Equidecomposability is Equivalence Relation

Theorem
The property of being equidecomposable is an equivalence relation on the power set $\powerset {\R^n}$.

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:


 * $\ds A = \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}$

and


 * $\ds B = \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}$

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:


 * $\ds B = \bigcup_{i \mathop = 1}^p \map {\xi_i} {Y_i}$

and:


 * $\ds C = \bigcup_{i \mathop = 1}^p \map {\tau_i} {Y_i}$

Consider the sets


 * $Z_{i, j} = \map {\nu_i} {X_i} \cap \map {\xi_j} {Y_j}$

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

We have:


 * $\ds \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \map {\paren {\mu_i \circ \nu_i^{-1} } } {Z_{i, j} } = \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \map {\paren {\mu_i \circ \nu_i^{-1} } } {\map {\nu_i} {X_i} \cap \map {\xi_j} {Y_j} } = \bigcup_{i \mathop = 1}^m \map {\paren {\mu_i \circ \nu_i^{-1} \circ \nu_i} } {X_i} = \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i} = A$


 * $\ds \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \map {\paren {\tau_j \circ \xi_j^{-1} } } {Z_{i, j} } = \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \map {\paren {\tau_j \circ \xi_j^{-1} } } {\map {\nu_i} {X_i} \cap \map {\xi_j} {Y_j} } = \bigcup_{j \mathop = 1}^p \map {\paren {\tau_j \circ \xi_j^{-1} \circ \xi_j} } {Y_j} = \bigcup_{j \mathop = 1}^p \map {\tau_j} {Y_j} = C$

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

Hence these two are equidecomposable.