Choose the first generator's minimal prime inside the target prime

Write $J=(x_2,\dots ,x_n)$. By [L1], the family $\{{\mathfrak{p}}_1^{\mathrm{\prime }},\dots ,{\mathfrak{p}}_r^{\mathrm{\prime }}\}$ is finite. Every ${\mathfrak{p}}_i^{\mathrm{\prime }}$ is minimal among primes containing $J$. Since $\mathfrak{p}$ contains $J$ but is not one of them, $\mathfrak{p}⊈{\mathfrak{p}}_i^{\mathrm{\prime }}$ for every $i$. Finite prime avoidance [L2] therefore supplies an element of $\mathfrak{p}$ outside their union. The given chain has $d\ge 1$, so its ${\mathfrak{p}}_1$ exists.

Starting at $j=d-1$ and descending to $j=1$, keep the current strict chain and the invariant that ${\mathfrak{p}}_{j+1}$ is contained in none of the ${\mathfrak{p}}_i^{\mathrm{\prime }}$. If ${\mathfrak{p}}_j$ already has this property, leave it unchanged. Otherwise choose $c\in {\mathfrak{p}}_{j+1}\setminus {\bigcup }_i{\mathfrak{p}}_i^{\mathrm{\prime }}$ by [L2]. Since ${\mathfrak{p}}_{j-1}⊊{\mathfrak{p}}_j\subseteq {\mathfrak{p}}_i^{\mathrm{\prime }}$ for at least one $i$, the element $c$ is outside ${\mathfrak{p}}_{j-1}$. In the Noetherian quotient $A=R\mathrm{/}{\mathfrak{p}}_{j-1}$, the radical of $(\bar{c})$ is the finite intersection of its minimal primes by [L1]. The prime ${\mathfrak{p}}_{j+1}\mathrm{/}{\mathfrak{p}}_{j-1}$ contains this intersection, so by primality it contains one of those minimal primes; call its inverse image $\mathfrak{q}$. Then ${\mathfrak{p}}_{j-1}⊊\mathfrak{q}$ because $c\in \mathfrak{q}$ but $c\mathrm{\notin }{\mathfrak{p}}_{j-1}$. By [L3] the selected prime has height at most $1$ in $A$, whereas ${\mathfrak{p}}_{j+1}\mathrm{/}{\mathfrak{p}}_{j-1}$ has the two-step chain through ${\mathfrak{p}}_j\mathrm{/}{\mathfrak{p}}_{j-1}$. Thus $\mathfrak{q}⊊{\mathfrak{p}}_{j+1}$. It also contains $c$, so it is contained in none of the ${\mathfrak{p}}_i^{\mathrm{\prime }}$. Replacing ${\mathfrak{p}}_j$ by $\mathfrak{q}$ preserves the chain length and establishes the invariant one level lower.

After the finite descent in step 2.1, ${\mathfrak{p}}_1$ is contained in none of the ${\mathfrak{p}}_i^{\mathrm{\prime }}$. When $d=1$, this says ${\mathfrak{p}}_1=\mathfrak{p}$, as already established in step 1.1.

Since ${\mathfrak{p}}_1$ is not contained in the finite union ${\bigcup }_i{\mathfrak{p}}_i^{\mathrm{\prime }}$, [L2] gives $b\in {\mathfrak{p}}_1\setminus {\bigcup }_i{\mathfrak{p}}_i^{\mathrm{\prime }}$. In particular $b\in \mathfrak{p}$, so $\mathfrak{p}$ contains $(b,x_2,\dots ,x_n)$.

Let $\mathfrak{q}\subseteq \mathfrak{p}$ be prime minimal over $(b,x_2,\dots ,x_n)$. Because $\mathfrak{q}$ contains $(x_2,\dots ,x_n)$, it contains one of the minimal primes ${\mathfrak{p}}_i^{\mathrm{\prime }}$. By the choice of $b$, one has $b\mathrm{\notin }{\mathfrak{p}}_i^{\mathrm{\prime }}$, so $\mathfrak{q}\mathrm{\ne }{\mathfrak{p}}_i^{\mathrm{\prime }}$. If $\mathfrak{q}\mathrm{\ne }\mathfrak{p}$, then $\mathfrak{p}⊋\mathfrak{q}⊋{\mathfrak{p}}_i^{\mathrm{\prime }}$ is a strict chain in the quotient ring $R\mathrm{/}(x_2,\dots ,x_n)$, so the prime $\mathfrak{p}\mathrm{/}(x_2,\dots ,x_n)$ has height at least $2$. But $\mathfrak{p}$ is minimal over $(x_1,\dots ,x_n)$, hence $\mathfrak{p}\mathrm{/}(x_2,\dots ,x_n)$ is minimal over the principal ideal generated by the image of $x_1$, contradicting [L3]. Therefore $\mathfrak{q}=\mathfrak{p}$.

Hence $\mathfrak{p}$ is minimal over $(b,x_2,\dots ,x_n)$, with $b$ chosen from ${\mathfrak{p}}_1$ of a chain of the same length.