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Theorem preleq 9506
Description: Equality of two unordered pairs when one member of each pair contains the other member. (Contributed by NM, 16-Oct-1996.) (Revised by AV, 15-Jun-2022.)
Hypothesis
Ref Expression
preleq.b 𝐵 ∈ V
Assertion
Ref Expression
preleq (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))

Proof of Theorem preleq
StepHypRef Expression
1 preleq.b . 2 𝐵 ∈ V
2 preleqg 9505 . 2 (((𝐴𝐵𝐵 ∈ V ∧ 𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
31, 2mp3anl2 1458 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1541  wcel 2111  Vcvv 3436  {cpr 4578
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703  ax-sep 5234  ax-nul 5244  ax-pr 5370  ax-reg 9478
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4284  df-if 4476  df-pw 4552  df-sn 4577  df-pr 4579  df-op 4583  df-br 5092  df-opab 5154  df-eprel 5516  df-fr 5569
This theorem is referenced by:  opthreg  9508  dfac2b  10022
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