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Theorem ssin0 41398
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 4206 . . . 4 ((𝐶𝐴𝐷𝐵) → (𝐶𝐷) ⊆ (𝐴𝐵))
213adant1 1125 . . 3 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐶𝐷) ⊆ (𝐴𝐵))
3 eqimss 4016 . . . 4 ((𝐴𝐵) = ∅ → (𝐴𝐵) ⊆ ∅)
433ad2ant1 1128 . . 3 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐴𝐵) ⊆ ∅)
52, 4sstrd 3970 . 2 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐶𝐷) ⊆ ∅)
6 ss0 4345 . 2 ((𝐶𝐷) ⊆ ∅ → (𝐶𝐷) = ∅)
75, 6syl 17 1 (((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → (𝐶𝐷) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  w3a 1082   = wceq 1536  cin 3928  wss 3929  c0 4284
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2792
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1084  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-clab 2799  df-cleq 2813  df-clel 2892  df-nfc 2962  df-rab 3146  df-v 3493  df-dif 3932  df-in 3936  df-ss 3945  df-nul 4285
This theorem is referenced by:  sge0resplit  42769
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