SAT is NP-complete

By [L1], $SAT$ is in $NP$. To prove NP-completeness via [L4], it remains to prove NP-hardness.

Because $L\in NP$, [L2] supplies a nondeterministic polynomial-time machine $N$ deciding $L$. For an input $x$, let $f(x)$ be the Cook-Levin formula ${\psi }_{N,x}$ for a polynomial bound on computations of $N$ on $x$.

By [L3], the formula $f(x)$ is satisfiable exactly when $N$ has an accepting bounded tableau on $x$; with the bounded-tableau convention that later rows may repeat a halting configuration, this happens exactly when $x\in L$. The same citation shows that $f$ is computable in polynomial time. Therefore $L{\le }_{p}SAT$.

Since $L\in NP$ was arbitrary, step 2.1 proves that every language in $NP$ reduces to $SAT$. Together with step 1.1, [L4] gives that $SAT$ is NP-complete.