Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y$:

$$
f(x-f(y))=1-x-y
$$
For $y=0$, we find $x=z+f(0)$, and we find $f(z)=f(x-f(y))=$ $1-x-y=1-z-f(0)$. Furthermore, if $z=0$, then $f(0)=1-f(0)$, so $f(0)=\frac{1}{2}$. Conversely, the function

$$
x \rightarrow \frac{1}{2}-x
$$

is indeed a solution to the problem.