Proof. [02RT]
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Proof.
Symmetry and reflexivity are straightforward. For transitivity assume that we have toric models , , that the first and second model are equivalent through and that the second and the third are equivalent through . Then, by Theorem 4.60, and are defined by SCR polyhedral complexes and respectively, with . Let . By Lemma 3.11, . Thus determines a model of . This model has morphisms and to and respectively. We put and . Now it is easy to verify that provides the transitivity property. ∎