Doubt regarding the implications of a 2-SAT constraint
$A\rightarrow B$ doesn't necessarily mean that $B\rightarrow A$. In your example, &quot;$x_1$ is false then $x_2$ must be true&quot; doesn't imply that if $x_2$ is true, $x_1$ must be false. Therefore, setting both $x_1$ and $x_2$ to true would not contradict the statement.
Also, as a side note, $\lnot x_1\rightarrow x_2$ is logically equivalent to $\lnot x_2 \rightarrow x_1$, so the two implications you described are identical.