Theorem irngssv

Description: An integral element is an element of the base set. (Contributed by Thierry Arnoux, 28-Jan-2025)

	Ref	Expression
Hypotheses	irngval.o	$$⊢O=R{\mathrm{evalSub}}_{1}S$$
	irngval.u	$$⊢U=R{\upharpoonright }_{𝑠}S$$
	irngval.b	$$⊢B={\mathrm{Base}}_R$$
	irngval.0	$$⊢\underset{\dot{}}{0}=0_R$$
	elirng.r	$$⊢\phi \to R\in \mathrm{CRing}$$
	elirng.s	$$⊢\phi \to S\in \mathrm{SubRing}\left(R\right)$$
Assertion	irngssv	$$⊢\phi \to R\mathrm{IntgRing}S\subseteq B$$

Proof

Step	Hyp	Ref	Expression
1		irngval.o	$$⊢O=R{\mathrm{evalSub}}_{1}S$$
2		irngval.u	$$⊢U=R{\upharpoonright }_{𝑠}S$$
3		irngval.b	$$⊢B={\mathrm{Base}}_R$$
4		irngval.0	$$⊢\underset{\dot{}}{0}=0_R$$
5		elirng.r	$$⊢\phi \to R\in \mathrm{CRing}$$
6		elirng.s	$$⊢\phi \to S\in \mathrm{SubRing}\left(R\right)$$
7	1 2 3 4 5 6	elirng	$$⊢\phi \to \left(x\in \left(R\mathrm{IntgRing}S\right)\leftrightarrow \left(x\in B\wedge \exists f\in {Monic}_{\mathrm{1p}}\left(U\right)O\left(f\right)\left(x\right)=\underset{\dot{}}{0}\right)\right)$$
8		simpl	$$⊢\left(x\in B\wedge \exists f\in {Monic}_{\mathrm{1p}}\left(U\right)O\left(f\right)\left(x\right)=\underset{\dot{}}{0}\right)\to x\in B$$
9	7 8	syl6bi	$$⊢\phi \to \left(x\in \left(R\mathrm{IntgRing}S\right)\to x\in B\right)$$
10	9	ssrdv	$$⊢\phi \to R\mathrm{IntgRing}S\subseteq B$$