Theorem elpr2 4618

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 4617	. 2 ⊢ ((𝐵 ∈ V ∧ 𝐶 ∈ V) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
4	1, 2, 3	mp2an 705	1 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861   = wceq 1570   ∈ wcel 2146  Vcvv 3457  {cpr 4593

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-un 3911  df-sn 4592  df-pr 4594

This theorem is used by:  elopg  5450  elxr  13157  nofv  27872  fprodex01  33239  opgpgvtx  48878