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Theorem preleq 9610
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 9609 . 2 (((𝐴𝐵𝐵 ∈ V ∧ 𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
31, 2mp3anl2 1452 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1533  wcel 2098  Vcvv 3468  {cpr 4625
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-12 2163  ax-ext 2697  ax-sep 5292  ax-nul 5299  ax-pr 5420  ax-reg 9586
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-ne 2935  df-ral 3056  df-rex 3065  df-rab 3427  df-v 3470  df-dif 3946  df-un 3948  df-in 3950  df-ss 3960  df-nul 4318  df-if 4524  df-pw 4599  df-sn 4624  df-pr 4626  df-op 4630  df-br 5142  df-opab 5204  df-eprel 5573  df-fr 5624
This theorem is referenced by:  opthreg  9612  dfac2b  10124
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