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Theorem inssdif0im 3264
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 3099 . . . . . 6 (x (AB) ↔ (x A x B))
21imbi1i 227 . . . . 5 ((x (AB) → x 𝐶) ↔ ((x A x B) → x 𝐶))
3 imanim 773 . . . . 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 2900 . . . . . 6 (x (B𝐶) ↔ (x B ¬ x 𝐶))
65anbi2i 433 . . . . 5 ((x A x (B𝐶)) ↔ (x A (x B ¬ x 𝐶)))
7 elin 3099 . . . . 5 (x (A ∩ (B𝐶)) ↔ (x A x (B𝐶)))
8 anass 383 . . . . 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 588 . . 3 ((x (AB) → x 𝐶) → ¬ x (A ∩ (B𝐶)))
1110alimi 1320 . 2 (x(x (AB) → x 𝐶) → x ¬ x (A ∩ (B𝐶)))
12 dfss2 2907 . 2 ((AB) ⊆ 𝐶x(x (AB) → x 𝐶))
13 eq0 3212 . 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 1224   = wceq 1226   wcel 1370  cdif 2887  cin 2889  wss 2890  c0 3197
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 532  ax-in2 533  ax-io 617  ax-5 1312  ax-7 1313  ax-gen 1314  ax-ie1 1359  ax-ie2 1360  ax-8 1372  ax-10 1373  ax-11 1374  ax-i12 1375  ax-bnd 1376  ax-4 1377  ax-17 1396  ax-i9 1400  ax-ial 1405  ax-i5r 1406  ax-ext 2000
This theorem depends on definitions:  df-bi 110  df-tru 1229  df-nf 1326  df-sb 1624  df-clab 2005  df-cleq 2011  df-clel 2014  df-nfc 2145  df-v 2533  df-dif 2893  df-in 2897  df-ss 2904  df-nul 3198
This theorem is referenced by:  disjdif  3269
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