# Question regarding 3SAT

I was thinking that in exact 3sat there's only one pattern that can be a contradiction, namely $$(a \vee b \vee c) \wedge (a \vee b \vee \neg c) \wedge (a \vee \neg b \vee c) \wedge (a \vee \neg b \vee \neg c) \wedge \\ (\neg a \vee b \vee c) \wedge (\neg a \vee b \vee \neg c) \wedge (\neg a \vee \neg b \vee c) \wedge (\neg a \vee \neg b \neg c)$$ $(a,b,c)$ are placeholders for e.g. $(a=x1, b=x33, c=x14)$. If there's no contradiction, then it must be satisfiable. Thus, if I can't find such a pattern, then it must be satisfiable. Meaning that if I find a term like $(x1 \vee x3 \vee \neg x5)$ then there must also exist a term $(\neg x1 \vee x3 \vee \neg x5)$, otherwise $(x1 \vee x3 \vee \neg x5)$ isn't part of a contradiction.

Obviously there must be some non-obvious error here, otherwise the problem wouldn't be hard. But I can't seem to find the fault?

It is not the only one. Each triplet reduces certain number of possibilities. You can use $a_1$ and $a_2$ to construct such triplets, that $c$ can't be zero, and then construct using $a_3$ and $a_4$ such triplets, that $c$ can't be one.
$(a_1 \lor a_2 \lor c) \land(a_1 \lor \neg a_2 \lor c) \land(\neg a_1 \lor a_2 \lor c) \land(\neg a_1 \lor \neg a_2 \lor c)$
$\land$
$(a_3 \lor a_4 \lor \neg c) \land(a_3 \lor \neg a_4 \lor \neg c) \land(\neg a_3 \lor a_4 \lor \neg c) \land(\neg a_3 \lor \neg a_4 \lor \neg c)$