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Theorem inssdif0 4308
Description: Intersection, subclass, and difference relationship. (Contributed by NM, 27-Oct-1996.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof shortened by Wolf Lammen, 30-Sep-2014.)
Assertion
Ref Expression
inssdif0 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵𝐶)) = ∅)

Proof of Theorem inssdif0
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 elin 3907 . . . . . 6 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
21imbi1i 349 . . . . 5 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) → 𝑥𝐶))
3 iman 401 . . . . 5 (((𝑥𝐴𝑥𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
42, 3bitri 274 . . . 4 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
5 eldif 3901 . . . . . 6 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵 ∧ ¬ 𝑥𝐶))
65anbi2i 622 . . . . 5 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
7 elin 3907 . . . . 5 (𝑥 ∈ (𝐴 ∩ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
8 anass 468 . . . . 5 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
96, 7, 83bitr4ri 303 . . . 4 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
104, 9xchbinx 333 . . 3 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1110albii 1825 . 2 (∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
12 dfss2 3911 . 2 ((𝐴𝐵) ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶))
13 eq0 4282 . 2 ((𝐴 ∩ (𝐵𝐶)) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1411, 12, 133bitr4i 302 1 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵𝐶)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 395  wal 1539   = wceq 1541  wcel 2109  cdif 3888  cin 3890  wss 3891  c0 4261
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-ext 2710
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1544  df-fal 1554  df-ex 1786  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-v 3432  df-dif 3894  df-in 3898  df-ss 3908  df-nul 4262
This theorem is referenced by:  disjdif  4410  inf3lem3  9349  ssfin4  10050  isnrm2  22490  1stccnp  22594  llycmpkgen2  22682  ufileu  23051  fclscf  23157  flimfnfcls  23160  inindif  30842  opnbnd  34493  diophrw  40561  setindtr  40826
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