Metamath Proof Explorer


Theorem mndlsmidm

Description: Subgroup sum is idempotent for monoids. This corresponds to the observation in Lang p. 6. (Contributed by AV, 27-Dec-2023)

Ref Expression
Hypotheses mndlsmidm.p ˙ = LSSum G
mndlsmidm.b B = Base G
Assertion mndlsmidm G Mnd B ˙ B = B

Proof

Step Hyp Ref Expression
1 mndlsmidm.p ˙ = LSSum G
2 mndlsmidm.b B = Base G
3 2 submid G Mnd B SubMnd G
4 1 smndlsmidm B SubMnd G B ˙ B = B
5 3 4 syl G Mnd B ˙ B = B