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

By [L1], the family $\{{\mathfrak{p}}_1^{\mathrm{\prime }},\dots ,{\mathfrak{p}}_r^{\mathrm{\prime }}\}$ is finite. Repeatedly applying the two-step prime-avoidance argument to the triples $\mathfrak{p}⊋{\mathfrak{p}}_j⊋{\mathfrak{p}}_{j-1}$ for $1\le j\le d-1$ produces a strict chain of the same length ending at $\mathfrak{p}$ whose first proper subprime is not contained in any ${\mathfrak{p}}_i^{\mathrm{\prime }}$. Relabel so that the resulting chain is again $\mathfrak{p}={\mathfrak{p}}_d⊋{\mathfrak{p}}_{d-1}⊋\cdots ⊋{\mathfrak{p}}_0$ with ${\mathfrak{p}}_1⊈{\mathfrak{p}}_i^{\mathrm{\prime }}$ for every $i$.

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 the first proper subprime of a chain of the same length.