Let $\mathcal R \subseteq S \times T$ be many-to-one. Then, from the definition of many-to-one: :$\mathcal R \subseteq S \times T: \forall x \in S: \left({x, y_1}\right) \in \mathcal R \land \left({x, y_2}\right) \in \mathcal R \implies y_1 = y_2$ Also, note that from Inverse of Many-to-One Relation is One-to-Many, $\mathcal R^{-1}$ is one-to-many. Let $\left({a, b}\right), \left({b, c}\right) \in \mathcal R^{-1} \circ \mathcal R$. {{begin-eqn}} {{eqn | o =        | r = \left({a, b}\right), \left({b, c}\right) \in \mathcal R^{-1} \circ \mathcal R       | c =  }} {{eqn | ll= \implies       | o =        | r = \exists p \in T: \left({a, p}\right) \in \mathcal R \land \left({p, b}\right) \in \mathcal R^{-1}       | c =  }} {{eqn | o = \land       | r = \exists q \in T: \left({b, q}\right) \in \mathcal R \land \left({q, c}\right) \in \mathcal R^{-1}       | c = Definition of Composite Relation }} {{eqn | ll= \implies       | o =        | r = \left({b, p}\right) \in \mathcal R \land \left({p, a}\right) \in \mathcal R^{-1}       | c =  }} {{eqn | o = \land       | r = \left({c, q}\right) \in \mathcal R \land \left({q, b}\right) \in \mathcal R^{-1}       | c = Definition of Inverse Relation }} {{eqn | ll= \implies       | o =        | r = \left({c, p}\right) \in \mathcal R \land \left({p, b}\right) \in \mathcal R^{-1}       | c = as $\mathcal R$ is many-to-one }} {{eqn | ll= \implies       | o =        | r = \left({a, p}\right) \in \mathcal R \land \left({p, c}\right) \in \mathcal R^{-1}       | c =  }} {{eqn | ll= \implies       | o =        | r = \left({a, c}\right) \in \mathcal R       | c = Definition of Composite Relation }} {{end-eqn}} Thus $\mathcal R^{-1} \circ \mathcal R$ is transitive. Now let $\left({p, q}\right), \left({q, r}\right) \in \mathcal R \circ \mathcal R^{-1}$. {{begin-eqn}} {{eqn | o =        | r = \left({p, q}\right), \left({q, r}\right) \in \mathcal R \circ \mathcal R^{-1}       | c =  }} {{eqn | ll= \implies       | o =        | r = \exists a \in S: \left({p, a}\right) \in \mathcal R^{-1} \land \left({a, q}\right) \in \mathcal R       | c =  }} {{eqn | o = \land       | r = \exists b \in S: \left({q, b}\right) \in \mathcal R^{-1} \land \left({b, r}\right) \in \mathcal R       | c = Definition of Composite Relation }} {{eqn | ll= \implies       | o =        | r = \left({a, p}\right) \in \mathcal R \land \left({q, a}\right) \in \mathcal R^{-1}       | c =  }} {{eqn | o = \land       | r = \left({b, q}\right) \in \mathcal R \land \left({r, b}\right) \in \mathcal R^{-1}       | c = Definition of Inverse Relation }} {{end-eqn}} But $\mathcal R$ is many-to-one.  This means that :$\forall x \in S: \left({x, y_1}\right) \in \mathcal R \land \left({x, y_2}\right) \in \mathcal R \implies y_1 = y_2$ So: {{begin-eqn}} {{eqn | l = \left({a, p}\right) \in \mathcal R \land \left({a, q}\right) \in \mathcal R       | o = \implies       | r = p = q       | c = as $\mathcal R$ is many-to-one }} {{eqn | l = \left({b, q}\right) \in \mathcal R \land \left({b, r}\right) \in \mathcal R       | o = \implies       | r = q = r       | c = as $\mathcal R$ is many-to-one }} {{eqn | o = \implies       | r = p = q = r       | c =  }} {{eqn | o = \implies       | r = \left({p, r}\right) \in \mathcal R \circ \mathcal R^{-1}       | c =  }} {{end-eqn}} Thus (trivially) $\mathcal R \circ \mathcal R^{-1}$ is transitive. {{qed}} Category:Composite Relations Category:Transitive Relations q2f53o6ksd17yxkbuerw6pcni4r6f4n