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Theorem minel 4380
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 4379 . . . 4 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21expcom 417 . . 3 (𝐴𝐵 → (𝐴𝐶 → (𝐶𝐵) ≠ ∅))
32necon2bd 2956 . 2 (𝐴𝐵 → ((𝐶𝐵) = ∅ → ¬ 𝐴𝐶))
43imp 410 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 399   = wceq 1543  wcel 2110  wne 2940  cin 3865  c0 4237
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 1976  ax-7 2016  ax-8 2112  ax-9 2120  ax-ext 2708
This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1546  df-fal 1556  df-ex 1788  df-sb 2071  df-clab 2715  df-cleq 2729  df-clel 2816  df-ne 2941  df-v 3410  df-dif 3869  df-in 3873  df-nul 4238
This theorem is referenced by:  peano5  7671  peano5OLD  7672  fnsuppres  7933  domunfican  8944  unwdomg  9200  dfac5  9742  ccatval2  14135  mreexexlem2d  17148  hauspwpwf1  22884
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