Theorem subrngrcl

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

	Ref	Expression
Assertion	subrngrcl	⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝑅 ∈ Rng )

Proof

Step	Hyp	Ref	Expression
1		eqid	⊢ ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
2	1	issubrng	⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) ↔ ( 𝑅 ∈ Rng ∧ ( 𝑅 ↾s 𝐴 ) ∈ Rng ∧ 𝐴 ⊆ ( Base ‘ 𝑅 ) ) )
3	2	simp1bi	⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝑅 ∈ Rng )