 4 recnd φ Y
9 7 8 subcld φ B Y
10 6 9 eqeltrid φ E
11 10 sqcld φ E 2
12 7 sqcld φ B 2
13 11 12 mulcld φ E 2 B 2
14 3 recnd φ X
15 1 recnd φ A
16 14 15 subcld φ X A
17 5 16 eqeltrid φ D
18 17 sqcld φ D 2
19 15 sqcld φ A 2
20 18 19 mulcld φ D 2 A 2
21 2cnd φ 2
22 17 15 mulcld φ D A
23 10 7 mulcld φ E B
24 22 23 mulcld φ D A E B
25 21 24 mulcld φ 2 D A E B
26 13 20 25 addsubassd φ E 2 B 2 + D 2 A 2 - 2 D A E B = E 2 B 2 + D 2 A 2 - 2 D A E B
27 20 25 subcld φ D 2 A 2 2 D A E B
28 13 27 addcomd φ E 2 B 2 + D 2 A 2 - 2 D A E B = D 2 A 2 - 2 D A E B + E 2 B 2
29 17 15 sqmuld φ D A 2 = D 2 A 2
30 29 eqcomd φ D 2 A 2 = D A 2
31 30 oveq1d φ D 2 A 2 2 D A E B = D A 2 2 D A E B
32 10 7 sqmuld φ E B 2 = E 2 B 2
33 32 eqcomd φ E 2 B 2 = E B 2
34 31 33 oveq12d φ D 2 A 2 - 2 D A E B + E 2 B 2 = D A 2 - 2 D A E B + E B 2
35 26 28 34 3eqtrd φ E 2 B 2 + D 2 A 2 - 2 D A E B = D A 2 - 2 D A E B + E B 2
36 binom2sub D A E B D A E B 2 = D A 2 - 2 D A E B + E B 2
37 22 23 36 syl2anc φ D A E B 2 = D A 2 - 2 D A E B + E B 2
38 35 37 eqtr4d φ E 2 B 2 + D 2 A 2 - 2 D A E B = D A E B 2