Theorem elprg 4581

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 2745	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqeq1 2745	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐶 ↔ 𝐴 = 𝐶))
3	1, 2	orbi12d 925	. 2 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐵 ∨ 𝑥 = 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
4		dfpr2 4579	. 2 ⊢ {𝐵, 𝐶} = {𝑥 ∣ (𝑥 = 𝐵 ∨ 𝑥 = 𝐶)}
5	3, 4	elab2g 3620	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

Syntax hints:   → wi 4   ↔ wb 208   ∨ wo 854   = wceq 1548   ∈ wcel 2121  {cpr 4560

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-8 2123  ax-9 2131  ax-ext 2713

This theorem depends on definitions:  df-bi 209  df-an 398  df-or 855  df-tru 1551  df-ex 1788  df-sb 2075  df-clab 2720  df-cleq 2733  df-clel 2816  df-v 3435  df-un 3890  df-sn 4559  df-pr 4561

This theorem is referenced by:  elpri  4582  elpr  4583  elpr2g  4584  nelpr2  4588  nelpr1  4589  eldifpr  4593  eltpg  4621  ifpr  4628  prid1g  4695  ssprss  4758  preq1b  4780  prel12g  4798  ordunpr  7770  hashtpg  14442  2nsgsimpgd  20074  cnsubrg  21406  atandm  26862  1egrvtxdg0  29602  eupth2lem1  30310  nelpr  32623  eliccioo  33013  linds2eq  33468  sfprmdvdsmersenne  48095  prelrrx2b  49219