Metamath Proof Explorer


Theorem isrisc

Description: The ring isomorphism relation. (Contributed by Jeff Madsen, 16-Jun-2011)

Ref Expression
Hypotheses isrisc.1 𝑅 ∈ V
isrisc.2 𝑆 ∈ V
Assertion isrisc ( 𝑅𝑟 𝑆 ↔ ( ( 𝑅 ∈ RingOps ∧ 𝑆 ∈ RingOps ) ∧ ∃ 𝑓 𝑓 ∈ ( 𝑅 RngIso 𝑆 ) ) )

Proof

Step Hyp Ref Expression
1 isrisc.1 𝑅 ∈ V
2 isrisc.2 𝑆 ∈ V
3 isriscg ( ( 𝑅 ∈ V ∧ 𝑆 ∈ V ) → ( 𝑅𝑟 𝑆 ↔ ( ( 𝑅 ∈ RingOps ∧ 𝑆 ∈ RingOps ) ∧ ∃ 𝑓 𝑓 ∈ ( 𝑅 RngIso 𝑆 ) ) ) )
4 1 2 3 mp2an ( 𝑅𝑟 𝑆 ↔ ( ( 𝑅 ∈ RingOps ∧ 𝑆 ∈ RingOps ) ∧ ∃ 𝑓 𝑓 ∈ ( 𝑅 RngIso 𝑆 ) ) )