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Theorem inssdif0 4315
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 3906 . . . . . 6 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
21imbi1i 349 . . . . 5 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) → 𝑥𝐶))
3 iman 401 . . . . 5 (((𝑥𝐴𝑥𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
42, 3bitri 275 . . . 4 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
5 eldif 3900 . . . . . 6 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵 ∧ ¬ 𝑥𝐶))
65anbi2i 624 . . . . 5 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
7 elin 3906 . . . . 5 (𝑥 ∈ (𝐴 ∩ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
8 anass 468 . . . . 5 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
96, 7, 83bitr4ri 304 . . . 4 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
104, 9xchbinx 334 . . 3 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1110albii 1821 . 2 (∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
12 df-ss 3907 . 2 ((𝐴𝐵) ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶))
13 eq0 4291 . 2 ((𝐴 ∩ (𝐵𝐶)) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1411, 12, 133bitr4i 303 1 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵𝐶)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wa 395  wal 1540   = wceq 1542  wcel 2114  cdif 3887  cin 3889  wss 3890  c0 4274
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3432  df-dif 3893  df-in 3897  df-ss 3907  df-nul 4275
This theorem is referenced by:  inindif  4316  disjdif  4413  inf3lem3  9545  ssfin4  10226  isnrm2  23336  1stccnp  23440  llycmpkgen2  23528  ufileu  23897  fclscf  24003  flimfnfcls  24006  opnbnd  36526  ttcwf2  36726  diophrw  43208  setindtr  43473
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