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 = M 𝑠 S
Assertion submmnd S SubMnd M H Mnd

Proof

Step Hyp Ref Expression
1 submmnd.h H = M 𝑠 S
2 submrcl S SubMnd M M Mnd
3 eqid Base M = Base M
4 eqid 0 M = 0 M
5 3 4 1 issubm2 M Mnd S SubMnd M S Base M 0 M S H Mnd
6 2 5 syl S SubMnd M S SubMnd M S Base M 0 M S H Mnd
7 6 ibi S SubMnd M S Base M 0 M S H Mnd
8 7 simp3d S SubMnd M H Mnd