Metamath Proof Explorer


Theorem cygznlem1

Description: Lemma for cygzn . (Contributed by Mario Carneiro, 21-Apr-2016)

Ref Expression
Hypotheses cygzn.b B = Base G
cygzn.n N = if B Fin B 0
cygzn.y Y = /N
cygzn.m · ˙ = G
cygzn.l L = ℤRHom Y
cygzn.e E = x B | ran n n · ˙ x = B
cygzn.g φ G CycGrp
cygzn.x φ X E
Assertion cygznlem1 φ K M L K = L M K · ˙ X = M · ˙ X

Proof

Step Hyp Ref Expression
1 cygzn.b B = Base G
2 cygzn.n N = if B Fin B 0
3 cygzn.y Y = /N
4 cygzn.m · ˙ = G
5 cygzn.l L = ℤRHom Y
6 cygzn.e E = x B | ran n n · ˙ x = B
7 cygzn.g φ G CycGrp
8 cygzn.x φ X E
9 hashcl B Fin B 0
10 9 adantl φ B Fin B 0
11 0nn0 0 0
12 11 a1i φ ¬ B Fin 0 0
13 10 12 ifclda φ if B Fin B 0 0
14 2 13 eqeltrid φ N 0
15 14 adantr φ K M N 0
16 simprl φ K M K
17 simprr φ K M M
18 3 5 zndvds N 0 K M L K = L M N K M
19 15 16 17 18 syl3anc φ K M L K = L M N K M
20 cyggrp G CycGrp G Grp
21 7 20 syl φ G Grp
22 eqid od G = od G
23 1 4 6 22 cyggenod2 G Grp X E od G X = if B Fin B 0
24 21 8 23 syl2anc φ od G X = if B Fin B 0
25 24 2 eqtr4di φ od G X = N
26 25 adantr φ K M od G X = N
27 26 breq1d φ K M od G X K M N K M
28 21 adantr φ K M G Grp
29 1 4 6 iscyggen X E X B ran n n · ˙ X = B
30 29 simplbi X E X B
31 8 30 syl φ X B
32 31 adantr φ K M X B
33 eqid 0 G = 0 G
34 1 22 4 33 odcong G Grp X B K M od G X K M K · ˙ X = M · ˙ X
35 28 32 16 17 34 syl112anc φ K M od G X K M K · ˙ X = M · ˙ X
36 19 27 35 3bitr2d φ K M L K = L M K · ˙ X = M · ˙ X