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Theorem minel 4007
 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 4006 . . . 4 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21expcom 451 . . 3 (𝐴𝐵 → (𝐴𝐶 → (𝐶𝐵) ≠ ∅))
32necon2bd 2806 . 2 (𝐴𝐵 → ((𝐶𝐵) = ∅ → ¬ 𝐴𝐶))
43imp 445 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 384   = wceq 1480   ∈ wcel 1987   ≠ wne 2790   ∩ cin 3555  ∅c0 3893 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ne 2791  df-v 3188  df-dif 3559  df-in 3563  df-nul 3894 This theorem is referenced by:  peano5  7039  fnsuppres  7270  domunfican  8180  unwdomg  8436  dfac5  8898  ccatval2  13304  mreexexlem2d  16229  hauspwpwf1  21704
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