Theorem elpr2 3420

Description: A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 14-Oct-2005.)

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		elprg 3418	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
2	1	ibi 174	. 2 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
3		elpr2.1	. . . . . 6 ⊢ 𝐵 ∈ V
4		eleq1 2141	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ∈ V ↔ 𝐵 ∈ V))
5	3, 4	mpbiri 166	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐴 ∈ V)
6		elpr2.2	. . . . . 6 ⊢ 𝐶 ∈ V
7		eleq1 2141	. . . . . 6 ⊢ (𝐴 = 𝐶 → (𝐴 ∈ V ↔ 𝐶 ∈ V))
8	6, 7	mpbiri 166	. . . . 5 ⊢ (𝐴 = 𝐶 → 𝐴 ∈ V)
9	5, 8	jaoi 668	. . . 4 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ V)
10		elprg 3418	. . . 4 ⊢ (𝐴 ∈ V → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
11	9, 10	syl 14	. . 3 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
12	11	ibir 175	. 2 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ {𝐵, 𝐶})
13	2, 12	impbii 124	1 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))

Syntax hints:   ↔ wb 103   ∨ wo 661   = wceq 1284   ∈ wcel 1433  Vcvv 2601  {cpr 3399

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-un 2977  df-sn 3404  df-pr 3405

This theorem is referenced by:  elxr  8850