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Theorem complss 3882
 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 3875 . 2 (𝐴𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2 sscon 3875 . . 3 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3 ddif 3873 . . 3 (V ∖ (V ∖ 𝐴)) = 𝐴
4 ddif 3873 . . 3 (V ∖ (V ∖ 𝐵)) = 𝐵
52, 3, 43sstr3g 3774 . 2 ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴𝐵)
61, 5impbii 199 1 (𝐴𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 196  Vcvv 3328   ∖ cdif 3700   ⊆ wss 3703 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1859  ax-4 1874  ax-5 1976  ax-6 2042  ax-7 2078  ax-9 2136  ax-10 2156  ax-11 2171  ax-12 2184  ax-13 2379  ax-ext 2728 This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1623  df-ex 1842  df-nf 1847  df-sb 2035  df-clab 2735  df-cleq 2741  df-clel 2744  df-nfc 2879  df-v 3330  df-dif 3706  df-in 3710  df-ss 3717 This theorem is referenced by:  compleq  3883
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