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Theorem complss 4120
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 4112 . 2 (𝐴𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2 sscon 4112 . . 3 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3 ddif 4110 . . 3 (V ∖ (V ∖ 𝐴)) = 𝐴
4 ddif 4110 . . 3 (V ∖ (V ∖ 𝐵)) = 𝐵
52, 3, 43sstr3g 4008 . 2 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴𝐵)
61, 5impbii 210 1 (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 207  Vcvv 3492  cdif 3930  wss 3933
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-v 3494  df-dif 3936  df-in 3940  df-ss 3949
This theorem is referenced by:  compleq  4121
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