Theorem subrngrcl

Description: Reverse closure for a subring predicate. (Contributed by AV, 14-Feb-2025)

	Ref	Expression
Assertion	subrngrcl	$$⊢A\in \mathrm{SubRng}\left(R\right)\to R\in \mathrm{Rng}$$

Proof

Step	Hyp	Ref	Expression
1		eqid	$$⊢{\mathrm{Base}}_R={\mathrm{Base}}_R$$
2	1	issubrng	$$⊢A\in \mathrm{SubRng}\left(R\right)\leftrightarrow \left(R\in \mathrm{Rng}\wedge R{\upharpoonright }_{𝑠}A\in \mathrm{Rng}\wedge A\subseteq {\mathrm{Base}}_R\right)$$
3	2	simp1bi	$$⊢A\in \mathrm{SubRng}\left(R\right)\to R\in \mathrm{Rng}$$