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Theorem ssnelpss 4114
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 2866 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2744 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 328 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 4088 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 536 . 2 (𝐴𝐵 → (¬ 𝐴 = 𝐵𝐴𝐵))
63, 5imbitrid 244 1 (𝐴𝐵 → ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → 𝐴𝐵))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 395   = wceq 1540  wcel 2108  wss 3951  wpss 3952
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-cleq 2729  df-clel 2816  df-ne 2941  df-pss 3971
This theorem is referenced by:  ssnelpssd  4115  ssexnelpss  4116  isfin4p1  10355  canthp1lem2  10693  nqpr  11054  uzindi  14023  nthruc  16288  nthruz  16289  vitali  25648  onpsstopbas  36431
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