Theorem subrgrcl

Description: Reverse closure for a subring predicate. (Contributed by Mario Carneiro, 3-Dec-2014)

	Ref	Expression
Assertion	subrgrcl	$$⊢A\in \mathrm{SubRing}\left(R\right)\to R\in \mathrm{Ring}$$

Proof

Step	Hyp	Ref	Expression
1		eqid	$$⊢{\mathrm{Base}}_R={\mathrm{Base}}_R$$
2		eqid	$$⊢1_R=1_R$$
3	1 2	issubrg	$$⊢A\in \mathrm{SubRing}\left(R\right)\leftrightarrow \left(\left(R\in \mathrm{Ring}\wedge R{\upharpoonright }_{𝑠}A\in \mathrm{Ring}\right)\wedge \left(A\subseteq {\mathrm{Base}}_R\wedge 1_R\in A\right)\right)$$
4	3	simplbi	$$⊢A\in \mathrm{SubRing}\left(R\right)\to \left(R\in \mathrm{Ring}\wedge R{\upharpoonright }_{𝑠}A\in \mathrm{Ring}\right)$$
5	4	simpld	$$⊢A\in \mathrm{SubRing}\left(R\right)\to R\in \mathrm{Ring}$$