Choose a parameter that misses the top-dimensional minimal components

By [L1], the minimal primes of $R$ form a finite set; because $d>0$, none equals $\mathfrak{m}$. Hence [L2] provides $x\in \mathfrak{m}$ outside every minimal prime.

Let $\mathfrak{p}\mathrm{/}(x)$ be a minimal prime of $R\mathrm{/}(x)$. By [L4], the prime $\mathfrak{p}$ is minimal over $(x)$ in $R$. Choose a minimal prime $\mathfrak{q}\subseteq \mathfrak{p}$. Since $x\mathrm{\notin }\mathfrak{q}$, its image in the domain $R\mathrm{/}\mathfrak{q}$ is a nonzerodivisor. The prime $\mathfrak{p}\mathrm{/}\mathfrak{q}$ is minimal over that image: an intervening prime would give a prime strictly below $\mathfrak{p}$ containing $x$. Thus [L3] gives ${\mathrm{ht}}_{R\mathrm{/}\mathfrak{q}}(\mathfrak{p}\mathrm{/}\mathfrak{q})=1$. Every chain in $R\mathrm{/}\mathfrak{p}$ extends upward to a chain in $R\mathrm{/}\mathfrak{q}$ longer by one step.

Since $\dim (R\mathrm{/}\mathfrak{q})\le d$, step 2.1 implies $\dim (R\mathrm{/}\mathfrak{p})\le d-1$. Taking the supremum over all minimal primes $\mathfrak{p}\mathrm{/}(x)$ of $R\mathrm{/}(x)$ yields $\dim (R\mathrm{/}(x))\le d-1$.

Thus one can choose a first parameter outside the minimal components, and every such choice lowers dimension by at least one.