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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 2858 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2739 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 327 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 4085 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 537 . 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 396   = wceq 1541  wcel 2106  wss 3948  wpss 3949
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703
This theorem depends on definitions:  df-bi 206  df-an 397  df-ex 1782  df-cleq 2724  df-clel 2810  df-ne 2941  df-pss 3967
This theorem is referenced by:  ssnelpssd  4112  ssexnelpss  4113  isfin4p1  10309  canthp1lem2  10647  nqpr  11008  uzindi  13946  nthruc  16194  nthruz  16195  vitali  25129  onpsstopbas  35310
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