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Theorem minel 4399
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 4398 . . . 4 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21expcom 414 . . 3 (𝐴𝐵 → (𝐴𝐶 → (𝐶𝐵) ≠ ∅))
32necon2bd 2959 . 2 (𝐴𝐵 → ((𝐶𝐵) = ∅ → ¬ 𝐴𝐶))
43imp 407 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 396   = wceq 1539  wcel 2106  wne 2943  cin 3885  c0 4256
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ne 2944  df-v 3431  df-dif 3889  df-in 3893  df-nul 4257
This theorem is referenced by:  peano5  7730  peano5OLD  7731  fnsuppres  7994  domunfican  9074  unwdomg  9330  dfac5  9894  ccatval2  14293  mreexexlem2d  17364  hauspwpwf1  23148
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