Metamath Proof Explorer


Theorem subsubg

Description: A subgroup of a subgroup is a subgroup. (Contributed by Mario Carneiro, 19-Jan-2015)

Ref Expression
Hypothesis subsubg.h H=G𝑠S
Assertion subsubg SSubGrpGASubGrpHASubGrpGAS

Proof

Step Hyp Ref Expression
1 subsubg.h H=G𝑠S
2 subgrcl SSubGrpGGGrp
3 2 adantr SSubGrpGASubGrpHGGrp
4 eqid BaseH=BaseH
5 4 subgss ASubGrpHABaseH
6 5 adantl SSubGrpGASubGrpHABaseH
7 1 subgbas SSubGrpGS=BaseH
8 7 adantr SSubGrpGASubGrpHS=BaseH
9 6 8 sseqtrrd SSubGrpGASubGrpHAS
10 eqid BaseG=BaseG
11 10 subgss SSubGrpGSBaseG
12 11 adantr SSubGrpGASubGrpHSBaseG
13 9 12 sstrd SSubGrpGASubGrpHABaseG
14 1 oveq1i H𝑠A=G𝑠S𝑠A
15 ressabs SSubGrpGASG𝑠S𝑠A=G𝑠A
16 14 15 eqtrid SSubGrpGASH𝑠A=G𝑠A
17 9 16 syldan SSubGrpGASubGrpHH𝑠A=G𝑠A
18 eqid H𝑠A=H𝑠A
19 18 subggrp ASubGrpHH𝑠AGrp
20 19 adantl SSubGrpGASubGrpHH𝑠AGrp
21 17 20 eqeltrrd SSubGrpGASubGrpHG𝑠AGrp
22 10 issubg ASubGrpGGGrpABaseGG𝑠AGrp
23 3 13 21 22 syl3anbrc SSubGrpGASubGrpHASubGrpG
24 23 9 jca SSubGrpGASubGrpHASubGrpGAS
25 1 subggrp SSubGrpGHGrp
26 25 adantr SSubGrpGASubGrpGASHGrp
27 simprr SSubGrpGASubGrpGASAS
28 7 adantr SSubGrpGASubGrpGASS=BaseH
29 27 28 sseqtrd SSubGrpGASubGrpGASABaseH
30 16 adantrl SSubGrpGASubGrpGASH𝑠A=G𝑠A
31 eqid G𝑠A=G𝑠A
32 31 subggrp ASubGrpGG𝑠AGrp
33 32 ad2antrl SSubGrpGASubGrpGASG𝑠AGrp
34 30 33 eqeltrd SSubGrpGASubGrpGASH𝑠AGrp
35 4 issubg ASubGrpHHGrpABaseHH𝑠AGrp
36 26 29 34 35 syl3anbrc SSubGrpGASubGrpGASASubGrpH
37 24 36 impbida SSubGrpGASubGrpHASubGrpGAS