Theorem isprrngo 33849

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 6191	. . . . . . 7 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = (1st ‘𝑅))
2		isprrng.1	. . . . . . 7 ⊢ 𝐺 = (1st ‘𝑅)
3	1, 2	syl6eqr 2674	. . . . . 6 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = 𝐺)
4	3	fveq2d 6195	. . . . 5 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = (GId‘𝐺))
5		isprrng.2	. . . . 5 ⊢ 𝑍 = (GId‘𝐺)
6	4, 5	syl6eqr 2674	. . . 4 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = 𝑍)
7	6	sneqd 4189	. . 3 ⊢ (𝑟 = 𝑅 → {(GId‘(1st ‘𝑟))} = {𝑍})
8		fveq2 6191	. . 3 ⊢ (𝑟 = 𝑅 → (PrIdl‘𝑟) = (PrIdl‘𝑅))
9	7, 8	eleq12d 2695	. 2 ⊢ (𝑟 = 𝑅 → ({(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟) ↔ {𝑍} ∈ (PrIdl‘𝑅)))
10		df-prrngo 33847	. 2 ⊢ PrRing = {𝑟 ∈ RingOps ∣ {(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟)}
11	9, 10	elrab2 3366	1 ⊢ (𝑅 ∈ PrRing ↔ (𝑅 ∈ RingOps ∧ {𝑍} ∈ (PrIdl‘𝑅)))

Syntax hints:   ↔ wb 196   ∧ wa 384   = wceq 1483   ∈ wcel 1990  {csn 4177  ‘cfv 5888  1^st c1st 7166  GIdcgi 27344  RingOpscrngo 33693  PrIdlcpridl 33807  PrRingcprrng 33845

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-rex 2918  df-rab 2921  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-uni 4437  df-br 4654  df-iota 5851  df-fv 5896  df-prrngo 33847

This theorem is referenced by:  prrngorngo  33850  smprngopr  33851  isdmn3  33873