Metamath Proof Explorer


Theorem subrg1cl

Description: A subring contains the multiplicative identity. (Contributed by Stefan O'Rear, 27-Nov-2014)

Ref Expression
Hypothesis subrg1cl.a 1 = ( 1r𝑅 )
Assertion subrg1cl ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 1𝐴 )

Proof

Step Hyp Ref Expression
1 subrg1cl.a 1 = ( 1r𝑅 )
2 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
3 2 1 issubrg ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) ↔ ( ( 𝑅 ∈ Ring ∧ ( 𝑅s 𝐴 ) ∈ Ring ) ∧ ( 𝐴 ⊆ ( Base ‘ 𝑅 ) ∧ 1𝐴 ) ) )
4 3 simprbi ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( 𝐴 ⊆ ( Base ‘ 𝑅 ) ∧ 1𝐴 ) )
5 4 simprd ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 1𝐴 )