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Theorem preleq 4361
Description: Equality of two unordered pairs when one member of each pair contains the other member. (Contributed by NM, 16-Oct-1996.)
Hypotheses
Ref Expression
preleq.1 𝐴 ∈ V
preleq.2 𝐵 ∈ V
preleq.3 𝐶 ∈ V
preleq.4 𝐷 ∈ V
Assertion
Ref Expression
preleq (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))

Proof of Theorem preleq
StepHypRef Expression
1 en2lp 4360 . . . . 5 ¬ (𝐷𝐶𝐶𝐷)
2 eleq12 2152 . . . . . 6 ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴𝐵𝐷𝐶))
32anbi1d 453 . . . . 5 ((𝐴 = 𝐷𝐵 = 𝐶) → ((𝐴𝐵𝐶𝐷) ↔ (𝐷𝐶𝐶𝐷)))
41, 3mtbiri 635 . . . 4 ((𝐴 = 𝐷𝐵 = 𝐶) → ¬ (𝐴𝐵𝐶𝐷))
54con2i 592 . . 3 ((𝐴𝐵𝐶𝐷) → ¬ (𝐴 = 𝐷𝐵 = 𝐶))
65adantr 270 . 2 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ¬ (𝐴 = 𝐷𝐵 = 𝐶))
7 preleq.1 . . . . 5 𝐴 ∈ V
8 preleq.2 . . . . 5 𝐵 ∈ V
9 preleq.3 . . . . 5 𝐶 ∈ V
10 preleq.4 . . . . 5 𝐷 ∈ V
117, 8, 9, 10preq12b 3609 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1211biimpi 118 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1312adantl 271 . 2 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
146, 13ecased 1285 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102  wo 664   = wceq 1289  wcel 1438  Vcvv 2619  {cpr 3442
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 579  ax-in2 580  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070  ax-setind 4343
This theorem depends on definitions:  df-bi 115  df-3an 926  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-ral 2364  df-v 2621  df-dif 2999  df-un 3001  df-sn 3447  df-pr 3448
This theorem is referenced by:  opthreg  4362
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