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Theorem ssin0 41194
Description: If two classes are disjoint, two respective subclasses are disjoint. (Contributed by Glauco Siliprandi, 17-Aug-2020.)
Assertion
Ref Expression
ssin0 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐶𝐷) = ∅)

Proof of Theorem ssin0
StepHypRef Expression
1 ss2in 4210 . . . 4 ((𝐶𝐴𝐷𝐵) → (𝐶𝐷) ⊆ (𝐴𝐵))
213adant1 1122 . . 3 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐶𝐷) ⊆ (𝐴𝐵))
3 eqimss 4020 . . . 4 ((𝐴𝐵) = ∅ → (𝐴𝐵) ⊆ ∅)
433ad2ant1 1125 . . 3 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐴𝐵) ⊆ ∅)
52, 4sstrd 3974 . 2 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐶𝐷) ⊆ ∅)
6 ss0 4349 . 2 ((𝐶𝐷) ⊆ ∅ → (𝐶𝐷) = ∅)
75, 6syl 17 1 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐶𝐷) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  w3a 1079   = wceq 1528  cin 3932  wss 3933  c0 4288
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-3an 1081  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-rab 3144  df-v 3494  df-dif 3936  df-in 3940  df-ss 3949  df-nul 4289
This theorem is referenced by:  sge0resplit  42565
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