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Theorem ssnelpss 3289
 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 2139 . . 3 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐵 = 𝐴)
2 eqcom 2042 . . 3 (𝐵 = 𝐴𝐴 = 𝐵)
31, 2sylnib 601 . 2 ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → ¬ 𝐴 = 𝐵)
4 dfpss2 3029 . . 3 (𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))
54baibr 829 . 2 (𝐴𝐵 → (¬ 𝐴 = 𝐵𝐴𝐵))
63, 5syl5ib 143 1 (𝐴𝐵 → ((𝐶𝐵 ∧ ¬ 𝐶𝐴) → 𝐴𝐵))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97   = wceq 1243   ∈ wcel 1393   ⊆ wss 2917   ⊊ wpss 2918 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-17 1419  ax-ial 1427  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-cleq 2033  df-clel 2036  df-ne 2206  df-pss 2933 This theorem is referenced by:  ssnelpssd  3290
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