Integer equals Ceiling iff between Number and One More

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:
 * $\left \lceil{x}\right \rceil = n \iff x \le n < x + 1$

Necessary Condition
Let $x \le n < x + 1$.

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

From $n < x + 1$:
 * $n - 1 < x$

Hence by Number greater than Integer iff Ceiling greater than Integer:
 * $n - 1 < \left \lceil{x}\right \rceil$

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

and so:
 * $n \le \left \lceil{x}\right \rceil$

Thus as:
 * $\left \lceil{x}\right \rceil \le n$

and:
 * $\left \lceil{x}\right \rceil \ge n$

it follows that:
 * $\left \lceil{x}\right \rceil = n$

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

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

By Number not greater than Integer iff Ceiling not greater than Integer:
 * $x \le n$

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

and so adding $1$ to both sides:
 * $\left \lceil{x}\right \rceil < x + 1$

and so by hypothesis:
 * $n < x + 1$

So:
 * $\left \lceil{x}\right \rceil = n \implies x \le n < x + 1$

Hence the result:
 * $\left \lceil{x}\right \rceil = n \iff x \le n < x + 1$