Theorem isprrngo 38002

Description: The predicate "is a prime ring". (Contributed by Jeff Madsen, 10-Jun-2010.)

Hypotheses

Ref	Expression
isprrng.1	⊢ 𝐺 = (1st ‘𝑅)
isprrng.2	⊢ 𝑍 = (GId‘𝐺)

Assertion

Ref	Expression
isprrngo	⊢ (𝑅 ∈ PrRing ↔ (𝑅 ∈ RingOps ∧ {𝑍} ∈ (PrIdl‘𝑅)))

Proof of Theorem isprrngo

Dummy variable 𝑟 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		fveq2 6915	. . . . . . 7 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = (1st ‘𝑅))
2		isprrng.1	. . . . . . 7 ⊢ 𝐺 = (1st ‘𝑅)
3	1, 2	eqtr4di 2798	. . . . . 6 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = 𝐺)
4	3	fveq2d 6919	. . . . 5 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = (GId‘𝐺))
5		isprrng.2	. . . . 5 ⊢ 𝑍 = (GId‘𝐺)
6	4, 5	eqtr4di 2798	. . . 4 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = 𝑍)
7	6	sneqd 4660	. . 3 ⊢ (𝑟 = 𝑅 → {(GId‘(1st ‘𝑟))} = {𝑍})
8		fveq2 6915	. . 3 ⊢ (𝑟 = 𝑅 → (PrIdl‘𝑟) = (PrIdl‘𝑅))
9	7, 8	eleq12d 2838	. 2 ⊢ (𝑟 = 𝑅 → ({(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟) ↔ {𝑍} ∈ (PrIdl‘𝑅)))
10		df-prrngo 38000	. 2 ⊢ PrRing = {𝑟 ∈ RingOps ∣ {(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟)}
11	9, 10	elrab2 3711	1 ⊢ (𝑅 ∈ PrRing ↔ (𝑅 ∈ RingOps ∧ {𝑍} ∈ (PrIdl‘𝑅)))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1537   ∈ wcel 2108  {csn 4648  ‘cfv 6568  1^st c1st 8022  GIdcgi 30514  RingOpscrngo 37846  PrIdlcpridl 37960  PrRingcprrng 37998

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-uni 4932  df-br 5167  df-iota 6520  df-fv 6576  df-prrngo 38000

This theorem is referenced by:  prrngorngo  38003  smprngopr  38004  isdmn3  38026