Union Distributes over Intersection

Theorem
Set union is distributive over set intersection:


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

General Result
Let $$S$$ and $$T$$ be sets.

Let $$\mathcal P \left({T}\right)$$ be the power set of $$T$$.

Let $$\mathbb T$$ be a subset of $$\mathcal P \left({T}\right)$$.

Then:
 * $$S \cup \bigcap \mathbb T = \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$

Proof
$$ $$ $$

Union Subset of Intersection
Let $$x \in S \cup \bigcap \mathbb T$$.

We need to show that $$x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$ and then by definition of subset we will have shown that $$S \cup \bigcap \mathbb T \subseteq \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$.

So, we have that $$x \in S \cup \bigcap \mathbb T$$.

By definition of set union, $$x \in S$$ or $$x \in \bigcap \mathbb T$$.

So there are two cases:


 * Suppose $$x \in S$$.

Then by definition of set union, $$\forall X \in \mathbb T: x \in S \cup X$$.

So:
 * $$x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$


 * Suppose $$x \in \bigcap \mathbb T$$.

Then by definition of set intersection, $$\forall X \in \mathbb T: x \in X$$.

So by definition of set union, $$\forall X \in \mathbb T: x \in S \cup X$$.

So:
 * $$x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$

In both cases we see that:
 * $$x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$

so by Proof by Cases, we have that:
 * $$S \cup \bigcap \mathbb T \subseteq \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$

Intersection Subset of Union
Let $$x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$.

We need to show that $$x \in S \cup \bigcap \mathbb T$$ and then by definition of subset we will have shown that $$\bigcap_{X \in \mathbb T} \left({S \cup X}\right) \subseteq S \cup \bigcap \mathbb T$$.

So, we have that $$x \in \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$.

By definition of set intersection, $$\forall X \in \mathbb T: x \in S \cup X$$.

Take any one of these instances of $$S \cup X$$.

Again there are two cases:


 * Suppose $$x \in S$$.

Then by definition of set union:
 * $$x \in S \cup \bigcap \mathbb T$$


 * Suppose $$x \in X$$.

Then by definition of set intersection and the set $$X$$, we have that $$x \in \bigcap \mathbb T$$.

So by definition of set union:
 * $$x \in S \cup \bigcap \mathbb T$$

In both cases we see that:
 * $$x \in S \cup \bigcap \mathbb T$$

so by Proof by Cases, we have that:
 * $$\bigcap_{X \in \mathbb T} \left({S \cup X}\right) \subseteq S \cup \bigcap \mathbb T$$

So we have that:
 * $$S \cup \bigcap \mathbb T \subseteq \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$

and
 * $$\bigcap_{X \in \mathbb T} \left({S \cup X}\right) \subseteq S \cup \bigcap \mathbb T$$

and so by definition of Equality of Sets:
 * $$S \cup \bigcap \mathbb T = \bigcap_{X \in \mathbb T} \left({S \cup X}\right)$$

Also see

 * Intersection Distributes over Union