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Theorem disjssun 3988
Description: Subset relation for disjoint classes. (Contributed by NM, 25-Oct-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
disjssun ((𝐴𝐵) = ∅ → (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴𝐶))

Proof of Theorem disjssun
StepHypRef Expression
1 uneq2 3723 . . . 4 ((𝐴𝐵) = ∅ → ((𝐴𝐶) ∪ (𝐴𝐵)) = ((𝐴𝐶) ∪ ∅))
2 indi 3832 . . . . 5 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
32equncomi 3721 . . . 4 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐶) ∪ (𝐴𝐵))
4 un0 3919 . . . . 5 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
54eqcomi 2619 . . . 4 (𝐴𝐶) = ((𝐴𝐶) ∪ ∅)
61, 3, 53eqtr4g 2669 . . 3 ((𝐴𝐵) = ∅ → (𝐴 ∩ (𝐵𝐶)) = (𝐴𝐶))
76eqeq1d 2612 . 2 ((𝐴𝐵) = ∅ → ((𝐴 ∩ (𝐵𝐶)) = 𝐴 ↔ (𝐴𝐶) = 𝐴))
8 df-ss 3554 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴 ∩ (𝐵𝐶)) = 𝐴)
9 df-ss 3554 . 2 (𝐴𝐶 ↔ (𝐴𝐶) = 𝐴)
107, 8, 93bitr4g 302 1 ((𝐴𝐵) = ∅ → (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 195   = wceq 1475  cun 3538  cin 3539  wss 3540  c0 3874
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875
This theorem is referenced by:  hashbclem  13093  alexsubALTlem2  21662  iccntr  22432  reconnlem1  22437  dvne0  23578
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