Union of Inverse of Relations is Inverse of their Union

Theorem
Let for $i \in \set {1, 2}$ $\RR_i \subseteq S_i \times T_i$ be relations on $S_i \times T_i$.

Let $\RR_i^{-1} \subseteq T_i \times S_i$ be the inverse of $\RR_i$.

Then $\RR_1^{-1} \cup \RR_2^{-1} = \paren {\RR_1 \cup \RR_2}^{-1}$

Proof
Let $\tuple {t, s} \in \RR_1^{-1} \cup \RR_2^{-1}$.

By definition of union:


 * $\tuple {t, s} \in \RR_1^{-1} \vee \tuple {t, s} \in \RR_2^{-1}$.

Assume $\tuple {t, s} \in \RR_i^{-1}$.

By definition of inverse:


 * $\tuple {s, t} \in \RR_i$.

By definition of union:


 * $\tuple {s, t} \in \RR_1 \vee \tuple {s, t} \in \RR_2 \iff \tuple {s, t} \in \RR_1 \cup \RR_2$.

By the definition of inverse:


 * $\tuple {t, s} \in \paren {\RR_1 \cup \RR_2}^{-1}$.