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Theorem minel 4489
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 4488 . . . 4 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21expcom 413 . . 3 (𝐴𝐵 → (𝐴𝐶 → (𝐶𝐵) ≠ ∅))
32necon2bd 2962 . 2 (𝐴𝐵 → ((𝐶𝐵) = ∅ → ¬ 𝐴𝐶))
43imp 406 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 395   = wceq 1537  wcel 2108  wne 2946  cin 3975  c0 4352
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ne 2947  df-v 3490  df-dif 3979  df-in 3983  df-nul 4353
This theorem is referenced by:  peano5  7932  peano5OLD  7933  fnsuppres  8232  domunfican  9389  unwdomg  9653  dfac5  10198  ccatval2  14626  mreexexlem2d  17703  hauspwpwf1  24016
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