Set Difference with Set Difference

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Theorem

The set difference with the set difference of two sets is the intersection of the two sets:

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


Proof 1

\(\ds S \setminus \paren {S \setminus T}\) \(=\) \(\ds \paren {S \setminus S} \cup \paren {S \cap T}\) Set Difference with Set Difference is Union of Set Difference with Intersection
\(\ds \) \(=\) \(\ds \O \cup \paren {S \cap T}\) Set Difference with Self is Empty Set
\(\ds \) \(=\) \(\ds S \cap T\) Union with Empty Set


Interchanging $S$ and $T$:

\(\ds T \setminus \paren {T \setminus S}\) \(=\) \(\ds T \cap S\)
\(\ds \) \(=\) \(\ds S \cap T\) Intersection is Commutative

$\blacksquare$


Proof 2

From the Axiom of Transitivity, all sets are classes.

The result then follows from Class Difference with Class Difference.


This needs considerable tedious hard slog to complete it.
In particular: Find the result that demonstrate the set difference of two sets is also a set, intersection as well. Probably via subset of set is set or something.
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