Since the characteristic of $L$ is $p$ we know that $M$ is an Artin-Schreier extension of $L$ (Fields, Lemma \ref{fields-lemma-Artin-Schreier}). Thus we may pick $z \in M$, $z \not \in L$ such that $\xi = z^p - z \in L$. Choose $n \geq 0$ such that $\pi^n\xi \in B$. We pick $z$ such that $n$ is minimal. If $n = 0$, then $M/L$ is unramified with respect to $B$ (Lemma \ref{lemma-pre-characteristic-p-case}) and we are done. Thus we have $n > 0$. \medskip\noindent Assumption (4) implies that $\kappa_A$ is perfect. Thus we may choose compatible ring maps $\overline{\sigma} : \kappa_A \to A/\pi^n A$ and $\overline{\sigma} : \kappa_B \to B/\pi^n B$ as in Lemma \ref{lemma-cohen}. We lift the second of these to a map of sets $\sigma : \kappa_B \to B$\footnote{If $B$ is complete, then we can choose $\sigma$ to be a ring map. If $A$ is also complete and $\sigma$ is a ring map, then $\sigma$ maps $\kappa_A$ into $A$.}. Then we can write $$ \xi = \sum\nolimits_{i = n, \ldots, 1} \sigma(\lambda_i) \pi^{-i} + b $$ for some $\lambda_i \in \kappa_B$ and $b \in B$. Let $$ I = \{i \in \{n, \ldots, 1\} \mid \lambda_i \in \kappa_A\} $$ and $$ J = \{j \in \{n, \ldots, 1\} \mid \lambda_i \not \in \kappa_A\} $$ We will argue by induction on the size of the finite set $J$. \medskip\noindent The case $J = \emptyset$. Here for all $i \in \{n, \ldots, 1\}$ we have $\sigma(\lambda_i) = a_i + \pi^n b_i$ for some $a_i \in A$ and $b_i \in B$ by our choice of $\sigma$. Thus $\xi = \pi^{-n} a + b$ for some $a \in A$ and $b \in B$. If $p | n$, then we write $a = a_0^p + \pi a_1$ for some $a_0, a_1 \in A$ (as the residue field of $A$ is perfect). We compute $$ (z - \pi^{-n/p}a_0)^p - (z - \pi^{-n/p}a_0) = \pi^{-(n - 1)}(a_1 + \pi^{n - 1 - n/p}a_0) + b' $$ for some $b' \in B$. This would contradict the minimality of $n$. Thus $p$ does not divide $n$. Consider the degree $p$ extension $K_1$ of $K$ given by $w^p - w = \pi^{-n}a$. By Lemma \ref{lemma-pre-characteristic-p-case} this extension is Galois and totally ramified with respect to $A$. Thus $L_1 = L \otimes_K K_1$ is a field and $A_1 \subset B_1$ is weakly unramified (Lemma \ref{lemma-weakly-unramified-goes-up-along-totally-ramified}). By Lemma \ref{lemma-pre-characteristic-p-case} the ring $M_1 = M \otimes_K K_1$ is either a product of $p$ copies of $L_1$ (in which case we are done) or a field extension of $L_1$ of degree $p$. Moreover, in the second case, either $C_1$ is weakly unramified over $B_1$ (in which case we are done) or $M_1/L_1$ is degree $p$, Galois, and totally ramified with respect to $B_1$. In this last case the extension $M_1/L_1$ is generated by the element $z - w$ and $$ (z - w)^p - (z - w) = z^p - z - (w^p - w) = b $$ with $b \in B$ (see above). Thus by Lemma \ref{lemma-pre-characteristic-p-case} once more the extension $M_1/L_1$ is unramified with respect to $B_1$ and we conclude that $K_1$ is a weak solution for $A \to C$. From now on we assume $J \not = \emptyset$. \medskip\noindent Suppose that $j', j \in J$ such that $j' = p^r j$ for some $r > 0$. Then we change our choice of $z$ into $$ z' = z - (\sigma(\lambda_j) \pi^{-j} + \sigma(\lambda_j^p) \pi^{-pj} + \ldots + \sigma(\lambda_j^{p^{r - 1}}) \pi^{-p^{r - 1}j}) $$ Then $\xi$ changes into $\xi' = (z')^p - (z')$ as follows $$ \xi' = \xi - \sigma(\lambda_j) \pi^{-j} + \sigma(\lambda_j^{p^r}) \pi^{-j'} + \text{something in }B $$ Writing $\xi' = \sum\nolimits_{i = n, \ldots, 1} \sigma(\lambda'_i) \pi^{-i} + b'$ as before we find that $\lambda'_i = \lambda_i$ for $i \not = j, j'$ and $\lambda'_j = 0$. Thus the set $J$ has gotten smaller. By induction on the size of $J$ we may assume no such pair $j, j'$ exists. (Please observe that in this procedure we may get thrown back into the case that $J = \emptyset$ we treated above.) \medskip\noindent For $j \in J$ write $\lambda_j = \mu_j^{p^{r_j}}$ for some $r_j \geq 0$ and $\mu_j \in \kappa_B$ which is not a $p$th power. This is possible by our assumption (4). Let $j \in J$ be the unique index such that $j p^{-r_j}$ is maximal. (The index is unique by the result of the preceding paragraph.) Choose $r > \max(r_j + 1)$ and such that $j p^{r - r_j} > n$ for $j \in J$. Choose a separable extension $K_1/K$ totally ramified with respect to $A$ of degree $p^r$ such that the corresponding discrete valuation ring $A_1 \subset K_1$ has uniformizer $\pi'$ with $(\pi')^{p^r} = \pi + \pi^{n + 1}a$ for some $a \in A_1$ (Lemma \ref{lemma-make-degree-q-extension}). Observe that $L_1 = L \otimes_K K_1$ is a field and that $L_1/L$ is totally ramified with respect to $B$ (Lemma \ref{lemma-weakly-unramified-goes-up-along-totally-ramified}). Computing in the integral closure $B_1$ we get $$ \xi = \sum\nolimits_{i \in I} \sigma(\lambda_i) (\pi')^{-i p^r} + \sum\nolimits_{j \in J} \sigma(\mu_j)^{p^{r_j}} (\pi')^{-j p^r} + b_1 $$ for some $b_1 \in B_1$. Note that $\sigma(\lambda_i)$ for $i \in I$ is a $q$th power modulo $\pi^n$, i.e., modulo $(\pi')^{n p^r}$. Hence we can rewrite the above as $$ \xi = \sum\nolimits_{i \in I} x_i^{p^r} (\pi')^{-i p^r} + \sum\nolimits_{j \in J} \sigma(\mu_j)^{p^{r_j}} (\pi')^{- j p^r} + b_1 $$ As in the previous paragraph we change our choice of $z$ into \begin{align*} z' & = z \\ & - \sum\nolimits_{i \in I} \left(x_i (\pi')^{-i} + \ldots + x_i^{p^{r - 1}} (\pi')^{-i p^{r - 1}}\right) \\ & - \sum\nolimits_{j \in J} \left( \sigma(\mu_j) (\pi')^{- j p^{r - r_j}} + \ldots + \sigma(\mu_j)^{p^{r_j - 1}} (\pi')^{- j p^{r - 1}} \right) \end{align*} to obtain $$ (z')^p - z' = \sum\nolimits_{i \in I} x_i (\pi')^{-i} + \sum\nolimits_{j \in J} \sigma(\mu_j) (\pi')^{- j p^{r - r_j}} + b_1' $$ for some $b'_1 \in B_1$. Since there is a unique $j$ such that $j p^{r - r_j}$ is maximal and since $j p^{r - r_j}$ is bigger than $i \in I$ and divisible by $p$, we see that $M_1 / L_1$ falls into case (C) of Lemma \ref{lemma-pre-characteristic-p-case}. This finishes the proof.