Theorem elop 3968

Description: An ordered pair has two elements. Exercise 3 of [TakeutiZaring] p. 15. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 26-Apr-2015.)

Hypotheses

Ref	Expression
elop.1	|- A e. _V
elop.2	|- B e. _V
elop.3	|- C e. _V

Assertion

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

Proof of Theorem elop

Step	Hyp	Ref	Expression
1		elop.2	. . . 4 |- B e. _V
2		elop.3	. . . 4 |- C e. _V
3	1, 2	dfop 3548	. . 3 |- <. B , C >. = { { B } , { B , C } }
4	3	eleq2i 2104	. 2 |- ( A e. <. B , C >. <-> A e. { { B } , { B , C } } )
5		elop.1	. . 3 |- A e. _V
6	5	elpr 3396	. 2 |- ( A e. { { B } , { B , C } } <-> ( A = { B } \/ A = { B , C } ) )
7	4, 6	bitri 173	1 |- ( A e. <. B , C >. <-> ( A = { B } \/ A = { B , C } ) )

Syntax hints:   <-> wb 98   \/ wo 629   = wceq 1243   e. wcel 1393   _Vcvv 2557   {csn 3375   {cpr 3376   <.cop 3378

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-3an 887  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-un 2922  df-sn 3381  df-pr 3382  df-op 3384

This theorem is referenced by:  relop  4486  bdop  9995