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Theorem nssss 5455
Description: Negation of subclass relationship. Compare nss 4046. (Contributed by NM, 30-Jun-2004.) (Proof shortened by Andrew Salmon, 25-Jul-2011.)
Assertion
Ref Expression
nssss 𝐴𝐵 ↔ ∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nssss
StepHypRef Expression
1 exanali 1861 . . 3 (∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵) ↔ ¬ ∀𝑥(𝑥𝐴𝑥𝐵))
2 ssextss 5453 . . 3 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
31, 2xchbinxr 335 . 2 (∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵) ↔ ¬ 𝐴𝐵)
43bicomi 223 1 𝐴𝐵 ↔ ∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 395  wal 1538  wex 1780  wss 3948
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702  ax-sep 5299  ax-pr 5427
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3475  df-un 3953  df-in 3955  df-ss 3965  df-pw 4604  df-sn 4629  df-pr 4631
This theorem is referenced by: (None)
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