Theorem elpr2 4621

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 4620	. 2 ⊢ ((𝐵 ∈ V ∧ 𝐶 ∈ V) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
4	1, 2, 3	mp2an 704	1 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))

Syntax hints:   ↔ wb 209   ∨ wo 860   = wceq 1567   ∈ wcel 2149  Vcvv 3463  {cpr 4596

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-v 3465  df-un 3918  df-sn 4595  df-pr 4597

This theorem is referenced by:  elopg  5449  elxr  13140  nofv  27786  fprodex01  33109  opgpgvtx  48708