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Theorem preleq 8459
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 4355 . . . . . 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 392 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} → (¬ (𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 = 𝐷𝐵 = 𝐶)))
8 en2lp 8455 . . . . 5 ¬ (𝐷𝐶𝐶𝐷)
9 eleq12 2694 . . . . . 6 ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴𝐵𝐷𝐶))
109anbi1d 740 . . . . 5 ((𝐴 = 𝐷𝐵 = 𝐶) → ((𝐴𝐵𝐶𝐷) ↔ (𝐷𝐶𝐶𝐷)))
118, 10mtbiri 317 . . . 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 446 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wo 383  wa 384   = wceq 1480  wcel 1992  Vcvv 3191  {cpr 4155
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1841  ax-6 1890  ax-7 1937  ax-9 2001  ax-10 2021  ax-11 2036  ax-12 2049  ax-13 2250  ax-ext 2606  ax-sep 4746  ax-nul 4754  ax-pr 4872  ax-reg 8442
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1038  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1883  df-eu 2478  df-mo 2479  df-clab 2613  df-cleq 2619  df-clel 2622  df-nfc 2756  df-ne 2797  df-ral 2917  df-rex 2918  df-rab 2921  df-v 3193  df-sbc 3423  df-dif 3563  df-un 3565  df-in 3567  df-ss 3574  df-nul 3897  df-if 4064  df-sn 4154  df-pr 4156  df-op 4160  df-br 4619  df-opab 4679  df-eprel 4990  df-fr 5038
This theorem is referenced by:  opthreg  8460  dfac2  8898
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