Metamath Proof Explorer


Theorem isriscg

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

Ref Expression
Assertion isriscg RASBR𝑟SRRingOpsSRingOpsffRRingOpsIsoS

Proof

Step Hyp Ref Expression
1 eleq1 r=RrRingOpsRRingOps
2 1 anbi1d r=RrRingOpssRingOpsRRingOpssRingOps
3 oveq1 r=RrRingOpsIsos=RRingOpsIsos
4 3 eleq2d r=RfrRingOpsIsosfRRingOpsIsos
5 4 exbidv r=RffrRingOpsIsosffRRingOpsIsos
6 2 5 anbi12d r=RrRingOpssRingOpsffrRingOpsIsosRRingOpssRingOpsffRRingOpsIsos
7 eleq1 s=SsRingOpsSRingOps
8 7 anbi2d s=SRRingOpssRingOpsRRingOpsSRingOps
9 oveq2 s=SRRingOpsIsos=RRingOpsIsoS
10 9 eleq2d s=SfRRingOpsIsosfRRingOpsIsoS
11 10 exbidv s=SffRRingOpsIsosffRRingOpsIsoS
12 8 11 anbi12d s=SRRingOpssRingOpsffRRingOpsIsosRRingOpsSRingOpsffRRingOpsIsoS
13 df-risc 𝑟=rs|rRingOpssRingOpsffrRingOpsIsos
14 6 12 13 brabg RASBR𝑟SRRingOpsSRingOpsffRRingOpsIsoS