Metamath Proof Explorer


Theorem submcmn

Description: A submonoid of a commutative monoid is also commutative. (Contributed by Mario Carneiro, 24-Apr-2016)

Ref Expression
Hypothesis subgabl.h 𝐻 = ( 𝐺s 𝑆 )
Assertion submcmn ( ( 𝐺 ∈ CMnd ∧ 𝑆 ∈ ( SubMnd ‘ 𝐺 ) ) → 𝐻 ∈ CMnd )

Proof

Step Hyp Ref Expression
1 subgabl.h 𝐻 = ( 𝐺s 𝑆 )
2 1 submmnd ( 𝑆 ∈ ( SubMnd ‘ 𝐺 ) → 𝐻 ∈ Mnd )
3 1 subcmn ( ( 𝐺 ∈ CMnd ∧ 𝐻 ∈ Mnd ) → 𝐻 ∈ CMnd )
4 2 3 sylan2 ( ( 𝐺 ∈ CMnd ∧ 𝑆 ∈ ( SubMnd ‘ 𝐺 ) ) → 𝐻 ∈ CMnd )