Metamath Proof Explorer


Theorem subm0

Description: Submonoids have the same identity. (Contributed by Mario Carneiro, 7-Mar-2015)

Ref Expression
Hypotheses submmnd.h H = M 𝑠 S
subm0.z 0 ˙ = 0 M
Assertion subm0 S SubMnd M 0 ˙ = 0 H

Proof

Step Hyp Ref Expression
1 submmnd.h H = M 𝑠 S
2 subm0.z 0 ˙ = 0 M
3 submrcl S SubMnd M M Mnd
4 1 submmnd S SubMnd M H Mnd
5 eqid Base M = Base M
6 5 submss S SubMnd M S Base M
7 2 subm0cl S SubMnd M 0 ˙ S
8 5 2 1 submnd0 M Mnd H Mnd S Base M 0 ˙ S 0 ˙ = 0 H
9 3 4 6 7 8 syl22anc S SubMnd M 0 ˙ = 0 H