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Theorem ssdifsn 4792
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 4134 . . 3 (𝐴 ⊆ (𝐵 ∖ {𝐶}) → 𝐴𝐵)
2 reldisj 4452 . . . 4 (𝐴𝐵 → ((𝐴 ∩ {𝐶}) = ∅ ↔ 𝐴 ⊆ (𝐵 ∖ {𝐶})))
32bicomd 222 . . 3 (𝐴𝐵 → (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴 ∩ {𝐶}) = ∅))
41, 3biadanii 821 . 2 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅))
5 disjsn 4716 . . 3 ((𝐴 ∩ {𝐶}) = ∅ ↔ ¬ 𝐶𝐴)
65anbi2i 624 . 2 ((𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
74, 6bitri 275 1 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205  wa 397   = wceq 1542  wcel 2107  cdif 3946  cin 3948  wss 3949  c0 4323  {csn 4629
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ral 3063  df-v 3477  df-dif 3952  df-in 3956  df-ss 3966  df-nul 4324  df-sn 4630
This theorem is referenced by:  naddcllem  8675  imadrhmcl  20413  isdrng4  32395  drngmxidl  32593  logdivsqrle  33662  elsetrecslem  47744
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