How do I prove the floor identity $⌊x + n⌋ = ⌊x⌋ + n$ in a more precise way? I am having trouble understanding the proof provided by the author for the property stated after "Goal:".
Except from the text here is a list of useful properties:

(PROPERTY 1a) $⌊x⌋ = n$ if and only if $n ≤ x < n + 1$ 
(1b) $⌈x⌉ = n$ if and only if $n − 1 < x ≤ n$ 
(1c) $⌊x⌋ = n$ if and only if $x − 1 < n ≤ x$ 
(1d) $⌈x⌉ = n$ if and only if $x ≤ n < x + 1$ 
(2) $x − 1 < ⌊x⌋ ≤ x ≤ ⌈x⌉ < x + 1$  (3a) $⌊−x⌋ = −⌈x⌉$ 
(3b) $⌈−x⌉ = −⌊x⌋$ 
(4a) $⌊x + n⌋ = ⌊x⌋ + n$ 
(4b) $⌈x + n⌉ = ⌈x⌉ + n$ 
Goal: As an exercise, prove the property $⌊x + n⌋ = ⌊x⌋ + n$
Proof: We will prove the property using a direct proof. Suppose that
  $⌊x⌋ = m$, where $m$ is a positive integer. By property (1a), it follows
  that  $m ≤ x < m + 1$. Adding $n$ to all three quantities in this chain of
  two inequalities shows that  $m + n ≤ x + n < m + n + 1$. Using property
  (1a) again, we see that $⌊x + n⌋ = m + n = ⌊x⌋ + n$.  This completes the
  proof. Proofs of the other properties are left as exercises.

From: Rosen, Kenneth. Discrete Mathematics and Its Applications (p. 159)
My understanding of the property we are tasked with proving is that it is an identity that is stating the proposition $∀x∀n(⌊x + n⌋ = ⌊x⌋ + n)$ where the domains of discourse for $x$ and $n$ are the set of all real numbers and the set of all integers, respectively. With that in mind, since this is a direct proof, shouldn't we start with arbitrary values of the domains (represented by real number $x$ and integer $n$, and NOT JUST POSITIVE INTEGER $n$) and try to show that their properties imply the equation $⌊x + n⌋ = ⌊x⌋ + n$? (Ignoring the positive-n issue, how could we even do this if the only properties we are allowed to assume 
about $x$ and $n$ are that $x$ is real and $n$ is an integer?) Also, why was he able to start with and assume $⌊x⌋ = m$ ("suppose $⌊x⌋ = m$") out of nowhere?
If anyone can provide a version of his proof that is more precise and doesn't skip steps of reasoning or explanations, that would answer my question.

Let $\lfloor x+n\rfloor=k$
Then by [1a] we have:
$$k \le x+n \lt k+1$$
$$k-n \le x \lt k+1-n$$
Using [1a] again:
$$\lfloor x\rfloor=k-n$$
$$\lfloor x\rfloor+n=k$$
