Metamath Proof Explorer


Theorem isriscg

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

Ref Expression
Assertion isriscg R A S B R 𝑟 S R RingOps S RingOps f f R RingOpsIso S

Proof

Step Hyp Ref Expression
1 eleq1 r = R r RingOps R RingOps
2 1 anbi1d r = R r RingOps s RingOps R RingOps s RingOps
3 oveq1 r = R r RingOpsIso s = R RingOpsIso s
4 3 eleq2d r = R f r RingOpsIso s f R RingOpsIso s
5 4 exbidv r = R f f r RingOpsIso s f f R RingOpsIso s
6 2 5 anbi12d r = R r RingOps s RingOps f f r RingOpsIso s R RingOps s RingOps f f R RingOpsIso s
7 eleq1 s = S s RingOps S RingOps
8 7 anbi2d s = S R RingOps s RingOps R RingOps S RingOps
9 oveq2 s = S R RingOpsIso s = R RingOpsIso S
10 9 eleq2d s = S f R RingOpsIso s f R RingOpsIso S
11 10 exbidv s = S f f R RingOpsIso s f f R RingOpsIso S
12 8 11 anbi12d s = S R RingOps s RingOps f f R RingOpsIso s R RingOps S RingOps f f R RingOpsIso S
13 df-risc 𝑟 = r s | r RingOps s RingOps f f r RingOpsIso s
14 6 12 13 brabg R A S B R 𝑟 S R RingOps S RingOps f f R RingOpsIso S