Metamath Proof Explorer


Theorem rrxmetlem

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

Ref Expression
Hypotheses rrxmval.1 X = h I | finSupp 0 h
rrxmval.d D = dist I
rrxmetlem.1 φ I V
rrxmetlem.2 φ F X
rrxmetlem.3 φ G X
rrxmetlem.4 φ A I
rrxmetlem.5 φ A Fin
rrxmetlem.6 φ supp 0 F supp 0 G A
Assertion rrxmetlem φ k supp 0 F supp 0 G F k G k 2 = k A F k G k 2

Proof

Step Hyp Ref Expression
1 rrxmval.1 X = h I | finSupp 0 h
2 rrxmval.d D = dist I
3 rrxmetlem.1 φ I V
4 rrxmetlem.2 φ F X
5 rrxmetlem.3 φ G X
6 rrxmetlem.4 φ A I
7 rrxmetlem.5 φ A Fin
8 rrxmetlem.6 φ supp 0 F supp 0 G A
9 8 6 sstrd φ supp 0 F supp 0 G I
10 9 sselda φ k supp 0 F supp 0 G k I
11 1 4 rrxf φ F : I
12 11 ffvelrnda φ k I F k
13 12 recnd φ k I F k
14 10 13 syldan φ k supp 0 F supp 0 G F k
15 1 5 rrxf φ G : I
16 15 ffvelrnda φ k I G k
17 16 recnd φ k I G k
18 10 17 syldan φ k supp 0 F supp 0 G G k
19 14 18 subcld φ k supp 0 F supp 0 G F k G k
20 19 sqcld φ k supp 0 F supp 0 G F k G k 2
21 6 ssdifd φ A supp 0 F supp 0 G I supp 0 F supp 0 G
22 21 sselda φ k A supp 0 F supp 0 G k I supp 0 F supp 0 G
23 simpr φ k I supp 0 F supp 0 G k I supp 0 F supp 0 G
24 23 eldifad φ k I supp 0 F supp 0 G k I
25 24 13 syldan φ k I supp 0 F supp 0 G F k
26 ssun1 F supp 0 supp 0 F supp 0 G
27 26 a1i φ F supp 0 supp 0 F supp 0 G
28 0red φ 0
29 11 27 3 28 suppssr φ k I supp 0 F supp 0 G F k = 0
30 ssun2 G supp 0 supp 0 F supp 0 G
31 30 a1i φ G supp 0 supp 0 F supp 0 G
32 15 31 3 28 suppssr φ k I supp 0 F supp 0 G G k = 0
33 29 32 eqtr4d φ k I supp 0 F supp 0 G F k = G k
34 25 33 subeq0bd φ k I supp 0 F supp 0 G F k G k = 0
35 34 sq0id φ k I supp 0 F supp 0 G F k G k 2 = 0
36 22 35 syldan φ k A supp 0 F supp 0 G F k G k 2 = 0
37 8 20 36 7 fsumss φ k supp 0 F supp 0 G F k G k 2 = k A F k G k 2