Metamath Proof Explorer


Theorem subgid

Description: A group is a subgroup of itself. (Contributed by Mario Carneiro, 7-Dec-2014)

Ref Expression
Hypothesis issubg.b 𝐵 = ( Base ‘ 𝐺 )
Assertion subgid ( 𝐺 ∈ Grp → 𝐵 ∈ ( SubGrp ‘ 𝐺 ) )

Proof

Step Hyp Ref Expression
1 issubg.b 𝐵 = ( Base ‘ 𝐺 )
2 id ( 𝐺 ∈ Grp → 𝐺 ∈ Grp )
3 ssidd ( 𝐺 ∈ Grp → 𝐵𝐵 )
4 1 ressid ( 𝐺 ∈ Grp → ( 𝐺s 𝐵 ) = 𝐺 )
5 4 2 eqeltrd ( 𝐺 ∈ Grp → ( 𝐺s 𝐵 ) ∈ Grp )
6 1 issubg ( 𝐵 ∈ ( SubGrp ‘ 𝐺 ) ↔ ( 𝐺 ∈ Grp ∧ 𝐵𝐵 ∧ ( 𝐺s 𝐵 ) ∈ Grp ) )
7 2 3 5 6 syl3anbrc ( 𝐺 ∈ Grp → 𝐵 ∈ ( SubGrp ‘ 𝐺 ) )