Number not greater than Integer iff Ceiling not 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:
 * $\left \lceil{x}\right \rceil \le n \iff x \le n$

Necessary Condition
Let $\left \lceil{x}\right \rceil \le n$.

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

Hence:
 * $x \le n$

Sufficient Condition
Let $x \le n$.

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

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

Hence:
 * $\left \lceil{x}\right \rceil - 1 \ge n$

and so by hypothesis:
 * $\left \lceil{x}\right \rceil - 1 \ge x$

This contradicts the result Number is between Ceiling and One Less:
 * $\left \lceil{x}\right \rceil - 1 < x$

Thus by Proof by Contradiction:
 * $\left \lceil{x}\right \rceil \le n$

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