Metamath Proof Explorer


Theorem submmulg

Description: A group multiple is the same if evaluated in a submonoid. (Contributed by Mario Carneiro, 15-Jun-2015)

Ref Expression
Hypotheses submmulgcl.t ˙ = G
submmulg.h H = G 𝑠 S
submmulg.t · ˙ = H
Assertion submmulg S SubMnd G N 0 X S N ˙ X = N · ˙ X

Proof

Step Hyp Ref Expression
1 submmulgcl.t ˙ = G
2 submmulg.h H = G 𝑠 S
3 submmulg.t · ˙ = H
4 simpl1 S SubMnd G N 0 X S N S SubMnd G
5 eqid + G = + G
6 2 5 ressplusg S SubMnd G + G = + H
7 4 6 syl S SubMnd G N 0 X S N + G = + H
8 7 seqeq2d S SubMnd G N 0 X S N seq 1 + G × X = seq 1 + H × X
9 8 fveq1d S SubMnd G N 0 X S N seq 1 + G × X N = seq 1 + H × X N
10 simpr S SubMnd G N 0 X S N N
11 eqid Base G = Base G
12 11 submss S SubMnd G S Base G
13 12 3ad2ant1 S SubMnd G N 0 X S S Base G
14 simp3 S SubMnd G N 0 X S X S
15 13 14 sseldd S SubMnd G N 0 X S X Base G
16 15 adantr S SubMnd G N 0 X S N X Base G
17 eqid seq 1 + G × X = seq 1 + G × X
18 11 5 1 17 mulgnn N X Base G N ˙ X = seq 1 + G × X N
19 10 16 18 syl2anc S SubMnd G N 0 X S N N ˙ X = seq 1 + G × X N
20 2 submbas S SubMnd G S = Base H
21 20 3ad2ant1 S SubMnd G N 0 X S S = Base H
22 14 21 eleqtrd S SubMnd G N 0 X S X Base H
23 22 adantr S SubMnd G N 0 X S N X Base H
24 eqid Base H = Base H
25 eqid + H = + H
26 eqid seq 1 + H × X = seq 1 + H × X
27 24 25 3 26 mulgnn N X Base H N · ˙ X = seq 1 + H × X N
28 10 23 27 syl2anc S SubMnd G N 0 X S N N · ˙ X = seq 1 + H × X N
29 9 19 28 3eqtr4d S SubMnd G N 0 X S N N ˙ X = N · ˙ X
30 simpl1 S SubMnd G N 0 X S N = 0 S SubMnd G
31 eqid 0 G = 0 G
32 2 31 subm0 S SubMnd G 0 G = 0 H
33 30 32 syl S SubMnd G N 0 X S N = 0 0 G = 0 H
34 15 adantr S SubMnd G N 0 X S N = 0 X Base G
35 11 31 1 mulg0 X Base G 0 ˙ X = 0 G
36 34 35 syl S SubMnd G N 0 X S N = 0 0 ˙ X = 0 G
37 22 adantr S SubMnd G N 0 X S N = 0 X Base H
38 eqid 0 H = 0 H
39 24 38 3 mulg0 X Base H 0 · ˙ X = 0 H
40 37 39 syl S SubMnd G N 0 X S N = 0 0 · ˙ X = 0 H
41 33 36 40 3eqtr4d S SubMnd G N 0 X S N = 0 0 ˙ X = 0 · ˙ X
42 simpr S SubMnd G N 0 X S N = 0 N = 0
43 42 oveq1d S SubMnd G N 0 X S N = 0 N ˙ X = 0 ˙ X
44 42 oveq1d S SubMnd G N 0 X S N = 0 N · ˙ X = 0 · ˙ X
45 41 43 44 3eqtr4d S SubMnd G N 0 X S N = 0 N ˙ X = N · ˙ X
46 simp2 S SubMnd G N 0 X S N 0
47 elnn0 N 0 N N = 0
48 46 47 sylib S SubMnd G N 0 X S N N = 0
49 29 45 48 mpjaodan S SubMnd G N 0 X S N ˙ X = N · ˙ X