Metamath Proof Explorer

Theorem sdrgsubrg

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

		Ref	Expression
	Assertion	sdrgsubrg	$⊢ A \in SubDRing (R) \to A \in SubRing (R)$

Step	Hyp	Ref	Expression
1		issdrg	$⊢ A \in SubDRing (R) \leftrightarrow (R \in DivRing \land A \in SubRing (R) \land R ↾_{𝑠} A \in DivRing)$
2	1	simp2bi	$⊢ A \in SubDRing (R) \to A \in SubRing (R)$