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

Theorem ssnelpss 4092
 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 2943 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2833 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 329 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 4066 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 537 . 2 (𝐴𝐵 → (¬ 𝐴 = 𝐵𝐴𝐵))
63, 5syl5ib 245 1 (𝐴𝐵 → ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → 𝐴𝐵))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   = wceq 1530   ∈ wcel 2107   ⊆ wss 3940   ⊊ wpss 3941 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 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-ext 2798 This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1774  df-cleq 2819  df-clel 2898  df-ne 3022  df-pss 3958 This theorem is referenced by:  ssnelpssd  4093  ssexnelpss  4094  isfin4p1  9731  canthp1lem2  10069  nqpr  10430  uzindi  13345  nthruc  15600  nthruz  15601  vitali  24148  onpsstopbas  33681
 Copyright terms: Public domain W3C validator