Union of Inverse is Inverse of Union
Theorem
Let for $i \in \left\{1,2\right\}$ $\mathcal R_i \subseteq S_i \times T_i$ be a relation on $S_i \times T_i$.
Let $\mathcal R_i^{-1} \subseteq T_i \times S_i$ be the inverse of $\mathcal R_i$.
Then $\mathcal R_1^{-1} \cup \mathcal R_2^{-1} = \left(\mathcal R_1 \cup \mathcal R_2\right)^{-1}$
Proof
Let $\left(t,s\right) \in \mathcal R_1^{-1} \cup \mathcal R_2^{-1}$.
By the definition of union: $\left(t,s\right) \in \mathcal R_1^{-1} \vee \left(t,s\right) \in \mathcal R_2^{-1}$.
Assume $\left(t,s\right) \in \mathcal R_i^{-1}$. By the definition of inverse: $\left(s,t\right) \in \mathcal R_i$.
By Disjunction Introduction: $\left(s,t\right) \in \mathcal R_1 \vee \left(s,t\right) \in \mathcal R_2 \iff \left(s,t\right) \in \mathcal R_1 \cup \mathcal R_2$.
And by the definition of inverse: $\left(t,s\right) \in \left(\mathcal R_1 \cup \mathcal R_2\right)^{-1}$.
$\blacksquare$