Theorem elprg 4646

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 2732	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqeq1 2732	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐶 ↔ 𝐴 = 𝐶))
3	1, 2	orbi12d 917	. 2 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐵 ∨ 𝑥 = 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
4		dfpr2 4644	. 2 ⊢ {𝐵, 𝐶} = {𝑥 ∣ (𝑥 = 𝐵 ∨ 𝑥 = 𝐶)}
5	3, 4	elab2g 3668	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

Syntax hints:   → wi 4   ↔ wb 205   ∨ wo 846   = wceq 1534   ∈ wcel 2099  {cpr 4627

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-v 3472  df-un 3950  df-sn 4626  df-pr 4628

This theorem is referenced by:  elpri  4647  elpr  4648  elpr2g  4649  elpr2OLD  4651  nelpr2  4652  nelpr1  4653  eldifpr  4657  eltpg  4686  ifpr  4692  prid1g  4761  ssprss  4824  preq1b  4844  prel12g  4861  ordunpr  7824  hashtpg  14473  2nsgsimpgd  20053  cnsubrg  21354  atandm  26802  1egrvtxdg0  29319  eupth2lem1  30022  nelpr  32321  eliccioo  32649  linds2eq  33091  sfprmdvdsmersenne  46934  prelrrx2b  47778

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