Now note that, (1) for all $A\in \text{ob}(\mathbf{C})$, $A\le A$ due to the existence of identity morphism for all $A\in \text{ob}(\mathbf{C})$ and (2) if $A\le B$ and $B\le C$ for some $A,B,C\in \text{ob}(\mathbf{C})$ then $A\le B$ since the composition of any two $\mathbf{C}$-morphisms is defined.