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Theorem ssdifsn 4761
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 4100 . . 3 (𝐴 ⊆ (𝐵 ∖ {𝐶}) → 𝐴𝐵)
2 reldisj 4419 . . . 4 (𝐴𝐵 → ((𝐴 ∩ {𝐶}) = ∅ ↔ 𝐴 ⊆ (𝐵 ∖ {𝐶})))
32bicomd 226 . . 3 (𝐴𝐵 → (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴 ∩ {𝐶}) = ∅))
41, 3biadanii 833 . 2 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅))
5 disjsn 4682 . . 3 ((𝐴 ∩ {𝐶}) = ∅ ↔ ¬ 𝐶𝐴)
65anbi2i 634 . 2 ((𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
74, 6bitri 278 1 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ ¬ 𝐶𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 209  wa 400   = wceq 1568  wcel 2150  cdif 3910  cin 3912  wss 3913  c0 4294  {csn 4594
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2152  ax-9 2160  ax-ext 2742
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1571  df-fal 1581  df-ex 1808  df-sb 2099  df-clab 2749  df-cleq 2762  df-clel 2845  df-ral 3087  df-v 3464  df-dif 3916  df-in 3920  df-ss 3930  df-nul 4295  df-sn 4595
This theorem is referenced by:  naddcllem  8665  isdomn6  20801  isdrng4  20828  imadrhmcl  20883  drngmxidl  33729  esplyind  33935  assafld  33997  logdivsqrle  35007  elsetrecslem  50426
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