MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  ssnelpss Structured version   Visualization version   GIF version

Theorem ssnelpss 3866
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 2874 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2777 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 317 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 3840 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 518 . 2 (𝐴𝐵 → (¬ 𝐴 = 𝐵𝐴𝐵))
63, 5syl5ib 234 1 (𝐴𝐵 → ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → 𝐴𝐵))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 382   = wceq 1630  wcel 2144  wss 3721  wpss 3722
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884  ax-5 1990  ax-6 2056  ax-7 2092  ax-9 2153  ax-ext 2750
This theorem depends on definitions:  df-bi 197  df-an 383  df-ex 1852  df-cleq 2763  df-clel 2766  df-ne 2943  df-pss 3737
This theorem is referenced by:  ssnelpssd  3867  ssexnelpss  3868  canthp1lem2  9676  nqpr  10037  uzindi  12988  nthruc  15186  nthruz  15187  vitali  23600  onpsstopbas  32760
  Copyright terms: Public domain W3C validator