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Theorem minel 4423
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 4422 . . . 4 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21expcom 418 . . 3 (𝐴𝐵 → (𝐴𝐶 → (𝐶𝐵) ≠ ∅))
32necon2bd 2976 . 2 (𝐴𝐵 → ((𝐶𝐵) = ∅ → ¬ 𝐴𝐶))
43imp 411 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 400   = wceq 1563  wcel 2145  wne 2960  cin 3906  c0 4288
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1566  df-fal 1576  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-v 3459  df-dif 3910  df-in 3914  df-nul 4289
This theorem is referenced by:  peano5  7878  fnsuppres  8175  domunfican  9269  unwdomg  9534  dfac5  10100  ccatval2  14603  mreexexlem2d  17689  hauspwpwf1  24101  noinfepfnregs  35435
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