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Theorem ssdifsn 3615
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 3168 . . 3 (𝐴 ⊆ (𝐵 ∖ {𝐶}) → 𝐴𝐵)
2 reldisj 3378 . . . 4 (𝐴𝐵 → ((𝐴 ∩ {𝐶}) = ∅ ↔ 𝐴 ⊆ (𝐵 ∖ {𝐶})))
32bicomd 140 . . 3 (𝐴𝐵 → (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴 ∩ {𝐶}) = ∅))
41, 3biadan2 449 . 2 (𝐴 ⊆ (𝐵 ∖ {𝐶}) ↔ (𝐴𝐵 ∧ (𝐴 ∩ {𝐶}) = ∅))
5 disjsn 3549 . . 3 ((𝐴 ∩ {𝐶}) = ∅ ↔ ¬ 𝐶𝐴)
65anbi2i 450 . 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 1312  wcel 1461  cdif 3032  cin 3034  wss 3035  c0 3327  {csn 3491
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 586  ax-in2 587  ax-io 681  ax-5 1404  ax-7 1405  ax-gen 1406  ax-ie1 1450  ax-ie2 1451  ax-8 1463  ax-10 1464  ax-11 1465  ax-i12 1466  ax-bndl 1467  ax-4 1468  ax-17 1487  ax-i9 1491  ax-ial 1495  ax-i5r 1496  ax-ext 2095
This theorem depends on definitions:  df-bi 116  df-tru 1315  df-fal 1318  df-nf 1418  df-sb 1717  df-clab 2100  df-cleq 2106  df-clel 2109  df-nfc 2242  df-ral 2393  df-v 2657  df-dif 3037  df-in 3041  df-ss 3048  df-nul 3328  df-sn 3497
This theorem is referenced by: (None)
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