Metamath Proof Explorer


Theorem subm0

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

Ref Expression
Hypotheses submmnd.h H=M𝑠S
subm0.z 0˙=0M
Assertion subm0 SSubMndM0˙=0H

Proof

Step Hyp Ref Expression
1 submmnd.h H=M𝑠S
2 subm0.z 0˙=0M
3 submrcl SSubMndMMMnd
4 1 submmnd SSubMndMHMnd
5 eqid BaseM=BaseM
6 5 submss SSubMndMSBaseM
7 2 subm0cl SSubMndM0˙S
8 5 2 1 submnd0 MMndHMndSBaseM0˙S0˙=0H
9 3 4 6 7 8 syl22anc SSubMndM0˙=0H