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 = ( G |`s S )
Assertion submcmn
|- ( ( G e. CMnd /\ S e. ( SubMnd ` G ) ) -> H e. CMnd )

Proof

Step Hyp Ref Expression
1 subgabl.h
 |-  H = ( G |`s S )
2 1 submmnd
 |-  ( S e. ( SubMnd ` G ) -> H e. Mnd )
3 1 subcmn
 |-  ( ( G e. CMnd /\ H e. Mnd ) -> H e. CMnd )
4 2 3 sylan2
 |-  ( ( G e. CMnd /\ S e. ( SubMnd ` G ) ) -> H e. CMnd )