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Theorem minel 3310
Description: A minimum element of a class has no elements in common with the class. (Contributed by NM, 22-Jun-1994.)
Assertion
Ref Expression
minel ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)

Proof of Theorem minel
StepHypRef Expression
1 inelcm 3309 . . . . 5 ((𝐴𝐶𝐴𝐵) → (𝐶𝐵) ≠ ∅)
21necon2bi 2275 . . . 4 ((𝐶𝐵) = ∅ → ¬ (𝐴𝐶𝐴𝐵))
3 imnan 634 . . . 4 ((𝐴𝐶 → ¬ 𝐴𝐵) ↔ ¬ (𝐴𝐶𝐴𝐵))
42, 3sylibr 141 . . 3 ((𝐶𝐵) = ∅ → (𝐴𝐶 → ¬ 𝐴𝐵))
54con2d 564 . 2 ((𝐶𝐵) = ∅ → (𝐴𝐵 → ¬ 𝐴𝐶))
65impcom 120 1 ((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101   = wceq 1259  wcel 1409  cin 2943  c0 3251
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 554  ax-in2 555  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ne 2221  df-v 2576  df-dif 2947  df-in 2951  df-nul 3252
This theorem is referenced by: (None)
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