Metamath Proof Explorer


Theorem isriscg

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

Ref Expression
Assertion isriscg RASBR𝑟SRRingOpsSRingOpsffRRngIsoS

Proof

Step Hyp Ref Expression
1 eleq1 r=RrRingOpsRRingOps
2 1 anbi1d r=RrRingOpssRingOpsRRingOpssRingOps
3 oveq1 r=RrRngIsos=RRngIsos
4 3 eleq2d r=RfrRngIsosfRRngIsos
5 4 exbidv r=RffrRngIsosffRRngIsos
6 2 5 anbi12d r=RrRingOpssRingOpsffrRngIsosRRingOpssRingOpsffRRngIsos
7 eleq1 s=SsRingOpsSRingOps
8 7 anbi2d s=SRRingOpssRingOpsRRingOpsSRingOps
9 oveq2 s=SRRngIsos=RRngIsoS
10 9 eleq2d s=SfRRngIsosfRRngIsoS
11 10 exbidv s=SffRRngIsosffRRngIsoS
12 8 11 anbi12d s=SRRingOpssRingOpsffRRngIsosRRingOpsSRingOpsffRRngIsoS
13 df-risc 𝑟=rs|rRingOpssRingOpsffrRngIsos
14 6 12 13 brabg RASBR𝑟SRRingOpsSRingOpsffRRngIsoS