Equidecomposability is Equivalence Relation

Theorem
The property of being equidecomposable is an equivalence relation on the power set $$\mathcal{P} \left({\mathbb{R}^n}\right) \ $$.

Relexivity
A set is necessarily equidecomposable with itself; the same decomposition and set of isometries suffice for $$A \ $$ as for $$A \ $$.

Symmetry
There is no order to the relation of being equidecomposable; symmetry follows.

Transitivity
Suppose $$A, B, C \subset \mathbb{R}^n \ $$ are sets such that $$A, B \ $$ are equidecomposable and $$B, C \ $$ are equidecomposable. Let $$X_1, \dots, X_m \ $$ be a decomposition of $$A, B \ $$ together with isometries $$\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m:\mathbb{R}^n \to \mathbb{R}^n \ $$ such that

$$A = \bigcup_{i=1}^m \mu_i(X_i) \ $$

and

$$B = \bigcup_{i=1}^m \nu_i(X_i) \ $$.

Further let $$Y_1, \dots, Y_p \ $$ together with $$\xi_1, \dots, \xi_p, \tau_1, \dots, \tau_p \ $$ be sets and isometries such that

$$B = \bigcup_{i=1}^p \xi_i(Y_i) \ $$

and

$$C = \bigcup_{i=1}^p \tau_i(Y_i) \ $$.

Consider the sets

$$Z_{i,j} = \nu_i(X_i) \cap \xi_j(Y_j) \ $$

where $$ 1 \leq i \leq m \ $$ and $$ 1 \leq j \leq p \ $$.

We have

$$\bigcup_{i=1}^m \bigcup_{j=1}^p (\mu_i \circ \nu_i^{-1})(Z_{i,j}) = \bigcup_{i=1}^m \bigcup_{j=1}^p (\mu_i \circ \nu_i^{-1} )( \nu_i(X_i) \cap \xi_j(Y_j) ) = \bigcup_{i=1}^m (\mu_i \circ \nu_i^{-1} \circ \nu_i) (X_i) = \bigcup_{i=1}^m \mu_i(X_i) = A \ $$,

$$\bigcup_{j=1}^p \bigcup_{i=1}^m (\tau_j \circ \xi_j^{-1})(Z_{i,j}) = \bigcup_{j=1}^p \bigcup_{i=1}^m (\tau_j \circ \xi_j^{-1}) (\nu_i(X_i) \cap \xi_j(Y_j)) = \bigcup_{j=1}^p (\tau_j \circ \xi_j^{-1} \circ \xi_j) (Y_j) = \bigcup_{j=1}^p \tau_j(Y_j) = C \ $$,

so $$Z_{i,j} \ $$ together with the isometries $$\mu_i \circ \nu_i^{-1}, \tau_j \circ \xi_j^{-1} \ $$ as a decomposition of $$A \ $$ and $$C \ $$, hence these two are equidecomposable.