Theorem subrgss

Description: A subring is a subset. (Contributed by Stefan O'Rear, 27-Nov-2014)

	Ref	Expression
Hypothesis	subrgss.1	$$⊢B={\mathrm{Base}}_R$$
Assertion	subrgss	$$⊢A\in \mathrm{SubRing}\left(R\right)\to A\subseteq B$$

Proof

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