Metamath Proof Explorer


Theorem submid

Description: Every monoid is trivially a submonoid of itself. (Contributed by Stefan O'Rear, 15-Aug-2015)

Ref Expression
Hypothesis submss.b 𝐵 = ( Base ‘ 𝑀 )
Assertion submid ( 𝑀 ∈ Mnd → 𝐵 ∈ ( SubMnd ‘ 𝑀 ) )

Proof

Step Hyp Ref Expression
1 submss.b 𝐵 = ( Base ‘ 𝑀 )
2 ssidd ( 𝑀 ∈ Mnd → 𝐵𝐵 )
3 eqid ( 0g𝑀 ) = ( 0g𝑀 )
4 1 3 mndidcl ( 𝑀 ∈ Mnd → ( 0g𝑀 ) ∈ 𝐵 )
5 1 ressid ( 𝑀 ∈ Mnd → ( 𝑀s 𝐵 ) = 𝑀 )
6 id ( 𝑀 ∈ Mnd → 𝑀 ∈ Mnd )
7 5 6 eqeltrd ( 𝑀 ∈ Mnd → ( 𝑀s 𝐵 ) ∈ Mnd )
8 eqid ( 𝑀s 𝐵 ) = ( 𝑀s 𝐵 )
9 1 3 8 issubm2 ( 𝑀 ∈ Mnd → ( 𝐵 ∈ ( SubMnd ‘ 𝑀 ) ↔ ( 𝐵𝐵 ∧ ( 0g𝑀 ) ∈ 𝐵 ∧ ( 𝑀s 𝐵 ) ∈ Mnd ) ) )
10 2 4 7 9 mpbir3and ( 𝑀 ∈ Mnd → 𝐵 ∈ ( SubMnd ‘ 𝑀 ) )