Metamath Proof Explorer


Theorem ricrcl

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

Ref Expression
Assertion ricrcl ( 𝑅𝑟 𝑆𝑆 ∈ Ring )

Proof

Step Hyp Ref Expression
1 brric ( 𝑅𝑟 𝑆 ↔ ( 𝑅 RingIso 𝑆 ) ≠ ∅ )
2 n0 ( ( 𝑅 RingIso 𝑆 ) ≠ ∅ ↔ ∃ 𝑓 𝑓 ∈ ( 𝑅 RingIso 𝑆 ) )
3 1 2 bitri ( 𝑅𝑟 𝑆 ↔ ∃ 𝑓 𝑓 ∈ ( 𝑅 RingIso 𝑆 ) )
4 rimrcl2 ( 𝑓 ∈ ( 𝑅 RingIso 𝑆 ) → 𝑆 ∈ Ring )
5 4 exlimiv ( ∃ 𝑓 𝑓 ∈ ( 𝑅 RingIso 𝑆 ) → 𝑆 ∈ Ring )
6 3 5 sylbi ( 𝑅𝑟 𝑆𝑆 ∈ Ring )