Set Difference of Intersection with Set is Empty Set

Theorem

The set difference of the intersection of two sets with one of those sets is the empty set.

Let $S, T$ be sets.

Then:

$\paren {S \cap T} \setminus S = \O$

$\paren {S \cap T} \setminus T = \O$

Proof

From Set Difference is Right Distributive over Set Intersection:

$\paren {R \cap S} \setminus T = \paren {R \setminus T} \cap paren {S \setminus T}$

Hence:

\(\ds \paren {S \cap T} \setminus S\)

\(=\)

\(\ds \paren {S \setminus S} \cap \paren {T \setminus S}\)

Set Difference is Right Distributive over Set Intersection

\(\ds \)

\(=\)

\(\ds \O \cap \paren {T \setminus S}\)

Set Difference with Self is Empty Set

\(\ds \)

\(=\)

\(\ds \O\)

Intersection with Empty Set

$\Box$

\(\ds \paren {S \cap T} \setminus T\)

\(=\)

\(\ds \paren {T \cap S} \setminus T\)

Intersection is Commutative

\(\ds \)

\(=\)

\(\ds \O\)

from above

$\blacksquare$