Theorem bj-sscon 37030

Description: Contraposition law for relative subclasses. Relative and generalized version of ssconb 4142, which it can shorten, as well as conss2 44462. (Contributed by BJ, 11-Nov-2021.) This proof does not rely, even indirectly, on ssconb 4142 nor conss2 44462. (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 4209	. . . 4 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	ineq1i 4216	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ((𝐵 ∩ 𝐴) ∩ 𝑉)
3	2	eqeq1i 2742	. 2 ⊢ (((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅ ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
4		bj-disj2r 37029	. 2 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅)
5		bj-disj2r 37029	. 2 ⊢ ((𝐵 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐴) ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
6	3, 4, 5	3bitr4i 303	1 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ (𝐵 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐴))

Syntax hints:   ↔ wb 206   = wceq 1540   ∖ cdif 3948   ∩ cin 3950   ⊆ wss 3951  ∅c0 4333

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 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rab 3437  df-v 3482  df-dif 3954  df-in 3958  df-ss 3968  df-nul 4334

This theorem is referenced by: (None)