Possible circular reasoning in textbook proof that $\lceil x+m\rceil=\lceil x\rceil +m$

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The goal is to prove that $\lceil x+m\rceil=\lceil x\rceil +m$, where $x$ is a real number and $m$ is an integer. The book outlines the following proof:

Write $x=n-\epsilon$, where $n$ is an integer and $0\leq\epsilon<1$; thus, $\lceil x\rceil=n$. Then $\lceil x+m\rceil=\lceil n-\epsilon+m\rceil=n+m=\lceil x\rceil+m$.

If we "read between the lines," it really seems like the following is being communicated:

\begin{align} \lceil x+m\rceil&=\lceil n-\epsilon+m\rceil\\ &= \lceil-\epsilon\rceil+(n+m)\tag{circular reasoning?}\\ &= 0+(n+m)\tag{since $\epsilon\in[0,1)$}\\ &= \lceil x\rceil+m \end{align}

Is the book's proof fine and I'm just not seeing something clearly or is there a subtle error somewhere (if so, what could be done to fix the proof?)?

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As is pointed out in the comments, there is no circular reasoning here. To clear things up substitute $u=n+m$ (which will be an integer) and then use the theorem you already used and accepted, that $\lceil n-\epsilon \rceil = n$ where $n$ is any integer (and replace $n$ by $u$)