Theorem complss 4102

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 4094	. 2 ⊢ (𝐴 ⊆ 𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		sscon 4094	. . 3 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3		ddif 4092	. . 3 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴
4		ddif 4092	. . 3 ⊢ (V ∖ (V ∖ 𝐵)) = 𝐵
5	2, 3, 4	3sstr3g 3986	. 2 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴 ⊆ 𝐵)
6	1, 5	impbii 211	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))

Syntax hints:   ↔ wb 208  Vcvv 3453   ∖ cdif 3899   ⊆ wss 3902

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1562  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3455  df-dif 3905  df-ss 3919

This theorem is referenced by:  compleq  4103