Equidecomposability is Equivalence Relation

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

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 $A, B, C \subset \R^n$ 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 $\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m:\R^n \to \R^n$ such that:


 * $\displaystyle A = \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right)$

and


 * $\displaystyle B = \bigcup_{i \mathop = 1}^m \nu_i \left({X_i}\right)$

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:


 * $\displaystyle B = \bigcup_{i \mathop = 1}^p \xi_i \left({Y_i}\right)$

and:


 * $\displaystyle C = \bigcup_{i \mathop = 1}^p \tau_i \left({Y_i}\right)$

Consider the sets


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

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

We have:


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


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

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

Hence these two are equidecomposable.