Theorem isprrngo 36135

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 6756	. . . . . . 7 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = (1st ‘𝑅))
2		isprrng.1	. . . . . . 7 ⊢ 𝐺 = (1st ‘𝑅)
3	1, 2	eqtr4di 2797	. . . . . 6 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = 𝐺)
4	3	fveq2d 6760	. . . . 5 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = (GId‘𝐺))
5		isprrng.2	. . . . 5 ⊢ 𝑍 = (GId‘𝐺)
6	4, 5	eqtr4di 2797	. . . 4 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = 𝑍)
7	6	sneqd 4570	. . 3 ⊢ (𝑟 = 𝑅 → {(GId‘(1st ‘𝑟))} = {𝑍})
8		fveq2 6756	. . 3 ⊢ (𝑟 = 𝑅 → (PrIdl‘𝑟) = (PrIdl‘𝑅))
9	7, 8	eleq12d 2833	. 2 ⊢ (𝑟 = 𝑅 → ({(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟) ↔ {𝑍} ∈ (PrIdl‘𝑅)))
10		df-prrngo 36133	. 2 ⊢ PrRing = {𝑟 ∈ RingOps ∣ {(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟)}
11	9, 10	elrab2 3620	1 ⊢ (𝑅 ∈ PrRing ↔ (𝑅 ∈ RingOps ∧ {𝑍} ∈ (PrIdl‘𝑅)))

Syntax hints:   ↔ wb 205   ∧ wa 395   = wceq 1539   ∈ wcel 2108  {csn 4558  ‘cfv 6418  1^st c1st 7802  GIdcgi 28753  RingOpscrngo 35979  PrIdlcpridl 36093  PrRingcprrng 36131

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-uni 4837  df-br 5071  df-iota 6376  df-fv 6426  df-prrngo 36133

This theorem is referenced by:  prrngorngo  36136  smprngopr  36137  isdmn3  36159