Theorem complss 4081

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 4073	. 2 ⊢ (𝐴 ⊆ 𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		sscon 4073	. . 3 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3		ddif 4071	. . 3 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴
4		ddif 4071	. . 3 ⊢ (V ∖ (V ∖ 𝐵)) = 𝐵
5	2, 3, 4	3sstr3g 3965	. 2 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴 ⊆ 𝐵)
6	1, 5	impbii 208	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))

Syntax hints:   ↔ wb 205  Vcvv 3432   ∖ cdif 3884   ⊆ wss 3887

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890  df-in 3894  df-ss 3904

This theorem is referenced by:  compleq  4082