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Theorem complss 3973
 Description: Complementation reverses inclusion. (Contributed by Andrew Salmon, 15-Jul-2011.) (Proof shortened by BJ, 19-Mar-2021.)
Assertion
Ref Expression
complss (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))

Proof of Theorem complss
StepHypRef Expression
1 sscon 3966 . 2 (𝐴𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2 sscon 3966 . . 3 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3 ddif 3964 . . 3 (V ∖ (V ∖ 𝐴)) = 𝐴
4 ddif 3964 . . 3 (V ∖ (V ∖ 𝐵)) = 𝐵
52, 3, 43sstr3g 3863 . 2 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴𝐵)
61, 5impbii 201 1 (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 198  Vcvv 3397   ∖ cdif 3788   ⊆ wss 3791 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2054  ax-9 2115  ax-10 2134  ax-11 2149  ax-12 2162  ax-ext 2753 This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-clab 2763  df-cleq 2769  df-clel 2773  df-nfc 2920  df-v 3399  df-dif 3794  df-in 3798  df-ss 3805 This theorem is referenced by:  compleq  3974
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