Proof of (3) $\Rightarrow$ (1). Assume (3). Let $k \subset k'$ be a finite purely inseparable extension. Set $A' = A \otimes_k k'$. This is a local ring with maximal ideal $\mathfrak m'$. Set $K' = A'/\mathfrak m'$. We get a commutative diagram $$ \xymatrix{ 0 \ar[r] & H_1(L_{K/k}) \otimes K' \ar[r] \ar[d]_\beta & \mathfrak m/\mathfrak m^2 \otimes K' \ar[r] \ar[d] & \Omega_{A/k} \otimes_A K' \ar[r] \ar[d]_{\cong} & \Omega_{K/k} \otimes K' \ar[r] \ar[d]_\alpha & 0 \\  & H_1(L_{K'/k'}) \ar[r] & \mathfrak m'/(\mathfrak m')^2 \ar[r] & \Omega_{A'/k'} \otimes_{A'} K' \ar[r] & \Omega_{K'/k'} \ar[r] & 0 } $$ with exact rows. The third vertical arrow is an isomorphism by base change for modules of differentials (Algebra, Lemma \ref{algebra-lemma-differentials-base-change}). Thus $\alpha$ is surjective. By Lemma \ref{lemma-gamma-commutative-diagram} we have $$ \dim \Ker(\alpha) - \dim \Ker(\beta) + \dim \Coker(\beta) = 0 $$ (and these dimensions are all finite). A diagram chase shows that $\dim \mathfrak m'/(\mathfrak m')^2 \leq \dim \mathfrak m/\mathfrak m^2$. However, since $A \to A'$ is finite flat we see that $\dim(A) = \dim(A')$, see Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-total}. Hence $A'$ is regular by definition. \medskip\noindent Equivalence of (3) and (4). Consider the Jacobi-Zariski sequences for rows of the commutative diagram $$ \xymatrix{ \mathbf{F}_p \ar[r] & A \ar[r] & K \\ \mathbf{F}_p \ar[r] \ar[u] & k \ar[r] \ar[u] & K \ar[u] } $$ to get a commutative diagram $$ \xymatrix{ 0 \ar[r] & \mathfrak m/\mathfrak m^2 \ar[r] & \Omega_{A/\mathbf{F}_p} \otimes_A K \ar[r] & \Omega_{K/\mathbf{F}_p} \ar[r] & 0 & \\ 0 \ar[r] & H_1(L_{K/k}) \ar[r] \ar[u] & \Omega_{k/\mathbf{F}_p} \otimes_k K \ar[r] \ar[u] & \Omega_{K/\mathbf{F}_p} \ar[r] \ar[u] & \Omega_{K/k} \ar[r] \ar[u] & 0 } $$ with exact rows. We have used that $H_1(L_{K/A}) = \mathfrak m/\mathfrak m^2$ and that $H_1(L_{K/\mathbf{F}_p}) = 0$ as $K/\mathbf{F}_p$ is separable, see Algebra, Proposition \ref{algebra-proposition-characterize-separable-field-extensions}. Thus it is clear that the kernels of $H_1(L_{K/k}) \to \mathfrak m/\mathfrak m^2$ and $\Omega_{k/\mathbf{F}_p} \otimes_k K \to \Omega_{A/\mathbf{F}_p} \otimes_A K$ have the same dimension. \medskip\noindent Proof of (2) $\Rightarrow$ (4) following Faltings, see \cite{Faltings-einfacher}. Let $a_1, \ldots, a_n \in k$ be elements such that $\text{d}a_1, \ldots, \text{d}a_n$ are linearly independent in $\Omega_{k/\mathbf{F}_p}$. Consider the field extension $k' = k(a_1^{1/p}, \ldots, a_n^{1/p})$. By Algebra, Lemma \ref{algebra-lemma-size-extension-pth-roots} we see that $k' = k[x_1, \ldots, x_n]/(x_1^p - a_1, \ldots, x_n^p - a_n)$. In particular we see that the naive cotangent complex of $k'/k$ is homotopic to the complex $\bigoplus_{j = 1, \ldots, n} k' \rightarrow \bigoplus_{i = 1, \ldots, n} k'$ with the zero differential as $\text{d}(x_j^p - a_j) = 0$ in $\Omega_{k[x_1, \ldots, x_n]/k}$. Set $A' = A \otimes_k k'$ and $K' = A'/\mathfrak m'$ as above. By Algebra, Lemma \ref{algebra-lemma-change-base-NL} we see that $\NL_{A'/A}$ is homotopy equivalent to the complex $\bigoplus_{j = 1, \ldots, n} A' \rightarrow \bigoplus_{i = 1, \ldots, n} A'$ with the zero differential, i.e., $H_1(L_{A'/A})$ and $\Omega_{A'/A}$ are free of rank $n$. The Jacobi-Zariski sequence for $\mathbf{F}_p \to A \to A'$ is $$ H_1(L_{A'/A}) \to \Omega_{A/\mathbf{F}_p} \otimes_A A' \to \Omega_{A'/\mathbf{F}_p} \to \Omega_{A'/A} \to 0 $$ Using the presentation $A[x_1, \ldots, x_n] \to A'$ with kernel $(x_j^p - a_j)$ we see, unwinding the maps in Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}, that the $j$th basis vector of $H_1(L_{A'/A})$ maps to $\text{d}a_j \otimes 1$ in $\Omega_{A/\mathbf{F}_p} \otimes A'$. As $\Omega_{A'/A}$ is free (hence flat) we get on tensoring with $K'$ an exact sequence $$ K'^{\oplus n} \to \Omega_{A/\mathbf{F}_p} \otimes_A K' \xrightarrow{\beta} \Omega_{A'/\mathbf{F}_p} \otimes_{A'} K' \to K'^{\oplus n} \to 0 $$ We conclude that the elements $\text{d}a_j \otimes 1$ generate $\Ker(\beta)$ and we have to show that are linearly independent, i.e., we have to show $\dim(\Ker(\beta)) = n$. Consider the following big diagram $$ \xymatrix{ 0 \ar[r] & \mathfrak m'/(\mathfrak m')^2 \ar[r] & \Omega_{A'/\mathbf{F}_p} \otimes K' \ar[r] & \Omega_{K'/\mathbf{F}_p} \ar[r] & 0 \\ 0 \ar[r] & \mathfrak m/\mathfrak m^2 \otimes K' \ar[r] \ar[u]^\alpha & \Omega_{A/\mathbf{F}_p} \otimes K' \ar[r] \ar[u]^\beta & \Omega_{K/\mathbf{F}_p} \otimes K' \ar[r] \ar[u]^\gamma & 0 } $$ By Lemma \ref{lemma-cartier-equality} and the Jacobi-Zariski sequence for $\mathbf{F}_p \to K \to K'$ we see that the kernel and cokernel of $\gamma$ have the same finite dimension. By assumption $A'$ is regular (and of the same dimension as $A$, see above) hence the kernel and cokernel of $\alpha$ have the same dimension. It follows that the kernel and cokernel of $\beta$ have the same dimension which is what we wanted to show. \medskip\noindent The implication (1) $\Rightarrow$ (2) is trivial. This finishes the proof of the proposition.