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Theorem inssdif0im 3285
Description: Intersection, subclass, and difference relationship. In classical logic the converse would also hold. (Contributed by Jim Kingdon, 3-Aug-2018.)
Assertion
Ref Expression
inssdif0im ((AB) ⊆ 𝐶 → (A ∩ (B𝐶)) = ∅)

Proof of Theorem inssdif0im
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 elin 3120 . . . . . 6 (x (AB) ↔ (x A x B))
21imbi1i 227 . . . . 5 ((x (AB) → x 𝐶) ↔ ((x A x B) → x 𝐶))
3 imanim 784 . . . . 5 (((x A x B) → x 𝐶) → ¬ ((x A x B) ¬ x 𝐶))
42, 3sylbi 114 . . . 4 ((x (AB) → x 𝐶) → ¬ ((x A x B) ¬ x 𝐶))
5 eldif 2921 . . . . . 6 (x (B𝐶) ↔ (x B ¬ x 𝐶))
65anbi2i 430 . . . . 5 ((x A x (B𝐶)) ↔ (x A (x B ¬ x 𝐶)))
7 elin 3120 . . . . 5 (x (A ∩ (B𝐶)) ↔ (x A x (B𝐶)))
8 anass 381 . . . . 5 (((x A x B) ¬ x 𝐶) ↔ (x A (x B ¬ x 𝐶)))
96, 7, 83bitr4ri 202 . . . 4 (((x A x B) ¬ x 𝐶) ↔ x (A ∩ (B𝐶)))
104, 9sylnib 600 . . 3 ((x (AB) → x 𝐶) → ¬ x (A ∩ (B𝐶)))
1110alimi 1341 . 2 (x(x (AB) → x 𝐶) → x ¬ x (A ∩ (B𝐶)))
12 dfss2 2928 . 2 ((AB) ⊆ 𝐶x(x (AB) → x 𝐶))
13 eq0 3233 . 2 ((A ∩ (B𝐶)) = ∅ ↔ x ¬ x (A ∩ (B𝐶)))
1411, 12, 133imtr4i 190 1 ((AB) ⊆ 𝐶 → (A ∩ (B𝐶)) = ∅)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4   wa 97  wal 1240   = wceq 1242   wcel 1390  cdif 2908  cin 2910  wss 2911  c0 3218
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-dif 2914  df-in 2918  df-ss 2925  df-nul 3219
This theorem is referenced by:  disjdif  3290
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