Metamath Proof Explorer


Theorem submmnd

Description: Submonoids are themselves monoids under the given operation. (Contributed by Mario Carneiro, 7-Mar-2015)

Ref Expression
Hypothesis submmnd.h H=M𝑠S
Assertion submmnd SSubMndMHMnd

Proof

Step Hyp Ref Expression
1 submmnd.h H=M𝑠S
2 submrcl SSubMndMMMnd
3 eqid BaseM=BaseM
4 eqid 0M=0M
5 3 4 1 issubm2 MMndSSubMndMSBaseM0MSHMnd
6 2 5 syl SSubMndMSSubMndMSBaseM0MSHMnd
7 6 ibi SSubMndMSBaseM0MSHMnd
8 7 simp3d SSubMndMHMnd