Metamath Proof Explorer


Theorem submgmid

Description: Every magma is trivially a submagma of itself. (Contributed by AV, 26-Feb-2020)

Ref Expression
Hypothesis submgmss.b 𝐵 = ( Base ‘ 𝑀 )
Assertion submgmid ( 𝑀 ∈ Mgm → 𝐵 ∈ ( SubMgm ‘ 𝑀 ) )

Proof

Step Hyp Ref Expression
1 submgmss.b 𝐵 = ( Base ‘ 𝑀 )
2 ssidd ( 𝑀 ∈ Mgm → 𝐵𝐵 )
3 1 ressid ( 𝑀 ∈ Mgm → ( 𝑀s 𝐵 ) = 𝑀 )
4 id ( 𝑀 ∈ Mgm → 𝑀 ∈ Mgm )
5 3 4 eqeltrd ( 𝑀 ∈ Mgm → ( 𝑀s 𝐵 ) ∈ Mgm )
6 eqid ( 𝑀s 𝐵 ) = ( 𝑀s 𝐵 )
7 1 6 issubmgm2 ( 𝑀 ∈ Mgm → ( 𝐵 ∈ ( SubMgm ‘ 𝑀 ) ↔ ( 𝐵𝐵 ∧ ( 𝑀s 𝐵 ) ∈ Mgm ) ) )
8 2 5 7 mpbir2and ( 𝑀 ∈ Mgm → 𝐵 ∈ ( SubMgm ‘ 𝑀 ) )