Theorem complss 3713

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 3706	. 2 ⊢ (𝐴 ⊆ 𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		sscon 3706	. . 3 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3		ddif 3704	. . 3 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴
4		ddif 3704	. . 3 ⊢ (V ∖ (V ∖ 𝐵)) = 𝐵
5	2, 3, 4	3sstr3g 3608	. 2 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴 ⊆ 𝐵)
6	1, 5	impbii 198	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))

Syntax hints:   ↔ wb 195  Vcvv 3173   ∖ cdif 3537   ⊆ wss 3540

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-dif 3543  df-in 3547  df-ss 3554

This theorem is referenced by:  compleq  3714