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Theorem nelss 3158
 Description: Demonstrate by witnesses that two classes lack a subclass relation. (Contributed by Stefan O'Rear, 5-Feb-2015.)
Assertion
Ref Expression
nelss ((𝐴𝐵 ∧ ¬ 𝐴𝐶) → ¬ 𝐵𝐶)

Proof of Theorem nelss
StepHypRef Expression
1 ssel 3091 . . 3 (𝐵𝐶 → (𝐴𝐵𝐴𝐶))
21com12 30 . 2 (𝐴𝐵 → (𝐵𝐶𝐴𝐶))
32con3dimp 624 1 ((𝐴𝐵 ∧ ¬ 𝐴𝐶) → ¬ 𝐵𝐶)
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ∈ wcel 1480   ⊆ wss 3071 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121 This theorem depends on definitions:  df-bi 116  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-in 3077  df-ss 3084 This theorem is referenced by: (None)
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