Set Difference Union Intersection

Theorem

 * $$S = \left({S \setminus T}\right) \cup \left({S \cap T}\right)$$

Proof 1
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Proof 2
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Proof 3
We have that:
 * From Set Difference Subset: $$S \setminus T \subseteq S$$
 * From Intersection Subset: $$S \cap T \subseteq S$$

Hence from Union Smallest it follows that $$\left({S \setminus T}\right) \cup \left({S \cap T}\right) \subseteq S$$.

Now let $$s \in S$$.

Either: or
 * $$s \in T$$, in which case $$s \in S \cap T$$ by definition of set intersection
 * $$s \notin T$$, in which case $$s \in S \setminus T$$ by definition of set difference.

That is, $$s \in \left({S \setminus T}\right) \cup \left({S \cap T}\right)$$ by definition of set union, and so $$S \subseteq \left({S \setminus T}\right) \cup \left({S \cap T}\right)$$.

Hence the result by Equality of Sets.