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Theorem inssdif0 3980
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 3829 . . . . . 6 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
21imbi1i 338 . . . . 5 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) → 𝑥𝐶))
3 iman 439 . . . . 5 (((𝑥𝐴𝑥𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
42, 3bitri 264 . . . 4 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
5 eldif 3617 . . . . . 6 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵 ∧ ¬ 𝑥𝐶))
65anbi2i 730 . . . . 5 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
7 elin 3829 . . . . 5 (𝑥 ∈ (𝐴 ∩ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
8 anass 682 . . . . 5 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
96, 7, 83bitr4ri 293 . . . 4 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
104, 9xchbinx 323 . . 3 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1110albii 1787 . 2 (∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
12 dfss2 3624 . 2 ((𝐴𝐵) ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶))
13 eq0 3962 . 2 ((𝐴 ∩ (𝐵𝐶)) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1411, 12, 133bitr4i 292 1 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵𝐶)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 196  wa 383  wal 1521   = wceq 1523  wcel 2030  cdif 3604  cin 3606  wss 3607  c0 3948
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-v 3233  df-dif 3610  df-in 3614  df-ss 3621  df-nul 3949
This theorem is referenced by:  disjdif  4073  inf3lem3  8565  ssfin4  9170  isnrm2  21210  1stccnp  21313  llycmpkgen2  21401  ufileu  21770  fclscf  21876  flimfnfcls  21879  inindif  29479  opnbnd  32445  diophrw  37639  setindtr  37908
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