SAT is the problem "Given a propositional logic statement, does the statement have an assignment of its variables that result in the statement being true".
3-SAT is a SAT problem, written as clauses with 3 variables or less. For example
((A or B or C) and (not B or not C)) has 2 clauses.
C are boolean variables; "
or" and "
and" are the standard logical operators. This problem has at least one solution
(A, B, C) = (true, false, true).
2-SAT is a SAT problem, written as clauses with 2 variables or less.
We can reduce 2-SAT to the problem of finding a cycle in a directed graph: We can create a vertex for each variable (and its negation). We write each clause
((A) or (B)) in implication form:
((A) or (B)) <=> ((not (not A)) or (B)) <=> ((not A) implies (B)). We add a directed edge for each "implies" clause. The 2-SAT problem will be satisfied if and only if for all variables
(not X) are not contained in a cycle. (Equivalently, there is a path from
(not X) and a path from
(not X) to
(X) if and only if there is a contradiction - that is, if the 2-SAT problem is not satisfied.)
Can a 3-SAT clause be "directly" (defined in ) reduced to a 2-SAT problem?
 2-SAT is equivalent to "for some X, does the implication graph contain both paths from
(not X) and from
(not X) to
 In other words, a 2-SAT problem is satisfied if and only if 2 vertices in its implication graph contradict (are in the same cycle).
How, then, can there be a "direct" reduction from a 3-SAT clause to 2-SAT?
 If there's a "direct" reduction from a 3-SAT clause to 2-SAT, then, for each clause
D = (A or B or C), there will exist 3 vertices
C in the 2-SAT implication graph such that the clause
D is satisfied if and only if
(not ((not A) and (not B) and (not C))) (which falsifies the clause).
 There are 3 variables involved there () in the satisfaction.
 A 2-SAT problem is satisfied, or not, based on 2 vertices ().
 You can't encode "unsatisfy if and only if 3 variables contradict" into directed edges ("directly"). (A 2-SAT problem could have multiple pairs of contradicting vertices; each contradiction corresponds to a pair of vertices. You can't "fit" 3 "into" a pair.)
 Given , if a 3-SAT clause can be "directly" reduced to 2-SAT, then, it must not be one-to-one with the variables.
Therefore, a 3-SAT clause cannot be "directly" reduced to 2-SAT.