Theorem elprg 4651

Description: A member of a pair of classes is one or the other of them, and conversely as soon as it is a set. Exercise 1 of [TakeutiZaring] p. 15, generalized. (Contributed by NM, 13-Sep-1995.)

Assertion

Ref	Expression
elprg	⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

Proof of Theorem elprg

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqeq1 2729	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqeq1 2729	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐶 ↔ 𝐴 = 𝐶))
3	1, 2	orbi12d 916	. 2 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐵 ∨ 𝑥 = 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
4		dfpr2 4649	. 2 ⊢ {𝐵, 𝐶} = {𝑥 ∣ (𝑥 = 𝐵 ∨ 𝑥 = 𝐶)}
5	3, 4	elab2g 3667	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

Syntax hints:   → wi 4   ↔ wb 205   ∨ wo 845   = wceq 1533   ∈ wcel 2098  {cpr 4631

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-v 3465  df-un 3950  df-sn 4630  df-pr 4632

This theorem is referenced by:  elpri  4652  elpr  4653  elpr2g  4654  nelpr2  4656  nelpr1  4657  eldifpr  4661  eltpg  4690  ifpr  4696  prid1g  4765  ssprss  4828  preq1b  4848  prel12g  4865  ordunpr  7828  hashtpg  14478  2nsgsimpgd  20063  cnsubrg  21364  atandm  26838  1egrvtxdg0  29381  eupth2lem1  30084  nelpr  32384  eliccioo  32711  linds2eq  33158  sfprmdvdsmersenne  47006  prelrrx2b  47899