Theorem eltp 1834

Description: A member of an unordered triple of classes is one of them. Special case of Exercise 1 of [TakeutiZaring] p. 17.

Hypothesis

Ref	Expression
eltp.1	|- A e. V

Assertion

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

Proof of Theorem eltp

Step	Hyp	Ref	Expression
1		df-tp 1814	. . . 4 |- {B, C, D} = ({B, C} u. {D})
2	1	eleq2i 1153	. . 3 |- (A e. {B, C, D} <-> A e. ({B, C} u. {D}))
3		elun 1601	. . 3 |- (A e. ({B, C} u. {D}) <-> (A e. {B, C} \/ A e. {D}))
4		eltp.1	. . . . 5 |- A e. V
5	4	elpr 1823	. . . 4 |- (A e. {B, C} <-> (A = B \/ A = C))
6	4	elsnc 1826	. . . 4 |- (A e. {D} <-> A = D)
7	5, 6	orbi12i 216	. . 3 |- ((A e. {B, C} \/ A e. {D}) <-> ((A = B \/ A = C) \/ A = D))
8	2, 3, 7	3bitr 155	. 2 |- (A e. {B, C, D} <-> ((A = B \/ A = C) \/ A = D))
9		df-3or 582	. 2 |- ((A = B \/ A = C \/ A = D) <-> ((A = B \/ A = C) \/ A = D))
10	8, 9	bitr4 154	1 |- (A e. {B, C, D} <-> (A = B \/ A = C \/ A = D))

Syntax hints:   <-> wb 127   \/ wo 195   \/ w3o 580   = wceq 1091   e. wcel 1092  Vcvv 1348   u. cun 1485  {csn 1808  {cpr 1809  {ctp 1813

This theorem is referenced by:  dftp2 1835  tpss 1855  fr3nr 2178

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-sn 1811  df-pr 1812  df-tp 1814

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