Theorem complss 4174

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

Assertion

Ref	Expression
complss	⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))

Proof of Theorem complss

Step	Hyp	Ref	Expression
1		sscon 4166	. 2 ⊢ (𝐴 ⊆ 𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		sscon 4166	. . 3 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3		ddif 4164	. . 3 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴
4		ddif 4164	. . 3 ⊢ (V ∖ (V ∖ 𝐵)) = 𝐵
5	2, 3, 4	3sstr3g 4053	. 2 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴 ⊆ 𝐵)
6	1, 5	impbii 209	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))

Syntax hints:   ↔ wb 206  Vcvv 3488   ∖ cdif 3973   ⊆ wss 3976

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490  df-dif 3979  df-ss 3993

This theorem is referenced by:  compleq  4175