Number greater than Integer iff Ceiling greater than Integer

Theorem
Let $x \in \R$ be a real number.

Let $\left \lceil{x}\right \rceil$ be the ceiling of $x$.

Let $n \in \Z$ be an integer.

Then:
 * $\left \lceil{x}\right \rceil > n \iff x > n$

Necessary Condition
Let $x > n$.

By Number is between Ceiling and One Less:
 * $\left \lceil{x}\right \rceil \ge x$

Hence:
 * $\left \lceil{x}\right \rceil > n$

Sufficient Condition
Let $\left \lceil{x}\right \rceil > n$.

We have that:
 * $\forall m, n \in \Z: m < n \iff m \le n - 1$

and so:
 * $(1): \quad \left \lceil{x}\right \rceil - 1 \ge n$

Then:

Hence the result:
 * $\left \lceil{x}\right \rceil > n \iff x > n$