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Metamath Proof Explorer

Theorem sdrgrcl

Description: Reverse closure for a sub-division-ring predicate. (Contributed by SN, 19-Feb-2025)

		Ref	Expression
	Assertion	sdrgrcl	$⊢ A \in SubDRing (R) \to R \in DivRing$

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	simp1bi	$⊢ A \in SubDRing (R) \to R \in DivRing$