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Theorem complss 4139
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 4131 . 2 (𝐴𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2 sscon 4131 . . 3 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3 ddif 4129 . . 3 (V ∖ (V ∖ 𝐴)) = 𝐴
4 ddif 4129 . . 3 (V ∖ (V ∖ 𝐵)) = 𝐵
52, 3, 43sstr3g 4019 . 2 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴𝐵)
61, 5impbii 208 1 (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 205  Vcvv 3466  cdif 3938  wss 3941
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2695
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802  df-v 3468  df-dif 3944  df-in 3948  df-ss 3958
This theorem is referenced by:  compleq  4140
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