Metamath Proof Explorer


Theorem rrxmetlem

Description: Lemma for rrxmet . (Contributed by Thierry Arnoux, 5-Jul-2019)

Ref Expression
Hypotheses rrxmval.1 X=hI|finSupp0h
rrxmval.d D=distmsup
rrxmetlem.1 φIV
rrxmetlem.2 φFX
rrxmetlem.3 φGX
rrxmetlem.4 φAI
rrxmetlem.5 φAFin
rrxmetlem.6 φsupp0Fsupp0GA
Assertion rrxmetlem φksupp0Fsupp0GFkGk2=kAFkGk2

Proof

Step Hyp Ref Expression
1 rrxmval.1 X=hI|finSupp0h
2 rrxmval.d D=distmsup
3 rrxmetlem.1 φIV
4 rrxmetlem.2 φFX
5 rrxmetlem.3 φGX
6 rrxmetlem.4 φAI
7 rrxmetlem.5 φAFin
8 rrxmetlem.6 φsupp0Fsupp0GA
9 8 6 sstrd φsupp0Fsupp0GI
10 9 sselda φksupp0Fsupp0GkI
11 1 4 rrxf φF:I
12 11 ffvelcdmda φkIFk
13 12 recnd φkIFk
14 10 13 syldan φksupp0Fsupp0GFk
15 1 5 rrxf φG:I
16 15 ffvelcdmda φkIGk
17 16 recnd φkIGk
18 10 17 syldan φksupp0Fsupp0GGk
19 14 18 subcld φksupp0Fsupp0GFkGk
20 19 sqcld φksupp0Fsupp0GFkGk2
21 6 ssdifd φAsupp0Fsupp0GIsupp0Fsupp0G
22 21 sselda φkAsupp0Fsupp0GkIsupp0Fsupp0G
23 simpr φkIsupp0Fsupp0GkIsupp0Fsupp0G
24 23 eldifad φkIsupp0Fsupp0GkI
25 24 13 syldan φkIsupp0Fsupp0GFk
26 ssun1 Fsupp0supp0Fsupp0G
27 26 a1i φFsupp0supp0Fsupp0G
28 0red φ0
29 11 27 3 28 suppssr φkIsupp0Fsupp0GFk=0
30 ssun2 Gsupp0supp0Fsupp0G
31 30 a1i φGsupp0supp0Fsupp0G
32 15 31 3 28 suppssr φkIsupp0Fsupp0GGk=0
33 29 32 eqtr4d φkIsupp0Fsupp0GFk=Gk
34 25 33 subeq0bd φkIsupp0Fsupp0GFkGk=0
35 34 sq0id φkIsupp0Fsupp0GFkGk2=0
36 22 35 syldan φkAsupp0Fsupp0GFkGk2=0
37 8 20 36 7 fsumss φksupp0Fsupp0GFkGk2=kAFkGk2