Theorem elpr2 3062

Description: A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15.

Hypotheses

Ref	Expression
elpr2.1	|- B e. _V
elpr2.2	|- C e. _V

Assertion

Ref	Expression
elpr2	|- (A e. {B, C} <-> (A = B \/ A = C))

Proof of Theorem elpr2

Step	Hyp	Ref	Expression
1		elprg 3060	. . 3 |- (A e. {B, C} -> (A e. {B, C} <-> (A = B \/ A = C)))
2	1	ibi 652	. 2 |- (A e. {B, C} -> (A = B \/ A = C))
3		elpr2.1	. . . . . 6 |- B e. _V
4		eleq1 1957	. . . . . 6 |- (A = B -> (A e. _V <-> B e. _V))
5	3, 4	mpbiri 211	. . . . 5 |- (A = B -> A e. _V)
6		elpr2.2	. . . . . 6 |- C e. _V
7		eleq1 1957	. . . . . 6 |- (A = C -> (A e. _V <-> C e. _V))
8	6, 7	mpbiri 211	. . . . 5 |- (A = C -> A e. _V)
9	5, 8	jaoi 368	. . . 4 |- ((A = B \/ A = C) -> A e. _V)
10		elprg 3060	. . . 4 |- (A e. _V -> (A e. {B, C} <-> (A = B \/ A = C)))
11	9, 10	syl 12	. . 3 |- ((A = B \/ A = C) -> (A e. {B, C} <-> (A = B \/ A = C)))
12	11	ibir 653	. 2 |- ((A = B \/ A = C) -> A e. {B, C})
13	2, 12	impbii 174	1 |- (A e. {B, C} <-> (A = B \/ A = C))

Syntax hints:   <-> wb 163   \/ wo 239   = wceq 1298   e. wcel 1300  _Vcvv 2292  {cpr 3045

This theorem is referenced by:  elxr 6706  nofv 13998

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294  df-un 2600  df-sn 3049  df-pr 3050

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