Reduce the principal ideal theorem to a Noetherian local domain

Let $\mathfrak{q}\subseteq \mathfrak{p}$ be a minimal prime of $R$. By [L1], $\mathrm{ht}(\mathfrak{q})=0$. The quotient $R\mathrm{/}\mathfrak{q}$ is a Noetherian domain by [L2] and [L3], and then [L4] makes $A=(R\mathrm{/}\mathfrak{q}{)}_{\mathfrak{p}\mathrm{/}\mathfrak{q}}$ a Noetherian local domain.

If $\mathfrak{q}=\mathfrak{p}$, then $\mathfrak{p}$ already has height $0$ by step 1.1 and there is nothing left to prove. Assume now that $\mathfrak{q}\mathrm{\ne }\mathfrak{p}$. Since $\mathfrak{p}$ is minimal over $(x)$, the element $x$ cannot lie in $\mathfrak{q}$; otherwise the prime $\mathfrak{q}$ would also contain $(x)$ and minimality would force $\mathfrak{q}=\mathfrak{p}$. Thus the image of $x$ in $R\mathrm{/}\mathfrak{q}$, and hence in $A$, is nonzero.

Let $\mathfrak{m}$ be the maximal ideal of $A$. By [L5], primes of $A$ correspond to primes of $R$ that lie between $\mathfrak{q}$ and $\mathfrak{p}$. Because $\mathfrak{p}$ is minimal over $(x)$, the prime $\mathfrak{p}\mathrm{/}\mathfrak{q}$ is minimal over the image of $(x)$ in $R\mathrm{/}\mathfrak{q}$, and after localizing there is no smaller prime of $A$ containing $x\mathrm{/}1$. Hence $\mathfrak{m}$ is minimal over $(x\mathrm{/}1)$.

Let ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_d=\mathfrak{p}$ be any strict prime chain. If $d=0$ there is nothing to bound. If $d>0$, then ${\mathfrak{p}}_0\mathrm{\ne }\mathfrak{p}$, so minimality of $\mathfrak{p}$ over $(x)$ gives $x\mathrm{\notin }{\mathfrak{p}}_0$. By [L2]–[L4], $A_0=(R\mathrm{/}{\mathfrak{p}}_0{)}_{\mathfrak{p}\mathrm{/}{\mathfrak{p}}_0}$ is a Noetherian local domain in which the image of $x$ is nonzero; [L5] also shows that its maximal ideal is minimal over that image. The original chain induces a strict chain of length $d$ ending at this maximal ideal. Therefore a height bound of $1$ in every reduced local-domain case forces $d\le 1$. Since the original chain was arbitrary, ${\mathrm{ht}}_R(\mathfrak{p})\le 1$.