Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 37 / 6 / a

Let $r$ be the number of true values among $a,b,c$. Count the ten clause occurrences for the two choices of $d$. With $d=0$, the three positive singletons contribute $r$, the three negated-pair clauses contribute $3-\left(\genfrac{}{}{0ex}{}{r}{2}\right)$, and the last three clauses all hold. With $d=1$, the four singleton clauses contribute $r+1$, the negated pairs again contribute $3-\left(\genfrac{}{}{0ex}{}{r}{2}\right)$, and the last three contribute $r$. ThereforeThe original three-literal clause is satisfied exactly when $r\ge 1$, and then a value of $d$ satisfies exactly seven gadget clauses. If $r=0$, no choice reaches seven. Thus the required equivalence holds, and seven is also an upper bound for every assignment to the gadget. This is the seven-clause gadget for MAX-2SAT.