If the characteristic of $\kappa_A$ is zero or if the residue characteristic is $p$, the ramification index is prime to $p$, and the residue field extension is separable, then this follows from Abhyankar's lemma (Lemma \ref{lemma-abhyankar}). Namely, suppose the ramification index is $e$. Choose a uniformizer $\pi \in A$. Let $K_1/K$ be the extension obtained by adjoining an $e$th root of $\pi$. By Lemma \ref{lemma-pull-root-uniformizer} we see that the integral closure $A_1$ of $A$ in $K_1$ is a discrete valuation ring with ramification index over $A$. Thus $A_1 \to (B_1)_\mathfrak m$ is formally smooth in the $\mathfrak m$-adic topology for all maximal ideals $\mathfrak m$ of $B_1$ by Lemma \ref{lemma-abhyankar} and a fortiori these are weakly unramified extensions of discrete valuation rings. \medskip\noindent From now on we let $p$ be a prime number and we assume that $\kappa_A$ has characteristic $p$. We first apply Lemma \ref{lemma-solution-after-strict-henselization} to reduce to the case that $A$ and $B$ have separably closed residue fields. Since $\kappa_A$ and $\kappa_B$ are replaced by their separable algebraic closures by this procedure we see that we obtain $$ \kappa_A \supset \bigcap\nolimits_{n \geq 1} \kappa_B^{p^n} $$ from the condition of the theorem. \medskip\noindent Let $\pi \in A$ be a uniformizer. Let $A^\wedge$ and $B^\wedge$ be the completions of $A$ and $B$. We have a commutative diagram $$ \xymatrix{ B \ar[r] & B^\wedge \\ A \ar[u] \ar[r] & A^\wedge \ar[u] } $$ of extensions of discrete valuation rings. Let $K^\wedge$ be the fraction field of $A^\wedge$. Suppose that we can find a finite extension $K^\wedge \subset M$ which is (a) a weak solution for $A^\wedge \to B^\wedge$ and (b) a compositum of a separable extension and an extension obtained by adjoining a $p$-power root of $\pi$. Then by Lemma \ref{lemma-approximate-separable-extension} we can find a finite extension $K \subset K_1$ such that $K^\wedge \otimes_K K_1 = M$. Let $A_1$, resp.\ $A_1^\wedge$ be the integral closure of $A$, resp.\ $A^\wedge$ in $K_1$, resp.\ $M$. Since $A \to A^\wedge$ is formally smooth in the $\mathfrak m^\wedge$-adic topology (Lemma \ref{lemma-extension-dvrs-formally-smooth}) we see that $A_1 \to A_1^\wedge$ is formally smooth in the $\mathfrak m_1^\wedge$-adic topology (Lemma \ref{lemma-formally-smooth-goes-up} and $A_1$ and $A_1^\wedge$ are discrete valuation rings by discussion in Remark \ref{remark-construction}). We conclude from Lemma \ref{lemma-solutions-go-down} part (2) that $K \subset K_1$ is a weak solution for $A \to B^\wedge$. Applying Lemma \ref{lemma-solutions-go-down} part (1) we see that $K \subset K_1$ is a weak solution for $A \to B$. \medskip\noindent Thus we may assume $A$ and $B$ are complete discrete valuation rings with separably closed residue fields of characteristic $p$ and with $\kappa_A \supset \bigcap\nolimits_{n \geq 1} \kappa_B^{p^n}$. We are also given a uniformizer $\pi \in A$ and we have to find a weak solution for $A \to B$ which is a compositum of a separable extension and a field obtained by taking $p$-power roots of $\pi$. Note that the second condition is automatic if $A$ has mixed characteristic. \medskip\noindent Set $k = \bigcap\nolimits_{n \geq 1} \kappa_B^{p^n}$. Observe that $k$ is an algebraically closed field of characteristic $p$. If $A$ has mixed characteristic let $\Lambda$ be a Cohen ring for $k$ and in the equicharacteristic case set $\Lambda = k[[t]]$. We can choose a ring map $\Lambda \to A$ which maps $t$ to $\pi$ in the equicharacteristic case. In the equicharacteristic case this follows from the Cohen structure theorem (Algebra, Theorem \ref{algebra-theorem-cohen-structure-theorem}) and in the mixed characteristic case this follows as $\mathbf{Z}_p \to \Lambda$ is formally smooth in the adic topology (Lemmas \ref{lemma-extension-dvrs-formally-smooth} and \ref{lemma-lift-continuous}). Applying Lemma \ref{lemma-solutions-go-down} we see that it suffices to prove the existence of a weak solution for $\Lambda \to B$ which in the equicharacteristic $p$ case is a compositum of a separable extension and a field obtained by taking $p$-power roots of $t$. However, since $\Lambda = k[[t]]$ in the equicharacteristic case and any extension of $k((t))$ is such a compositum, we can now drop this requirement! \medskip\noindent Thus we arrive at the situation where $A$ and $B$ are complete, the residue field $k$ of $A$ is algebraically closed of characteristic $p > 0$, we have $k = \bigcap \kappa_B^{p^n}$, and in the mixed characteristic case $p$ is a uniformizer of $A$ (i.e., $A$ is a Cohen ring for $k$). If $A$ has mixed characteristic choose a Cohen ring $\Lambda$ for $\kappa_B$ and in the equicharacteristic case set $\Lambda = \kappa_B[[t]]$. Arguing as above we may choose a ring map $A \to \Lambda$ lifting $k \to \kappa_B$ and mapping a uniformizer to a uniformizer. Since $k \subset \kappa_B$ is separable the ring map $A \to \Lambda$ is formally smooth in the adic topology (Lemma \ref{lemma-extension-dvrs-formally-smooth}). Hence we can find a ring map $\Lambda \to B$ such that the composition $A \to \Lambda \to B$ is the given ring map $A \to B$ (see Lemma \ref{lemma-lift-continuous}). Since $\Lambda$ and $B$ are complete discrete valuation rings with the same residue field, $B$ is finite over $\Lambda$ (Algebra, Lemma \ref{algebra-lemma-finite-over-complete-ring}). This reduces us to the special case discussed in Lemma \ref{lemma-special-case}.