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	⊢ ( 0g ‘ 𝑀 ) = ( 0g ‘ 𝑀 )
5	3 4 1	issubm2	⊢ ( 𝑀 ∈ Mnd → ( 𝑆 ∈ ( SubMnd ‘ 𝑀 ) ↔ ( 𝑆 ⊆ ( Base ‘ 𝑀 ) ∧ ( 0g ‘ 𝑀 ) ∈ 𝑆 ∧ 𝐻 ∈ Mnd ) ) )
6	2 5	syl	⊢ ( 𝑆 ∈ ( SubMnd ‘ 𝑀 ) → ( 𝑆 ∈ ( SubMnd ‘ 𝑀 ) ↔ ( 𝑆 ⊆ ( Base ‘ 𝑀 ) ∧ ( 0g ‘ 𝑀 ) ∈ 𝑆 ∧ 𝐻 ∈ Mnd ) ) )
7	6	ibi	⊢ ( 𝑆 ∈ ( SubMnd ‘ 𝑀 ) → ( 𝑆 ⊆ ( Base ‘ 𝑀 ) ∧ ( 0g ‘ 𝑀 ) ∈ 𝑆 ∧ 𝐻 ∈ Mnd ) )
8	7	simp3d	⊢ ( 𝑆 ∈ ( SubMnd ‘ 𝑀 ) → 𝐻 ∈ Mnd )