Theorem bj-sscon 34780

Description: Contraposition law for relative subclasses. Relative and generalized version of ssconb 4045, which it can shorten, as well as conss2 41555. (Contributed by BJ, 11-Nov-2021.) This proof does not rely, even indirectly, on ssconb 4045 nor conss2 41555. (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-sscon	⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ (𝐵 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐴))

Proof of Theorem bj-sscon

Step	Hyp	Ref	Expression
1		incom 4108	. . . 4 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	ineq1i 4115	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ((𝐵 ∩ 𝐴) ∩ 𝑉)
3	2	eqeq1i 2763	. 2 ⊢ (((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅ ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
4		bj-disj2r 34779	. 2 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅)
5		bj-disj2r 34779	. 2 ⊢ ((𝐵 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐴) ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
6	3, 4, 5	3bitr4i 306	1 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ (𝐵 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐴))

Syntax hints:   ↔ wb 209   = wceq 1538   ∖ cdif 3857   ∩ cin 3859   ⊆ wss 3860  ∅c0 4227

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2729

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-ral 3075  df-rab 3079  df-v 3411  df-dif 3863  df-in 3867  df-ss 3877  df-nul 4228

This theorem is referenced by: (None)