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