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Theorem ssnelpss 4111
Description: A subclass missing a member is a proper subclass. (Contributed by NM, 12-Jan-2002.)
Assertion
Ref Expression
ssnelpss (𝐴𝐵 → ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → 𝐴𝐵))

Proof of Theorem ssnelpss
StepHypRef Expression
1 nelneq2 2854 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2735 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 327 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 4085 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 535 . 2 (𝐴𝐵 → (¬ 𝐴 = 𝐵𝐴𝐵))
63, 5imbitrid 243 1 (𝐴𝐵 → ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → 𝐴𝐵))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 394   = wceq 1533  wcel 2098  wss 3949  wpss 3950
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2699
This theorem depends on definitions:  df-bi 206  df-an 395  df-ex 1774  df-cleq 2720  df-clel 2806  df-ne 2938  df-pss 3968
This theorem is referenced by:  ssnelpssd  4112  ssexnelpss  4113  isfin4p1  10348  canthp1lem2  10686  nqpr  11047  uzindi  13989  nthruc  16238  nthruz  16239  vitali  25570  onpsstopbas  35955
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