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Theorem minel 4419
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 4418 . . . 4 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21expcom 419 . . 3 (𝐴𝐵 → (𝐴𝐶 → (𝐶𝐵) ≠ ∅))
32necon2bd 2971 . 2 (𝐴𝐵 → ((𝐶𝐵) = ∅ → ¬ 𝐴𝐶))
43imp 412 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3  wi 4  wa 401   = wceq 1570  wcel 2145  wne 2955  cin 3898  c0 4279
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-v 3452  df-dif 3902  df-in 3906  df-nul 4280
This theorem is used by:  peano5  7890  fnsuppres  8189  domunfican  9291  unwdomg  9556  dfac5  10131  ccatval2  14643  mreexexlem2d  17733  hauspwpwf1  24213  noinfepfnregs  35658
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