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Theorem preleq 3454
Description: Equality of two unordered pairs when one member of each pair contains the other member.
Hypotheses
Ref Expression
preleq.1 AV
preleq.2 BV
preleq.3 CV
preleq.4 DV
Assertion
Ref Expression
preleq (((ABCD) ∧ {A, B} = {C, D}) → (A = CB = D))

Proof of Theorem preleq
StepHypRef Expression
1 preleq.1 . . . . . . . 8 AV
2 preleq.2 . . . . . . . 8 BV
3 preleq.3 . . . . . . . 8 CV
4 preleq.4 . . . . . . . 8 DV
51, 2, 3, 4preq12b 1874 . . . . . . 7 ({A, B} = {C, D} ↔ ((A = CB = D) ∨ (A = DB = C)))
65biimp 133 . . . . . 6 ({A, B} = {C, D} → ((A = CB = D) ∨ (A = DB = C)))
76ord 202 . . . . 5 ({A, B} = {C, D} → (¬ (A = CB = D) → (A = DB = C)))
8 en2lp 3453 . . . . . 6 ¬ (DCCD)
9 eleq12 1151 . . . . . . 7 ((A = DB = C) → (ABDC))
109anbi1d 469 . . . . . 6 ((A = DB = C) → ((ABCD) ↔ (DCCD)))
118, 10mtbiri 539 . . . . 5 ((A = DB = C) → ¬ (ABCD))
127, 11syl6 23 . . . 4 ({A, B} = {C, D} → (¬ (A = CB = D) → ¬ (ABCD)))
1312a3d 70 . . 3 ({A, B} = {C, D} → ((ABCD) → (A = CB = D)))
1413imp 277 . 2 (({A, B} = {C, D} ∧ (ABCD)) → (A = CB = D))
1514ancoms 334 1 (((ABCD) ∧ {A, B} = {C, D}) → (A = CB = D))
Colors of variables: wff set class
Syntax hints:  ¬ wn 1   → wi 2   ∨ wo 195   ∧ wa 196   = wceq 1091   ∈ wcel 1092  Vcvv 1348  {cpr 1809
This theorem is referenced by:  opthreg 3455  aceq6b 3565
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-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077  ax-reg 1078
This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-opab 2098  df-eprel 2122  df-fr 2169
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