Set Union is Self-Distributive/General Result

Theorem
Let $\left\langle{\mathbb S_i}\right\rangle_{i \in I}$ be an $I$-indexed family of sets of sets.

Then:
 * $\displaystyle \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$

Proof
By the definition of set union, we have:
 * $\displaystyle \forall i \in I: \forall S \in \mathbb S_i: S \in \bigcup_{j \mathop \in I} \mathbb S_j$

Hence, by Subset of Union: General Result:
 * $\displaystyle \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$

By Union is Smallest Superset: General Result, it follows that:
 * $\displaystyle \forall i \in I: \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$

Therefore, by Union is Smallest Superset: Family of Sets, we conclude that:
 * $\displaystyle \bigcup_{i \mathop \in I} \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$

By Subset of Union: General Result and then Subset of Union: Family of Sets, we have:
 * $\displaystyle \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$

Because $\subseteq$ is transitive, we can apply the definition set union to conclude that:
 * $\displaystyle \forall S \in \bigcup_{i \mathop \in I} \mathbb S_i: S \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$

Hence, by Union is Smallest Superset: General Result:
 * $\displaystyle \bigcup \bigcup_{i \mathop \in I} \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$

By definition of set equality, the result follows.