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Theorem preleq 8397
 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 preleq.1 . . . . . . 7 𝐴 ∈ V
2 preleq.2 . . . . . . 7 𝐵 ∈ V
3 preleq.3 . . . . . . 7 𝐶 ∈ V
4 preleq.4 . . . . . . 7 𝐷 ∈ V
51, 2, 3, 4preq12b 4322 . . . . . 6 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
65biimpi 205 . . . . 5 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
76ord 391 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} → (¬ (𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 = 𝐷𝐵 = 𝐶)))
8 en2lp 8393 . . . . 5 ¬ (𝐷𝐶𝐶𝐷)
9 eleq12 2678 . . . . . 6 ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴𝐵𝐷𝐶))
109anbi1d 737 . . . . 5 ((𝐴 = 𝐷𝐵 = 𝐶) → ((𝐴𝐵𝐶𝐷) ↔ (𝐷𝐶𝐶𝐷)))
118, 10mtbiri 316 . . . 4 ((𝐴 = 𝐷𝐵 = 𝐶) → ¬ (𝐴𝐵𝐶𝐷))
127, 11syl6 34 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐷} → (¬ (𝐴 = 𝐶𝐵 = 𝐷) → ¬ (𝐴𝐵𝐶𝐷)))
1312con4d 113 . 2 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴𝐵𝐶𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
1413impcom 445 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 382   ∧ wa 383   = wceq 1475   ∈ wcel 1977  Vcvv 3173  {cpr 4127 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709  ax-nul 4717  ax-pr 4833  ax-reg 8380 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-ral 2901  df-rex 2902  df-rab 2905  df-v 3175  df-sbc 3403  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-br 4584  df-opab 4644  df-eprel 4949  df-fr 4997 This theorem is referenced by:  opthreg  8398  dfac2  8836
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