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Theorem minel 4397
Description: A minimum element of a class has no elements in common with the class. (Contributed by NM, 22-Jun-1994.) (Proof shortened by JJ, 14-Jul-2021.)
Assertion
Ref Expression
minel ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)

Proof of Theorem minel
StepHypRef Expression
1 inelcm 4396 . . . 4 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21expcom 415 . . 3 (𝐴𝐵 → (𝐴𝐶 → (𝐶𝐵) ≠ ∅))
32necon2bd 2952 . 2 (𝐴𝐵 → ((𝐶𝐵) = ∅ → ¬ 𝐴𝐶))
43imp 408 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 397   = wceq 1548  wcel 2121  wne 2936  cin 3884  c0 4264
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-8 2123  ax-9 2131  ax-ext 2713
This theorem depends on definitions:  df-bi 209  df-an 398  df-tru 1551  df-fal 1561  df-ex 1788  df-sb 2075  df-clab 2720  df-cleq 2733  df-clel 2816  df-ne 2937  df-v 3435  df-dif 3888  df-in 3892  df-nul 4265
This theorem is referenced by:  peano5  7837  fnsuppres  8135  domunfican  9226  unwdomg  9493  dfac5  10046  ccatval2  14535  mreexexlem2d  17606  hauspwpwf1  23974  noinfepfnregs  35328
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