Metamath Proof Explorer


Theorem sdrgsubrg

Description: A sub-division-ring is a subring. (Contributed by SN, 19-Feb-2025)

Ref Expression
Assertion sdrgsubrg ( 𝐴 ∈ ( SubDRing ‘ 𝑅 ) → 𝐴 ∈ ( SubRing ‘ 𝑅 ) )

Proof

Step Hyp Ref Expression
1 issdrg ( 𝐴 ∈ ( SubDRing ‘ 𝑅 ) ↔ ( 𝑅 ∈ DivRing ∧ 𝐴 ∈ ( SubRing ‘ 𝑅 ) ∧ ( 𝑅s 𝐴 ) ∈ DivRing ) )
2 1 simp2bi ( 𝐴 ∈ ( SubDRing ‘ 𝑅 ) → 𝐴 ∈ ( SubRing ‘ 𝑅 ) )