Theorem complss 4090

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 4082	. 2 ⊢ (𝐴 ⊆ 𝐵 → (V ∖ 𝐵) ⊆ (V ∖ 𝐴))
2		sscon 4082	. . 3 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → (V ∖ (V ∖ 𝐴)) ⊆ (V ∖ (V ∖ 𝐵)))
3		ddif 4080	. . 3 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴
4		ddif 4080	. . 3 ⊢ (V ∖ (V ∖ 𝐵)) = 𝐵
5	2, 3, 4	3sstr3g 3965	. 2 ⊢ ((V ∖ 𝐵) ⊆ (V ∖ 𝐴) → 𝐴 ⊆ 𝐵)
6	1, 5	impbii 212	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (V ∖ 𝐵) ⊆ (V ∖ 𝐴))

This proof depends on syntax axioms:   ↔ wb 209  Vcvv 3454   ⊆ wss 3900   ∖ cdif 4048

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-ss 3901  df-dif 4050

This theorem is used by:  compleq  4091