I have actors A and B who have a preference relations $\succeq_A$ and $\succeq_B$ on a set $X$. Both are complete and transitive. Actor A will report Actor B's preferences as his own if Actor B specifies a strict preference $\succ$ for any items $x, y \in X$. However, if Actor B is indifferent between any items $x, y \in X$, then Actor A will reveal his preferences, leading to a preference relation $\succeq_C$.
I know that a complete relation is defined as any relation that for any $x, y \in X$, we have that $x\succeq y$, $y\succeq x$ or both.
A transitive relation is also defined as any relation that for any $x, y, z \in X$, if $x\succeq y$ and $y\succeq z$, then $x\succeq z$.
How would I go about proving that $\succeq_c$ is both complete and transitive?
To show that the relation is complete, let $x, y \in X$. Because $\succeq_B$ is complete, you know that either $x \succeq_B y$ or $y \succeq_B x$. If the latter is true, we can swap the labels on $x$ and $y$. So we can assume without loss of generality that $x \succeq_B y$. There are still two cases: Either $x \succ_B y$ or $x \sim_B y$. If the former is true, then $x \succ_C y$. If the latter is true, then by the completeness of $\succeq_A$, we have either $x \succeq_A y$ or $y \succeq_A x$. So either $x \succeq_C y$ or $y \succeq_C x$. In all cases, either $x \succeq_C y$ or $y \succeq_C x$. So $\succeq_C$ is complete.
The proof for transitivity follows a similar pattern. for any $x, y, z \in X$ with $x \succeq_C y$ and $y \succeq_C z$, if you examine all the possible cases, and recall that $x \sim y$ means ($x \succeq y$ and $y \succeq x$), then in each case you find that it always ends up with $x \succeq_C z$.