ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  ssdifsn GIF version

Theorem ssdifsn 3687
Description: Subset of a set with an element removed. (Contributed by Emmett Weisz, 7-Jul-2021.) (Proof shortened by JJ, 31-May-2022.)
Assertion
Ref Expression
ssdifsn (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))

Proof of Theorem ssdifsn
StepHypRef Expression
1 difss2 3235 . . 3 (𝐴 ⊆ (𝐵 ∖ {𝐶}) → 𝐴𝐵)
2 reldisj 3445 . . . 4 (𝐴𝐵 → ((𝐴 ∩ {𝐶}) = ∅ ↔ 𝐴 ⊆ (𝐵 ∖ {𝐶})))
32bicomd 140 . . 3 (𝐴𝐵 → (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴 ∩ {𝐶}) = ∅))
41, 3biadan2 452 . 2 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅))
5 disjsn 3621 . . 3 ((𝐴 ∩ {𝐶}) = ∅ ↔ ¬ 𝐶𝐴)
65anbi2i 453 . 2 ((𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
74, 6bitri 183 1 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wa 103  wb 104   = wceq 1335  wcel 2128  cdif 3099  cin 3101  wss 3102  c0 3394  {csn 3560
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139
This theorem depends on definitions:  df-bi 116  df-tru 1338  df-fal 1341  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-ral 2440  df-v 2714  df-dif 3104  df-in 3108  df-ss 3115  df-nul 3395  df-sn 3566
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator