Metamath Proof Explorer


Theorem smndlsmidm

Description: The direct product is idempotent for submonoids. (Contributed by AV, 27-Dec-2023)

Ref Expression
Hypothesis lsmub1.p ˙ = LSSum G
Assertion smndlsmidm U SubMnd G U ˙ U = U

Proof

Step Hyp Ref Expression
1 lsmub1.p ˙ = LSSum G
2 elfvdm U SubMnd G G dom SubMnd
3 eqid Base G = Base G
4 3 submss U SubMnd G U Base G
5 eqid + G = + G
6 3 5 1 lsmvalx G dom SubMnd U Base G U Base G U ˙ U = ran x U , y U x + G y
7 2 4 4 6 syl3anc U SubMnd G U ˙ U = ran x U , y U x + G y
8 5 submcl U SubMnd G x U y U x + G y U
9 8 3expb U SubMnd G x U y U x + G y U
10 9 ralrimivva U SubMnd G x U y U x + G y U
11 eqid x U , y U x + G y = x U , y U x + G y
12 11 fmpo x U y U x + G y U x U , y U x + G y : U × U U
13 10 12 sylib U SubMnd G x U , y U x + G y : U × U U
14 13 frnd U SubMnd G ran x U , y U x + G y U
15 7 14 eqsstrd U SubMnd G U ˙ U U
16 3 1 lsmub1x U Base G U SubMnd G U U ˙ U
17 4 16 mpancom U SubMnd G U U ˙ U
18 15 17 eqssd U SubMnd G U ˙ U = U