Metamath Proof Explorer


Theorem ghmcmn

Description: The image of a commutative monoid G under a group homomorphism F is a commutative monoid. (Contributed by Thierry Arnoux, 26-Jan-2020)

Ref Expression
Hypotheses ghmabl.x X=BaseG
ghmabl.y Y=BaseH
ghmabl.p +˙=+G
ghmabl.q ˙=+H
ghmabl.f φxXyXFx+˙y=Fx˙Fy
ghmabl.1 φF:XontoY
ghmcmn.3 φGCMnd
Assertion ghmcmn φHCMnd

Proof

Step Hyp Ref Expression
1 ghmabl.x X=BaseG
2 ghmabl.y Y=BaseH
3 ghmabl.p +˙=+G
4 ghmabl.q ˙=+H
5 ghmabl.f φxXyXFx+˙y=Fx˙Fy
6 ghmabl.1 φF:XontoY
7 ghmcmn.3 φGCMnd
8 cmnmnd GCMndGMnd
9 7 8 syl φGMnd
10 5 1 2 3 4 6 9 mhmmnd φHMnd
11 simp-6l φiYjYaXFa=ibXFb=jφ
12 11 7 syl φiYjYaXFa=ibXFb=jGCMnd
13 simp-4r φiYjYaXFa=ibXFb=jaX
14 simplr φiYjYaXFa=ibXFb=jbX
15 1 3 cmncom GCMndaXbXa+˙b=b+˙a
16 12 13 14 15 syl3anc φiYjYaXFa=ibXFb=ja+˙b=b+˙a
17 16 fveq2d φiYjYaXFa=ibXFb=jFa+˙b=Fb+˙a
18 11 5 syl3an1 φiYjYaXFa=ibXFb=jxXyXFx+˙y=Fx˙Fy
19 18 13 14 mhmlem φiYjYaXFa=ibXFb=jFa+˙b=Fa˙Fb
20 18 14 13 mhmlem φiYjYaXFa=ibXFb=jFb+˙a=Fb˙Fa
21 17 19 20 3eqtr3d φiYjYaXFa=ibXFb=jFa˙Fb=Fb˙Fa
22 simpllr φiYjYaXFa=ibXFb=jFa=i
23 simpr φiYjYaXFa=ibXFb=jFb=j
24 22 23 oveq12d φiYjYaXFa=ibXFb=jFa˙Fb=i˙j
25 23 22 oveq12d φiYjYaXFa=ibXFb=jFb˙Fa=j˙i
26 21 24 25 3eqtr3d φiYjYaXFa=ibXFb=ji˙j=j˙i
27 foelcdmi F:XontoYjYbXFb=j
28 6 27 sylan φjYbXFb=j
29 28 ad5ant13 φiYjYaXFa=ibXFb=j
30 26 29 r19.29a φiYjYaXFa=ii˙j=j˙i
31 foelcdmi F:XontoYiYaXFa=i
32 6 31 sylan φiYaXFa=i
33 32 adantr φiYjYaXFa=i
34 30 33 r19.29a φiYjYi˙j=j˙i
35 34 anasss φiYjYi˙j=j˙i
36 35 ralrimivva φiYjYi˙j=j˙i
37 2 4 iscmn HCMndHMndiYjYi˙j=j˙i
38 10 36 37 sylanbrc φHCMnd