Metamath Proof Explorer


Theorem subsubmgm

Description: A submagma of a submagma is a submagma. (Contributed by AV, 26-Feb-2020)

Ref Expression
Hypothesis subsubmgm.h H = G 𝑠 S
Assertion subsubmgm S SubMgm G A SubMgm H A SubMgm G A S

Proof

Step Hyp Ref Expression
1 subsubmgm.h H = G 𝑠 S
2 eqid Base H = Base H
3 2 submgmss A SubMgm H A Base H
4 3 adantl S SubMgm G A SubMgm H A Base H
5 1 submgmbas S SubMgm G S = Base H
6 5 adantr S SubMgm G A SubMgm H S = Base H
7 4 6 sseqtrrd S SubMgm G A SubMgm H A S
8 eqid Base G = Base G
9 8 submgmss S SubMgm G S Base G
10 9 adantr S SubMgm G A SubMgm H S Base G
11 7 10 sstrd S SubMgm G A SubMgm H A Base G
12 1 oveq1i H 𝑠 A = G 𝑠 S 𝑠 A
13 ressabs S SubMgm G A S G 𝑠 S 𝑠 A = G 𝑠 A
14 12 13 syl5eq S SubMgm G A S H 𝑠 A = G 𝑠 A
15 7 14 syldan S SubMgm G A SubMgm H H 𝑠 A = G 𝑠 A
16 eqid H 𝑠 A = H 𝑠 A
17 16 submgmmgm A SubMgm H H 𝑠 A Mgm
18 17 adantl S SubMgm G A SubMgm H H 𝑠 A Mgm
19 15 18 eqeltrrd S SubMgm G A SubMgm H G 𝑠 A Mgm
20 submgmrcl S SubMgm G G Mgm
21 20 adantr S SubMgm G A SubMgm H G Mgm
22 eqid G 𝑠 A = G 𝑠 A
23 8 22 issubmgm2 G Mgm A SubMgm G A Base G G 𝑠 A Mgm
24 21 23 syl S SubMgm G A SubMgm H A SubMgm G A Base G G 𝑠 A Mgm
25 11 19 24 mpbir2and S SubMgm G A SubMgm H A SubMgm G
26 25 7 jca S SubMgm G A SubMgm H A SubMgm G A S
27 simprr S SubMgm G A SubMgm G A S A S
28 5 adantr S SubMgm G A SubMgm G A S S = Base H
29 27 28 sseqtrd S SubMgm G A SubMgm G A S A Base H
30 14 adantrl S SubMgm G A SubMgm G A S H 𝑠 A = G 𝑠 A
31 22 submgmmgm A SubMgm G G 𝑠 A Mgm
32 31 ad2antrl S SubMgm G A SubMgm G A S G 𝑠 A Mgm
33 30 32 eqeltrd S SubMgm G A SubMgm G A S H 𝑠 A Mgm
34 1 submgmmgm S SubMgm G H Mgm
35 34 adantr S SubMgm G A SubMgm G A S H Mgm
36 2 16 issubmgm2 H Mgm A SubMgm H A Base H H 𝑠 A Mgm
37 35 36 syl S SubMgm G A SubMgm G A S A SubMgm H A Base H H 𝑠 A Mgm
38 29 33 37 mpbir2and S SubMgm G A SubMgm G A S A SubMgm H
39 26 38 impbida S SubMgm G A SubMgm H A SubMgm G A S