Let A, B, C and D be sets. Prove that A / B and C / D are disjoint if and only if A ∩ C ⊆ B ∪ D

Step-by-step explanation:

We have to prove both implications of the affirmation.

1) Let's assume that A / B and C / D are disjoint, we have to prove that A ∩ C ⊆ B ∪ D.

We'll prove it by reducing to absurd.

Let's suppose that A ∩ C ⊄ B ∪ D. That means that there is an element x that belongs to A ∩ C but not to B ∪ D.

As x belongs to A ∩ C, x ∈ A and x ∈ C.

As x doesn't belong to B ∪ D, x ∉ B and x ∉ D.

With this, we can say that x ∈ A / B and x ∈ C / D.

Therefore, x ∈ (A / B) ∩ (C / D), absurd!

It's absurd because we were assuming that A / B and C / D were disjoint, therefore their intersection must be empty.

The absurd came from assuming that A ∩ C ⊄ B ∪ D.

That proves that A ∩ C ⊆ B ∪ D.

2) Let's assume that A ∩ C ⊆ B ∪ D, we have to prove that A / B and C / D are disjoint (i. e. A / B ∩ C / D is empty)

We'll prove it again by reducing to absurd.

Let's suppose that A / B ∩ C / D is not empty. That means there is an element x that belongs to A / B ∩ C / D. Therefore, x ∈ A / B and x ∈ C / D.

As x ∈ A / B, x belongs to A but x doesn't belong to B.

As x ∈ C / D, x belongs to C but x doesn't belong to D.

With this, we can say that x ∈ A ∩ C and x ∉ B ∪ D.

So, there is an element that belongs to A ∩ C but not to B∪D, absurd!

It's absurd because we were assuming that A ∩ C ⊆ B ∪ D, therefore every element of A ∩ C must belong to B ∪ D.

The absurd came from assuming that A / B ∩ C / D is not empty.

That proves that A / B ∩ C / D is empty, i. e. A / B and C / D are disjoint.