Theorem riclcl

Description: Ring isomorphism implies the left side is a ring. (Contributed by AV, 23-Jul-2026)

	Ref	Expression
Assertion	riclcl	$$⊢R{\simeq }_{𝑟}S\to R\in \mathrm{Ring}$$

Proof

Step	Hyp	Ref	Expression
1		brric	$$⊢R{\simeq }_{𝑟}S\leftrightarrow R\mathrm{RingIso}S\ne \varnothing$$
2		n0	$$⊢R\mathrm{RingIso}S\ne \varnothing \leftrightarrow \exists ff\in \left(R\mathrm{RingIso}S\right)$$
3	1 2	bitri	$$⊢R{\simeq }_{𝑟}S\leftrightarrow \exists ff\in \left(R\mathrm{RingIso}S\right)$$
4		rimrcl1	$$⊢f\in \left(R\mathrm{RingIso}S\right)\to R\in \mathrm{Ring}$$
5	4	exlimiv	$$⊢\exists ff\in \left(R\mathrm{RingIso}S\right)\to R\in \mathrm{Ring}$$
6	3 5	sylbi	$$⊢R{\simeq }_{𝑟}S\to R\in \mathrm{Ring}$$