Metamath Proof Explorer


Theorem complss

Description: Complementation reverses inclusion. (Contributed by Andrew Salmon, 15-Jul-2011) (Proof shortened by BJ, 19-Mar-2021)

Ref Expression
Assertion complss ⊢ A ⊆ B ↔ V ∖ B ⊆ V ∖ A

Proof

Step Hyp Ref Expression
1 sscon ⊢ A ⊆ B → V ∖ B ⊆ V ∖ A
2 sscon ⊢ V ∖ B ⊆ V ∖ A → V ∖ V ∖ A ⊆ V ∖ V ∖ B
3 ddif ⊢ V ∖ V ∖ A = A
4 ddif ⊢ V ∖ V ∖ B = B
5 2 3 4 3sstr3g ⊢ V ∖ B ⊆ V ∖ A → A ⊆ B
6 1 5 impbii ⊢ A ⊆ B ↔ V ∖ B ⊆ V ∖ A