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Theorem preleq 8552
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 4413 . . . . . 6 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
65biimpi 206 . . . . 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 8548 . . . . 5 ¬ (𝐷𝐶𝐶𝐷)
9 eleq12 2720 . . . . . 6 ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴𝐵𝐷𝐶))
109anbi1d 741 . . . . 5 ((𝐴 = 𝐷𝐵 = 𝐶) → ((𝐴𝐵𝐶𝐷) ↔ (𝐷𝐶𝐶𝐷)))
118, 10mtbiri 316 . . . 4 ((𝐴 = 𝐷𝐵 = 𝐶) → ¬ (𝐴𝐵𝐶𝐷))
127, 11syl6 35 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐷} → (¬ (𝐴 = 𝐶𝐵 = 𝐷) → ¬ (𝐴𝐵𝐶𝐷)))
1312con4d 114 . 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 1523  wcel 2030  Vcvv 3231  {cpr 4212
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631  ax-sep 4814  ax-nul 4822  ax-pr 4936  ax-reg 8538
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1056  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-eu 2502  df-mo 2503  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-ne 2824  df-ral 2946  df-rex 2947  df-rab 2950  df-v 3233  df-sbc 3469  df-dif 3610  df-un 3612  df-in 3614  df-ss 3621  df-nul 3949  df-if 4120  df-sn 4211  df-pr 4213  df-op 4217  df-br 4686  df-opab 4746  df-eprel 5058  df-fr 5102
This theorem is referenced by:  opthreg  8553  dfac2  8991
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