if there is an edge $a \Rightarrow b$, then there also is an edge $\lnot b \Rightarrow \lnot a$.

Also note, that if $x$ is reachable from $\lnot x$, and $\lnot x$ is reachable from $x$, then the problem has no solution.

Whatever value we choose for the variable $x$, it will always end in a contradiction - if $x$ will be assigned $\text{true}$ then the implicationtell us that $\lnot x$ should also be $\text{true}$ and visa versa.

Whatever value we choose for the variable $x$, it will always end in a contradiction - if $x$ will be assigned $\text{true}$ then the implicationtells us that $\lnot x$ should also be $\text{true}$ and visa versa.

It turns out, that this condition is not only necessary, but also sufficient.

We will prove this in a few paragraphs below.

First recall, if a vertex is reachable from a second one, and the second one is reachable from the first one, then these two vertices are in the same strongly connected component.