Theorem elpr2 3598

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 3596	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
2	1	ibi 175	. 2 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
3		elpr2.1	. . . . . 6 ⊢ 𝐵 ∈ V
4		eleq1 2229	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ∈ V ↔ 𝐵 ∈ V))
5	3, 4	mpbiri 167	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐴 ∈ V)
6		elpr2.2	. . . . . 6 ⊢ 𝐶 ∈ V
7		eleq1 2229	. . . . . 6 ⊢ (𝐴 = 𝐶 → (𝐴 ∈ V ↔ 𝐶 ∈ V))
8	6, 7	mpbiri 167	. . . . 5 ⊢ (𝐴 = 𝐶 → 𝐴 ∈ V)
9	5, 8	jaoi 706	. . . 4 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ V)
10		elprg 3596	. . . 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 176	. 2 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ {𝐵, 𝐶})
13	2, 12	impbii 125	1 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))

Syntax hints:   ↔ wb 104   ∨ wo 698   = wceq 1343   ∈ wcel 2136  Vcvv 2726  {cpr 3577

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-un 3120  df-sn 3582  df-pr 3583

This theorem is referenced by:  elxr  9712