Complete separated adic pairs are Henselian

Let $x\in I$ and $r\in A$. Then $rx\in I$, so $1-rx\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$. By [L1], the element $1-rx$ is a unit. This is the Jacobson-radical criterion for $x$, so $x\in J(A)$. Hence $I\subseteq J(A)$.

Let $f\in A[T]$ be monic and let $\bar{f}=g_0{h}_0$ be a coprime monic factorization in $(A\mathrm{/}I)[T]$. Choose monic lifts $g_1,h_1$ of $g_0,h_0$. By [L2], choose $a,b\in A[T]$ with $a{g}_1+b{h}_1\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$. Repeatedly applying [L3] produces monic pairs $(g_r,h_r)$ with $f-g_r{h}_r\in I^r[T]$ for every $r\ge 1$.

By [L4], the coefficient sequences of $(g_r)$ and $(h_r)$ are Cauchy and converge to monic polynomials $g,h\in A[T]$ with $f=gh$. Their reductions are still $g_0,h_0$. Thus the required lifted factorization exists.

If $f=g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ is another monic lift of the same residue factorization, then [L5] gives $g\equiv g^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^r[T])$ and $h\equiv h^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^r[T])$ for every $r\ge 1$. Separatedness forces $g=g^{\mathrm{\prime }}$ and $h=h^{\mathrm{\prime }}$. Hence the lift is unique.

Steps 1.1-3.1 verify both clauses of the definition, so $(A,I)$ is a Henselian pair.