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Theorem ssdifsn 4718
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 4064 . . 3 (𝐴 ⊆ (𝐵 ∖ {𝐶}) → 𝐴𝐵)
2 reldisj 4382 . . . 4 (𝐴𝐵 → ((𝐴 ∩ {𝐶}) = ∅ ↔ 𝐴 ⊆ (𝐵 ∖ {𝐶})))
32bicomd 222 . . 3 (𝐴𝐵 → (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴 ∩ {𝐶}) = ∅))
41, 3biadanii 818 . 2 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅))
5 disjsn 4644 . . 3 ((𝐴 ∩ {𝐶}) = ∅ ↔ ¬ 𝐶𝐴)
65anbi2i 622 . 2 ((𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
74, 6bitri 274 1 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205  wa 395   = wceq 1539  wcel 2108  cdif 3880  cin 3882  wss 3883  c0 4253  {csn 4558
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-v 3424  df-dif 3886  df-in 3890  df-ss 3900  df-nul 4254  df-sn 4559
This theorem is referenced by:  logdivsqrle  32530  naddcllem  33758  elsetrecslem  46290
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