3-SAT is NP-complete

A truth assignment for a $3$-CNF formula can be checked clause by clause in polynomial time, exactly as for SAT. So $3$-SAT lies in $NP$.

By [L1], it is enough to reduce $SAT$ to $3$-SAT. Start from $\phi$ and apply [L2] to obtain an equisatisfiable CNF formula $T(\phi )$ whose clauses have size at most three.

Replace every $2$-clause $({\mathrm{\ell }}_1\vee {\mathrm{\ell }}_2)$ by $({\mathrm{\ell }}_1\vee {\mathrm{\ell }}_2\vee z)\wedge ({\mathrm{\ell }}_1\vee {\mathrm{\ell }}_2\vee \mathrm{\neg }z)$ with a fresh variable $z$, and replace every $1$-clause $(\mathrm{\ell })$ by $(\mathrm{\ell }\vee u\vee v)\wedge (\mathrm{\ell }\vee u\vee \mathrm{\neg }v)\wedge (\mathrm{\ell }\vee \mathrm{\neg }u\vee v)\wedge (\mathrm{\ell }\vee \mathrm{\neg }u\vee \mathrm{\neg }v)$ with fresh variables $u,v$. These replacements preserve satisfiability and produce only $3$-clauses.

Steps 1.2 and 2.1 give a polynomial-time many-one reduction from $SAT$ to $3$-SAT, because [L2] is linear-size and the padding in step 2.1 adds only constantly many clauses per short clause. Therefore $3$-SAT is NP-hard. Combined with step 1.1, [L3] yields that $3$-SAT is NP-complete.