Intersection Distributes over Union

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

Generalized Result
$$\forall n \in \mathbb{N}^*: S \cap \bigcup_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \cap T_i}\right)$$

Generalized Proof
For all $$n \in \mathbb{N}$$, let $$P \left({n}\right)$$ be the proposition: $$S \cap \bigcup_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \cap T_i}\right)$$.


 * $$P(1)$$ is true, as this just says $$S \cap T_1 = S \cap T_1$$.


 * $$P(2)$$ is the case $$S \cap \left({T_1 \cup T_2}\right) = \left({S \cap T_1}\right) \cup \left({S \cap T_2}\right)$$ which has been proved. This is our basis for the induction.


 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:

$$S \cap \bigcup_{i = 1}^k T_i = \bigcup_{i = 1}^k \left({S \cap T_i}\right)$$

Then we need to show:

$$S \cap \bigcup_{i = 1}^{k+1} T_i = \bigcup_{i = 1}^{k+1} \left({S \cap T_i}\right)$$

This is our induction step:

So $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.