Equidecomposability Unaffected by Union

Theorem
Suppose $$\left\{{S_1, \dots, S_m}\right\}, \left\{{T_1, \dots, T_m }\right\} \ $$ are collections of point sets in $$\mathbb{R}^n \ $$ such that for each $$k \in \left\{{1, \dots, m}\right\}, S_k \ $$ and $$T_k \ $$ are equidecomposable.

Then the two sets $$S = \bigcup_{i=1}^m S_i \ $$ is equidecomposable with $$T = \bigcup_{i=1}^m T_i \ $$.

Proof
We have for each $$k \in \left\{{1, \dots, m}\right\} \ $$ a decomposition $$\left\{{S_{k,1}, \dots, S_{k,l_k}}\right\} \ $$ and set of isometries $$\phi_{i,j}:\mathbb{R}^n \to \mathbb{R}^n \ $$ such that

$$S_k = \bigcup_{a=1}^{l_k} \phi_{k,a}(S_{k,a}) \ $$

and similarly for $$T_k \ $$ and some isometries $$\theta_{i,j}:\mathbb{R}^n \to \mathbb{R}^n \ $$.

So then

$$S = \bigcup_{k=1}^m \bigcup_{i=1}^{l_k} \phi_{k,i}(S_{k,i}) \ $$

and

$$T = \bigcup_{k=1}^m \bigcup_{i=1}^{l_k} \theta_{k,i}(T_{k,i}) \ $$

but since be definition, each of the $$S_{a,b}, T_{a,b} \ $$ are congruent, they yield equivalent decompositions of $$S \ $$ and $$T \ $$.