Metamath Proof Explorer


Theorem 2itscplem1

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

Ref Expression
Hypotheses 2itscp.a φA
2itscp.b φB
2itscp.x φX
2itscp.y φY
2itscp.d D=XA
2itscp.e E=BY
Assertion 2itscplem1 φE2B2+D2A2-2DAEB=DAEB2

Proof

Step Hyp Ref Expression
1 2itscp.a φA
2 2itscp.b φB
3 2itscp.x φX
4 2itscp.y φY
5 2itscp.d D=XA
6 2itscp.e E=BY
7 2 recnd φB
8 4 recnd φY
9 7 8 subcld φBY
10 6 9 eqeltrid φE
11 10 sqcld φE2
12 7 sqcld φB2
13 11 12 mulcld φE2B2
14 3 recnd φX
15 1 recnd φA
16 14 15 subcld φXA
17 5 16 eqeltrid φD
18 17 sqcld φD2
19 15 sqcld φA2
20 18 19 mulcld φD2A2
21 2cnd φ2
22 17 15 mulcld φDA
23 10 7 mulcld φEB
24 22 23 mulcld φDAEB
25 21 24 mulcld φ2DAEB
26 13 20 25 addsubassd φE2B2+D2A2-2DAEB=E2B2+D2A2-2DAEB
27 20 25 subcld φD2A22DAEB
28 13 27 addcomd φE2B2+D2A2-2DAEB=D2A2-2DAEB+E2B2
29 17 15 sqmuld φDA2=D2A2
30 29 eqcomd φD2A2=DA2
31 30 oveq1d φD2A22DAEB=DA22DAEB
32 10 7 sqmuld φEB2=E2B2
33 32 eqcomd φE2B2=EB2
34 31 33 oveq12d φD2A2-2DAEB+E2B2=DA2-2DAEB+EB2
35 26 28 34 3eqtrd φE2B2+D2A2-2DAEB=DA2-2DAEB+EB2
36 binom2sub DAEBDAEB2=DA2-2DAEB+EB2
37 22 23 36 syl2anc φDAEB2=DA2-2DAEB+EB2
38 35 37 eqtr4d φE2B2+D2A2-2DAEB=DAEB2