How to prove property of greatest integer function: 
How can we prove that
  $$\lfloor x + y \rfloor = \lfloor y + x -  \lfloor x \rfloor \rfloor + \lfloor x \rfloor$$
  for all real $x$, where $ \lfloor x \rfloor$ denotes greatest integer less than or equal to $x$?

I was able to prove that $\lfloor x + y \rfloor =  \lfloor x \rfloor + \lfloor y \rfloor $ or $\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + 1$ by using the property that $\lfloor x \rfloor = m$ means $m \le x \lt m + 1$ but cannot prove above one by any means. I have tried a lot, can any one please help and please give a simple proof?
 A: If $n$ is an integer and $t$ any real number, then it is straightforward to show that$\lfloor t + n \rfloor = \lfloor t \rfloor + n.$
Therefore, since $\lfloor x \rfloor$ is an integer,
\begin{align*}
\quad &\lfloor y + x -  \lfloor x \rfloor \rfloor + \lfloor x \rfloor \\
= &\lfloor y + x \rfloor - \lfloor x \rfloor + \lfloor x \rfloor \\
= &\lfloor x + y \rfloor.
\end{align*}
A: If you introduced the notation {$x$} = $x - [x]$ and note $0 \le \{x\} < 1$ this becomes easy.
$0 \le \{x\} + \{y\} < 2$.
Case 1:  $0 \le \{x\} + \{y\} < 1$.
Note for integers $n, m$ and $0 \le r < 1$ that: $[n + m + r] = n+m$ and $[n + m] = n + m$.  As $[x]$ and $[y]$ are integers:
Then $[x + y] = [[x]+ [y] + \{x\} + \{y\}] = [[x] + [y]] = [x] + [y]$.
Furthermore $[y] = [[y] + \{y\} + \{x\}]= [y + x - [x]]$ so $[x + y]=[y + x - [x]] + [x]$.
Case 2:  $1 \le \{x\} + \{y\} < 2$
Then $[x + y] = [[x]+ [y] + \{x\} + \{y\}] = [[x] + [y]] + 1 = [x] + [y]+ 1$
And note $[y + x - [x]] = [y + \{x\}] = [[y] + \{y\} + \{x\}] = [[y] + 1] = [y]+1$.
So $[x +y] = [x]+[y] + 1 = [x] + [y + x - [x]]$.
