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Theorem preleq 9588
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 9587 . 2 (((𝐴𝐵𝐵 ∈ V ∧ 𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
31, 2mp3anl2 1485 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401   = wceq 1570  wcel 2146  Vcvv 3457  {cpr 4593
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406  ax-reg 9557
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-pw 4566  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-eprel 5563  df-fr 5616
This theorem is used by:  opthreg  9590  dfac2b  10126
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