Let $C$ be a Boolean circuit $($using AND, NOT, OR gates$),$ which has input

$(x_1,$cdots,$x_n)$ and one output $y.$ The circuit is satisfiable if for some input,

the output $y$ produced by $C$ is $1.$ Suppose that we have a deterministic

polynomial time algorithm that decides whether $C$ is satisfiable.

Now, let $C_1$ and $C_2$ be two Boolean circuits $($using AND, NOT, OR gates$),$

each of which has input $(x_1,$cdots,$x_n)$ and one output $y.$ We say that the two

circuits are equivalent if for any input, the output produced by $C_1$ equals the

the output produced by $C_2.$

Show that we also have a deterministic polynomial time algorithm that

decides whether $C_1$ and $C_2$ are equivalent.