Theorem isprrngo 38251

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 6834	. . . . . . 7 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = (1st ‘𝑅))
2		isprrng.1	. . . . . . 7 ⊢ 𝐺 = (1st ‘𝑅)
3	1, 2	eqtr4di 2789	. . . . . 6 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = 𝐺)
4	3	fveq2d 6838	. . . . 5 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = (GId‘𝐺))
5		isprrng.2	. . . . 5 ⊢ 𝑍 = (GId‘𝐺)
6	4, 5	eqtr4di 2789	. . . 4 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = 𝑍)
7	6	sneqd 4592	. . 3 ⊢ (𝑟 = 𝑅 → {(GId‘(1st ‘𝑟))} = {𝑍})
8		fveq2 6834	. . 3 ⊢ (𝑟 = 𝑅 → (PrIdl‘𝑟) = (PrIdl‘𝑅))
9	7, 8	eleq12d 2830	. 2 ⊢ (𝑟 = 𝑅 → ({(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟) ↔ {𝑍} ∈ (PrIdl‘𝑅)))
10		df-prrngo 38249	. 2 ⊢ PrRing = {𝑟 ∈ RingOps ∣ {(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟)}
11	9, 10	elrab2 3649	1 ⊢ (𝑅 ∈ PrRing ↔ (𝑅 ∈ RingOps ∧ {𝑍} ∈ (PrIdl‘𝑅)))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1541   ∈ wcel 2113  {csn 4580  ‘cfv 6492  1^st c1st 7931  GIdcgi 30565  RingOpscrngo 38095  PrIdlcpridl 38209  PrRingcprrng 38247

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2715  df-cleq 2728  df-clel 2811  df-rab 3400  df-v 3442  df-dif 3904  df-un 3906  df-ss 3918  df-nul 4286  df-if 4480  df-sn 4581  df-pr 4583  df-op 4587  df-uni 4864  df-br 5099  df-iota 6448  df-fv 6500  df-prrngo 38249

This theorem is referenced by:  prrngorngo  38252  smprngopr  38253  isdmn3  38275