Metamath Proof Explorer


Theorem subcmn

Description: A submonoid of a commutative monoid is also commutative. (Contributed by Mario Carneiro, 10-Jan-2015)

Ref Expression
Hypothesis subgabl.h H = G 𝑠 S
Assertion subcmn G CMnd H Mnd H CMnd

Proof

Step Hyp Ref Expression
1 subgabl.h H = G 𝑠 S
2 eqidd G CMnd H Mnd Base H = Base H
3 eqid Base H = Base H
4 eqid 0 H = 0 H
5 3 4 mndidcl H Mnd 0 H Base H
6 n0i 0 H Base H ¬ Base H =
7 5 6 syl H Mnd ¬ Base H =
8 reldmress Rel dom 𝑠
9 8 ovprc2 ¬ S V G 𝑠 S =
10 1 9 eqtrid ¬ S V H =
11 10 fveq2d ¬ S V Base H = Base
12 base0 = Base
13 11 12 eqtr4di ¬ S V Base H =
14 7 13 nsyl2 H Mnd S V
15 14 adantl G CMnd H Mnd S V
16 eqid + G = + G
17 1 16 ressplusg S V + G = + H
18 15 17 syl G CMnd H Mnd + G = + H
19 simpr G CMnd H Mnd H Mnd
20 simpl G CMnd H Mnd G CMnd
21 eqid Base G = Base G
22 1 21 ressbasss Base H Base G
23 22 sseli x Base H x Base G
24 22 sseli y Base H y Base G
25 21 16 cmncom G CMnd x Base G y Base G x + G y = y + G x
26 20 23 24 25 syl3an G CMnd H Mnd x Base H y Base H x + G y = y + G x
27 2 18 19 26 iscmnd G CMnd H Mnd H CMnd