This is proved in exactly the same manner as Theorem \ref{theorem-normalized-base-change-with-reduced-fibre} with a few minor modifications. The most important change is to use More on Algebra, Lemma \ref{more-algebra-lemma-epp-essentially-finite-type-separable} instead of More on Algebra, Proposition \ref{more-algebra-proposition-epp-essentially-finite-type}. During the proof we will repeatedly use that formation of the set of points where a (flat, finitely presented) morphism like $X \to \Spec(A)$ is smooth commutes with base change, see Morphisms, Lemma \ref{morphisms-lemma-set-points-where-fibres-smooth}. \medskip\noindent Since $X$ is flat over $A$ every generic point $\eta$ of $X$ maps to the generic point of $\Spec(A)$. After replacing $X$ by its reduction we may assume $X$ is reduced. In this case $X_K$ is geometrically reduced over $K$ by Varieties, Lemma \ref{varieties-lemma-generic-points-geometrically-reduced}. Hence $X_K \to \Spec(K)$ is smooth on a dense open by Varieties, Lemma \ref{varieties-lemma-geometrically-reduced-dense-smooth-open}. Thus the smooth locus $U \subset X$ of the morphism $X \to \Spec(A)$ is open (by Morphisms, Definition \ref{morphisms-definition-smooth}) and is dense in the generic fibre. This reduces us to the situation of the following paragraph. \medskip\noindent Assume $X$ is normal and the smooth locus $U \subset X$ of $X \to \Spec(A)$ is dense in the generic fibre. This implies that $U$ is dense in all but finitely many fibres, see Lemma \ref{lemma-nowhere-dense-generic-fibre}. Let $x_1, \ldots, x_r \in X \setminus U$ be the finitely many generic points of irreducible components of $X \setminus U$ which are moreover generic points of irreducible components of fibres of $X \to \Spec(A)$. Set $\mathcal{O}_i = \mathcal{O}_{X, x_i}$. Observe that the fraction field of $\mathcal{O}_i$ is the residue field of a generic point of $X$. Let $A_i$ be the localization of $A$ at the maximal ideal corresponding to the image of $x_i$ in $\Spec(A)$. We may apply More on Algebra, Lemma \ref{more-algebra-lemma-epp-essentially-finite-type-separable} and we find finite separable extensions $K \subset K_i$ which are solutions for $A_i \to \mathcal{O}_i$. Let $K \subset L$ be a finite separable extension dominating all of the extensions $K \subset K_i$. Then $K \subset L$ is still a solution for $A_i \to \mathcal{O}_i$ by More on Algebra, Lemma \ref{more-algebra-lemma-formally-smooth-goes-up}. \medskip\noindent Consider the diagram (\ref{equation-normalized-base-change}) with the extension $L/K$ we just produced. Note that $U_B \subset X_B$ is smooth over $B$, hence normal (for example use Algebra, Lemma \ref{algebra-lemma-normal-goes-up}). Thus $Y \to X_B$ is an isomorphism over $U_B$. Let $y \in Y$ be a generic point of an irreducible component of a fibre of $Y \to \Spec(B)$ lying over the maximal ideal $\mathfrak m \subset B$. Assume that $y \not \in U_B$. Then $y$ maps to one of the points $x_i$. It follows that $\mathcal{O}_{Y, y}$ is a local ring of the integral closure of $\mathcal{O}_i$ in $R(X) \otimes_K L$ (details omitted). Hence because $K \subset L$ is a solution for $A_i \to \mathcal{O}_i$ we see that $B_\mathfrak m \to \mathcal{O}_{Y, y}$ is formally smooth (this is the definition of being a "solution"). In other words, $\mathfrak m\mathcal{O}_{Y, y} = \mathfrak m_y$ and the residue field extension is separable. Hence the local ring of the fibre at $y$ is $\kappa(y)$. This implies the fibre is smooth over $\kappa(\mathfrak m)$ at $y$ for example by Algebra, Lemma \ref{algebra-lemma-separable-smooth}. This finishes the proof.