Theorem elpr2 3640

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 3638	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))
2	1	ibi 176	. 2 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
3		elpr2.1	. . . . . 6 ⊢ 𝐵 ∈ V
4		eleq1 2256	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ∈ V ↔ 𝐵 ∈ V))
5	3, 4	mpbiri 168	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐴 ∈ V)
6		elpr2.2	. . . . . 6 ⊢ 𝐶 ∈ V
7		eleq1 2256	. . . . . 6 ⊢ (𝐴 = 𝐶 → (𝐴 ∈ V ↔ 𝐶 ∈ V))
8	6, 7	mpbiri 168	. . . . 5 ⊢ (𝐴 = 𝐶 → 𝐴 ∈ V)
9	5, 8	jaoi 717	. . . 4 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ V)
10		elprg 3638	. . . 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 177	. 2 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ {𝐵, 𝐶})
13	2, 12	impbii 126	1 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))

Syntax hints:   ↔ wb 105   ∨ wo 709   = wceq 1364   ∈ wcel 2164  Vcvv 2760  {cpr 3619

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-un 3157  df-sn 3624  df-pr 3625

This theorem is referenced by:  elxr  9842