Theorem elpr2 4628

Description: A member of a pair of sets is one or the other of them, and conversely. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 14-Oct-2005.) (Proof shortened by JJ, 23-Jul-2021.)

Hypotheses

Ref	Expression
elpr2.1	⊢ 𝐵 ∈ V
elpr2.2	⊢ 𝐶 ∈ V

Assertion

Ref	Expression
elpr2	⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))

Proof of Theorem elpr2

Step	Hyp	Ref	Expression
1		elpr2.1	. 2 ⊢ 𝐵 ∈ V
2		elpr2.2	. 2 ⊢ 𝐶 ∈ V
3		elpr2g 4627	. 2 ⊢ ((𝐵 ∈ V ∧ 𝐶 ∈ V) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
4	1, 2, 3	mp2an 692	1 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))

Syntax hints:   ↔ wb 206   ∨ wo 847   = wceq 1540   ∈ wcel 2108  Vcvv 3459  {cpr 4603

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 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-v 3461  df-un 3931  df-sn 4602  df-pr 4604

This theorem is referenced by:  elopg  5441  elxr  13132  nofv  27621  fprodex01  32804  opgpgvtx  48059