Theorem bj-sscon 35146

Description: Contraposition law for relative subclasses. Relative and generalized version of ssconb 4068, which it can shorten, as well as conss2 41950. (Contributed by BJ, 11-Nov-2021.) This proof does not rely, even indirectly, on ssconb 4068 nor conss2 41950. (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 4131	. . . 4 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	ineq1i 4139	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ((𝐵 ∩ 𝐴) ∩ 𝑉)
3	2	eqeq1i 2743	. 2 ⊢ (((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅ ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
4		bj-disj2r 35145	. 2 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅)
5		bj-disj2r 35145	. 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 3880   ∩ cin 3882   ⊆ wss 3883  ∅c0 4253

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rab 3072  df-v 3424  df-dif 3886  df-in 3890  df-ss 3900  df-nul 4254

This theorem is referenced by: (None)