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Theorem inssdif0 4283
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 3897 . . . . . 6 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
21imbi1i 353 . . . . 5 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) → 𝑥𝐶))
3 iman 405 . . . . 5 (((𝑥𝐴𝑥𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
42, 3bitri 278 . . . 4 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ ((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶))
5 eldif 3891 . . . . . 6 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵 ∧ ¬ 𝑥𝐶))
65anbi2i 625 . . . . 5 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
7 elin 3897 . . . . 5 (𝑥 ∈ (𝐴 ∩ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
8 anass 472 . . . . 5 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ (𝑥𝐴 ∧ (𝑥𝐵 ∧ ¬ 𝑥𝐶)))
96, 7, 83bitr4ri 307 . . . 4 (((𝑥𝐴𝑥𝐵) ∧ ¬ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
104, 9xchbinx 337 . . 3 ((𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1110albii 1821 . 2 (∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶) ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
12 dfss2 3901 . 2 ((𝐴𝐵) ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ (𝐴𝐵) → 𝑥𝐶))
13 eq0 4258 . 2 ((𝐴 ∩ (𝐵𝐶)) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
1411, 12, 133bitr4i 306 1 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵𝐶)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 209  wa 399  wal 1536   = wceq 1538  wcel 2111  cdif 3878  cin 3880  wss 3881  c0 4243
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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-v 3443  df-dif 3884  df-in 3888  df-ss 3898  df-nul 4244
This theorem is referenced by:  disjdif  4379  inf3lem3  9077  ssfin4  9721  isnrm2  21963  1stccnp  22067  llycmpkgen2  22155  ufileu  22524  fclscf  22630  flimfnfcls  22633  inindif  30287  opnbnd  33786  diophrw  39700  setindtr  39965
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