Theorem subrngin

Description: The intersection of two subrings is a subring. (Contributed by AV, 15-Feb-2025)

	Ref	Expression
Assertion	subrngin	|- ( ( A e. ( SubRng ` R ) /\ B e. ( SubRng ` R ) ) -> ( A i^i B ) e. ( SubRng ` R ) )

Proof

Step	Hyp	Ref	Expression
1		intprg	|- ( ( A e. ( SubRng ` R ) /\ B e. ( SubRng ` R ) ) -> |^| { A , B } = ( A i^i B ) )
2		prssi	|- ( ( A e. ( SubRng ` R ) /\ B e. ( SubRng ` R ) ) -> { A , B } C_ ( SubRng ` R ) )
3		prnzg	|- ( A e. ( SubRng ` R ) -> { A , B } =/= (/) )
4	3	adantr	|- ( ( A e. ( SubRng ` R ) /\ B e. ( SubRng ` R ) ) -> { A , B } =/= (/) )
5		subrngint	|- ( ( { A , B } C_ ( SubRng ` R ) /\ { A , B } =/= (/) ) -> |^| { A , B } e. ( SubRng ` R ) )
6	2 4 5	syl2anc	|- ( ( A e. ( SubRng ` R ) /\ B e. ( SubRng ` R ) ) -> |^| { A , B } e. ( SubRng ` R ) )
7	1 6	eqeltrrd	|- ( ( A e. ( SubRng ` R ) /\ B e. ( SubRng ` R ) ) -> ( A i^i B ) e. ( SubRng ` R ) )