Theorem isprrngo 35831

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 6674	. . . . . . 7 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = (1st ‘𝑅))
2		isprrng.1	. . . . . . 7 ⊢ 𝐺 = (1st ‘𝑅)
3	1, 2	eqtr4di 2791	. . . . . 6 ⊢ (𝑟 = 𝑅 → (1st ‘𝑟) = 𝐺)
4	3	fveq2d 6678	. . . . 5 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = (GId‘𝐺))
5		isprrng.2	. . . . 5 ⊢ 𝑍 = (GId‘𝐺)
6	4, 5	eqtr4di 2791	. . . 4 ⊢ (𝑟 = 𝑅 → (GId‘(1st ‘𝑟)) = 𝑍)
7	6	sneqd 4528	. . 3 ⊢ (𝑟 = 𝑅 → {(GId‘(1st ‘𝑟))} = {𝑍})
8		fveq2 6674	. . 3 ⊢ (𝑟 = 𝑅 → (PrIdl‘𝑟) = (PrIdl‘𝑅))
9	7, 8	eleq12d 2827	. 2 ⊢ (𝑟 = 𝑅 → ({(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟) ↔ {𝑍} ∈ (PrIdl‘𝑅)))
10		df-prrngo 35829	. 2 ⊢ PrRing = {𝑟 ∈ RingOps ∣ {(GId‘(1st ‘𝑟))} ∈ (PrIdl‘𝑟)}
11	9, 10	elrab2 3591	1 ⊢ (𝑅 ∈ PrRing ↔ (𝑅 ∈ RingOps ∧ {𝑍} ∈ (PrIdl‘𝑅)))

Syntax hints:   ↔ wb 209   ∧ wa 399   = wceq 1542   ∈ wcel 2114  {csn 4516  ‘cfv 6339  1^st c1st 7712  GIdcgi 28425  RingOpscrngo 35675  PrIdlcpridl 35789  PrRingcprrng 35827

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-ext 2710

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-3an 1090  df-tru 1545  df-ex 1787  df-sb 2075  df-clab 2717  df-cleq 2730  df-clel 2811  df-rab 3062  df-v 3400  df-un 3848  df-in 3850  df-ss 3860  df-sn 4517  df-pr 4519  df-op 4523  df-uni 4797  df-br 5031  df-iota 6297  df-fv 6347  df-prrngo 35829

This theorem is referenced by:  prrngorngo  35832  smprngopr  35833  isdmn3  35855