Set Difference with Set Difference is Union of Set Difference with Intersection/Corollary

Theorem

Let $S$ and $T$ be sets.

Then:

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

where $S \setminus T$ denotes set difference.

Proof

From Set Difference with Set Difference is Union of Set Difference with Intersection:

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

where $R, S, T$ are sets.

Hence:

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

\(=\)

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

Set Difference with Set Difference is Union of Set Difference with Intersection

\(\ds \)

\(=\)

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

Set Intersection is Idempotent

\(\ds \)

\(=\)

\(\ds T\)

Set Difference Union First Set is First Set

$\blacksquare$