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

$$
f(x-f(y))=1-x-y
$$
By taking $x=f(y)$ in the initial equation, we obtain $f(y)=1-$ $f(0)-y$ thus $f$ is of the form $x \mapsto c-x$ with $c=1-f(0)$, and by substituting back into the initial equation for all real $x, y$:

$$
c-(x-(c-y))=1-x-y
$$

from which $c=\frac{1}{2}$, the unique solution is $x \mapsto \frac{1}{2}-x$, which is indeed a solution.