Theorem bj-sscon 36213

Description: Contraposition law for relative subclasses. Relative and generalized version of ssconb 4136, which it can shorten, as well as conss2 43504. (Contributed by BJ, 11-Nov-2021.) This proof does not rely, even indirectly, on ssconb 4136 nor conss2 43504. (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 4200	. . . 4 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	ineq1i 4207	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ((𝐵 ∩ 𝐴) ∩ 𝑉)
3	2	eqeq1i 2735	. 2 ⊢ (((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅ ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
4		bj-disj2r 36212	. 2 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅)
5		bj-disj2r 36212	. 2 ⊢ ((𝐵 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐴) ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
6	3, 4, 5	3bitr4i 302	1 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ (𝐵 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐴))

Syntax hints:   ↔ wb 205   = wceq 1539   ∖ cdif 3944   ∩ cin 3946   ⊆ wss 3947  ∅c0 4321

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1542  df-fal 1552  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-ral 3060  df-rab 3431  df-v 3474  df-dif 3950  df-in 3954  df-ss 3964  df-nul 4322

This theorem is referenced by: (None)