Theorem elpr2g 4615

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 3464	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → 𝐴 ∈ V)
2	1	a1i 11	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} → 𝐴 ∈ V))
3		elex 3464	. . . 4 ⊢ (𝐵 ∈ 𝑉 → 𝐵 ∈ V)
4		eleq1a 2827	. . . 4 ⊢ (𝐵 ∈ V → (𝐴 = 𝐵 → 𝐴 ∈ V))
5	3, 4	syl 17	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐴 = 𝐵 → 𝐴 ∈ V))
6		elex 3464	. . . 4 ⊢ (𝐶 ∈ 𝑊 → 𝐶 ∈ V)
7		eleq1a 2827	. . . 4 ⊢ (𝐶 ∈ V → (𝐴 = 𝐶 → 𝐴 ∈ V))
8	6, 7	syl 17	. . 3 ⊢ (𝐶 ∈ 𝑊 → (𝐴 = 𝐶 → 𝐴 ∈ V))
9	5, 8	jaao 953	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ V))
10		elprg 4612	. . 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 380	1 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396   ∨ wo 845   = wceq 1541   ∈ wcel 2106  Vcvv 3446  {cpr 4593

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3448  df-un 3918  df-sn 4592  df-pr 4594

This theorem is referenced by:  elpr2  4616