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Theorem ssnelpss 4106
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 2852 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2733 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 328 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 4080 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 536 . 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 395   = wceq 1533  wcel 2098  wss 3943  wpss 3944
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 2697
This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-cleq 2718  df-clel 2804  df-ne 2935  df-pss 3962
This theorem is referenced by:  ssnelpssd  4107  ssexnelpss  4108  isfin4p1  10312  canthp1lem2  10650  nqpr  11011  uzindi  13953  nthruc  16202  nthruz  16203  vitali  25497  onpsstopbas  35823
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