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Theorem ssnelpss 3680
 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 2713 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2617 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 317 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 3654 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 943 . 2 (𝐴𝐵 → (¬ 𝐴 = 𝐵𝐴𝐵))
63, 5syl5ib 233 1 (𝐴𝐵 → ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → 𝐴𝐵))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 383   = wceq 1475   ∈ wcel 1977   ⊆ wss 3540   ⊊ wpss 3541 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-cleq 2603  df-clel 2606  df-ne 2782  df-pss 3556 This theorem is referenced by:  ssnelpssd  3681  ssexnelpss  3682  canthp1lem2  9354  nqpr  9715  uzindi  12643  nthruc  14819  nthruz  14820  vitali  23188  onpsstopbas  31599
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