Theorem opthpr 2485

Description: A way to represent ordered pairs using unordered pairs with distinct members.

Hypotheses

Ref	Expression
preq12b.1	|- A e. V
preq12b.2	|- B e. V
preq12b.3	|- C e. V
preq12b.4	|- D e. V

Assertion

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

Proof of Theorem opthpr

Step	Hyp	Ref	Expression
1		idd 61	. . . 4 |- (A =/= D -> ((A = C /\ B = D) -> (A = C /\ B = D)))
2		df-ne 1587	. . . . . 6 |- (A =/= D <-> -. A = D)
3		pm2.21 76	. . . . . 6 |- (-. A = D -> (A = D -> (B = C -> (A = C /\ B = D))))
4	2, 3	sylbi 199	. . . . 5 |- (A =/= D -> (A = D -> (B = C -> (A = C /\ B = D))))
5	4	imp3a 361	. . . 4 |- (A =/= D -> ((A = D /\ B = C) -> (A = C /\ B = D)))
6	1, 5	jaod 424	. . 3 |- (A =/= D -> (((A = C /\ B = D) \/ (A = D /\ B = C)) -> (A = C /\ B = D)))
7		orc 269	. . 3 |- ((A = C /\ B = D) -> ((A = C /\ B = D) \/ (A = D /\ B = C)))
8	6, 7	impbid1 517	. 2 |- (A =/= D -> (((A = C /\ B = D) \/ (A = D /\ B = C)) <-> (A = C /\ B = D)))
9		preq12b.1	. . 3 |- A e. V
10		preq12b.2	. . 3 |- B e. V
11		preq12b.3	. . 3 |- C e. V
12		preq12b.4	. . 3 |- D e. V
13	9, 10, 11, 12	preq12b 2483	. 2 |- ({A, B} = {C, D} <-> ((A = C /\ B = D) \/ (A = D /\ B = C)))
14	8, 13	syl5bb 532	1 |- (A =/= D -> ({A, B} = {C, D} <-> (A = C /\ B = D)))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   \/ wo 222   /\ wa 223   = wceq 956   e. wcel 958   =/= wne 1585  Vcvv 1811  {cpr 2410

This theorem is referenced by:  brdom7disj 4804  brdom6disj 4805

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-ne 1587  df-v 1812  df-un 2050  df-sn 2412  df-pr 2413

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