Theorem elpr2 3753

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 3751	. . 3 ⊢ (A ∈ {B, C} → (A ∈ {B, C} ↔ (A = B ∨ A = C)))
2	1	ibi 232	. 2 ⊢ (A ∈ {B, C} → (A = B ∨ A = C))
3		elpr2.1	. . . . . 6 ⊢ B ∈ V
4		eleq1 2413	. . . . . 6 ⊢ (A = B → (A ∈ V ↔ B ∈ V))
5	3, 4	mpbiri 224	. . . . 5 ⊢ (A = B → A ∈ V)
6		elpr2.2	. . . . . 6 ⊢ C ∈ V
7		eleq1 2413	. . . . . 6 ⊢ (A = C → (A ∈ V ↔ C ∈ V))
8	6, 7	mpbiri 224	. . . . 5 ⊢ (A = C → A ∈ V)
9	5, 8	jaoi 368	. . . 4 ⊢ ((A = B ∨ A = C) → A ∈ V)
10		elprg 3751	. . . 4 ⊢ (A ∈ V → (A ∈ {B, C} ↔ (A = B ∨ A = C)))
11	9, 10	syl 15	. . 3 ⊢ ((A = B ∨ A = C) → (A ∈ {B, C} ↔ (A = B ∨ A = C)))
12	11	ibir 233	. 2 ⊢ ((A = B ∨ A = C) → A ∈ {B, C})
13	2, 12	impbii 180	1 ⊢ (A ∈ {B, C} ↔ (A = B ∨ A = C))

This proof depends on syntax axioms:   ↔ wb 176   ∨ wo 357   = wceq 1642   ∈ wcel 1710  Vcvv 2860  {cpr 3739

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-un 3215  df-sn 3742  df-pr 3743

This theorem is used by:  elopk  4130