Metamath Proof Explorer
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 ) |