2-SAT is unsatisfiable iff it contains unsatisfiable XOR-2-SAT. So, first, we just need to combine every clause that contains share same variables (both of them). Then, all of those which remain uncombined are uninteresting (don't affect satisfiability).Example:$\Phi = (x \lor y) \land (\overline x \lor \overline y) \land (x \lor \overline z)\land(\overline x \lor z)\land (\overline y \lor z)\land (y \lor \overline z)\land (z \lor t)$Now I can replace this formula by equisatisfiable (under log-space reduction):$\Phi' = (x \oplus y)\land(x \oplus \overline z)\land(y \oplus \overline z)$As you can see, clause $(z\lor t)$ was removed as uninteresting.Ah, well, I need to see if there are cycles in implication graph. So, finding a cycle in directed graph that contains two given vertices is $\mathsf{NL}$-complete problem, right?
Reduction from 2-SAT to 2-XOR-SAT?
complexity theory;satisfiability