Theorem elprg 4602

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 2733	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqeq1 2733	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐶 ↔ 𝐴 = 𝐶))
3	1, 2	orbi12d 918	. 2 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐵 ∨ 𝑥 = 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
4		dfpr2 4600	. 2 ⊢ {𝐵, 𝐶} = {𝑥 ∣ (𝑥 = 𝐵 ∨ 𝑥 = 𝐶)}
5	3, 4	elab2g 3638	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

Syntax hints:   → wi 4   ↔ wb 206   ∨ wo 847   = wceq 1540   ∈ wcel 2109  {cpr 4581

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-v 3440  df-un 3910  df-sn 4580  df-pr 4582

This theorem is referenced by:  elpri  4603  elpr  4604  elpr2g  4605  nelpr2  4607  nelpr1  4608  eldifpr  4612  eltpg  4640  ifpr  4647  prid1g  4714  ssprss  4778  preq1b  4800  prel12g  4818  ordunpr  7765  hashtpg  14410  2nsgsimpgd  20001  cnsubrg  21352  atandm  26802  1egrvtxdg0  29475  eupth2lem1  30180  nelpr  32493  eliccioo  32884  linds2eq  33328  sfprmdvdsmersenne  47588  prelrrx2b  48700