Metamath Proof Explorer


Theorem 4sqlem2

Description: Lemma for 4sq . Change bound variables in S . (Contributed by Mario Carneiro, 14-Jul-2014)

Ref Expression
Hypothesis 4sq.1 S=n|xyzwn=x2+y2+z2+w2
Assertion 4sqlem2 ASabcdA=a2+b2+c2+d2

Proof

Step Hyp Ref Expression
1 4sq.1 S=n|xyzwn=x2+y2+z2+w2
2 1 eleq2i ASAn|xyzwn=x2+y2+z2+w2
3 id A=a2+b2+c2+d2A=a2+b2+c2+d2
4 ovex a2+b2+c2+d2V
5 3 4 eqeltrdi A=a2+b2+c2+d2AV
6 5 a1i abcdA=a2+b2+c2+d2AV
7 6 rexlimdvva abcdA=a2+b2+c2+d2AV
8 7 rexlimivv abcdA=a2+b2+c2+d2AV
9 oveq1 x=ax2=a2
10 9 oveq1d x=ax2+y2=a2+y2
11 10 oveq1d x=ax2+y2+z2+w2=a2+y2+z2+w2
12 11 eqeq2d x=an=x2+y2+z2+w2n=a2+y2+z2+w2
13 12 2rexbidv x=azwn=x2+y2+z2+w2zwn=a2+y2+z2+w2
14 oveq1 y=by2=b2
15 14 oveq2d y=ba2+y2=a2+b2
16 15 oveq1d y=ba2+y2+z2+w2=a2+b2+z2+w2
17 16 eqeq2d y=bn=a2+y2+z2+w2n=a2+b2+z2+w2
18 17 2rexbidv y=bzwn=a2+y2+z2+w2zwn=a2+b2+z2+w2
19 13 18 cbvrex2vw xyzwn=x2+y2+z2+w2abzwn=a2+b2+z2+w2
20 oveq1 z=cz2=c2
21 20 oveq1d z=cz2+w2=c2+w2
22 21 oveq2d z=ca2+b2+z2+w2=a2+b2+c2+w2
23 22 eqeq2d z=cn=a2+b2+z2+w2n=a2+b2+c2+w2
24 oveq1 w=dw2=d2
25 24 oveq2d w=dc2+w2=c2+d2
26 25 oveq2d w=da2+b2+c2+w2=a2+b2+c2+d2
27 26 eqeq2d w=dn=a2+b2+c2+w2n=a2+b2+c2+d2
28 23 27 cbvrex2vw zwn=a2+b2+z2+w2cdn=a2+b2+c2+d2
29 eqeq1 n=An=a2+b2+c2+d2A=a2+b2+c2+d2
30 29 2rexbidv n=Acdn=a2+b2+c2+d2cdA=a2+b2+c2+d2
31 28 30 bitrid n=Azwn=a2+b2+z2+w2cdA=a2+b2+c2+d2
32 31 2rexbidv n=Aabzwn=a2+b2+z2+w2abcdA=a2+b2+c2+d2
33 19 32 bitrid n=Axyzwn=x2+y2+z2+w2abcdA=a2+b2+c2+d2
34 8 33 elab3 An|xyzwn=x2+y2+z2+w2abcdA=a2+b2+c2+d2
35 2 34 bitri ASabcdA=a2+b2+c2+d2