Theorem elpr2g 4617

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.) Generalize from sethood hypothesis to sethood antecedent. (Revised by BJ, 25-May-2024.)

Assertion

Ref	Expression
elpr2g	⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

Proof of Theorem elpr2g

Step	Hyp	Ref	Expression
1		elex 3478	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → 𝐴 ∈ V)
2	1	a1i 11	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} → 𝐴 ∈ V))
3		elex 3478	. . . 4 ⊢ (𝐵 ∈ 𝑉 → 𝐵 ∈ V)
4		eleq1a 2860	. . . 4 ⊢ (𝐵 ∈ V → (𝐴 = 𝐵 → 𝐴 ∈ V))
5	3, 4	syl 18	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐴 = 𝐵 → 𝐴 ∈ V))
6		elex 3478	. . . 4 ⊢ (𝐶 ∈ 𝑊 → 𝐶 ∈ V)
7		eleq1a 2860	. . . 4 ⊢ (𝐶 ∈ V → (𝐴 = 𝐶 → 𝐴 ∈ V))
8	6, 7	syl 18	. . 3 ⊢ (𝐶 ∈ 𝑊 → (𝐴 = 𝐶 → 𝐴 ∈ V))
9	5, 8	jaao 969	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ V))
10		elprg 4614	. . 3 ⊢ (𝐴 ∈ V → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
11	10	a1i 11	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ V → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))))
12	2, 9, 11	pm5.21ndd 382	1 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ 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:  elpr2  4618