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Theorem ssnelpss 3696
 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 2723 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2628 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 318 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 3670 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 944 . 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 384   = wceq 1480   ∈ wcel 1987   ⊆ wss 3555   ⊊ wpss 3556 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-ext 2601 This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1702  df-cleq 2614  df-clel 2617  df-ne 2791  df-pss 3571 This theorem is referenced by:  ssnelpssd  3697  ssexnelpss  3698  canthp1lem2  9419  nqpr  9780  uzindi  12721  nthruc  14905  nthruz  14906  vitali  23288  onpsstopbas  32071
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