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Theorem complss 4052
 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 4044 . 2 (𝐴𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2 sscon 4044 . . 3 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3 ddif 4042 . . 3 (V ∖ (V ∖ 𝐴)) = 𝐴
4 ddif 4042 . . 3 (V ∖ (V ∖ 𝐵)) = 𝐵
52, 3, 43sstr3g 3936 . 2 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴𝐵)
61, 5impbii 212 1 (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 209  Vcvv 3409   ∖ cdif 3855   ⊆ wss 3858 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2729 This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-v 3411  df-dif 3861  df-in 3865  df-ss 3875 This theorem is referenced by:  compleq  4053
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