Theorem bj-sscon 37146

Description: Contraposition law for relative subclasses. Relative and generalized version of ssconb 4091, which it can shorten, as well as conss2 44599. (Contributed by BJ, 11-Nov-2021.) This proof does not rely, even indirectly, on ssconb 4091 nor conss2 44599. (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 4158	. . . 4 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	ineq1i 4165	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ((𝐵 ∩ 𝐴) ∩ 𝑉)
3	2	eqeq1i 2738	. 2 ⊢ (((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅ ↔ ((𝐵 ∩ 𝐴) ∩ 𝑉) = ∅)
4		bj-disj2r 37145	. 2 ⊢ ((𝐴 ∩ 𝑉) ⊆ (𝑉 ∖ 𝐵) ↔ ((𝐴 ∩ 𝐵) ∩ 𝑉) = ∅)
5		bj-disj2r 37145	. 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 1541   ∖ cdif 3895   ∩ cin 3897   ⊆ wss 3898  ∅c0 4282

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ral 3049  df-rab 3397  df-v 3439  df-dif 3901  df-in 3905  df-ss 3915  df-nul 4283

This theorem is referenced by: (None)