Select generators witnessing the converse height theorem

If $r=0$, the empty list works.

Assume $r\ge 1$. The prime ideals of height $0$ are exactly the minimal primes of $R$, hence are finite by [L1]. None of them contains $\mathfrak{a}$, because $\mathrm{ht}(\mathfrak{a})=r\ge 1$. Therefore [L2] provides $a_1\in \mathfrak{a}$ outside every height-zero prime. Any prime minimal over $(a_1)$ must then have height at least $1$, while [L3] gives height at most $1$. Thus $(a_1)$ has height exactly $1$.

Suppose $1<i\le r$ and $a_1,\dots ,a_{i-1}\in \mathfrak{a}$ have already been chosen so that $J=(a_1,\dots ,a_{i-1})$ has height exactly $i-1$. The minimal primes over $J$ are finite by [L1]. Among them, collect those of height $i-1$; none can contain $\mathfrak{a}$, because $\mathfrak{a}$ has height $r\ge i$. By [L2], choose $a_i\in \mathfrak{a}$ outside all of those primes. Any prime $\mathfrak{q}$ minimal over $J+(a_i)$ contains some minimal prime $\mathfrak{p}$ over $J$. If $\mathrm{ht}\mathfrak{q}\le i-1$, then $\mathrm{ht}\mathfrak{p}\ge i-1$ and strict containment $\mathfrak{p}⊊\mathfrak{q}$ would force $\mathrm{ht}\mathfrak{q}\ge i$; hence $\mathfrak{q}=\mathfrak{p}$ is one of the excluded height-$(i-1)$ primes, impossible since it contains $a_i$. Thus every such $\mathfrak{q}$ has height at least $i$. By [L3] it has height at most $i$, so $J+(a_i)$ has height exactly $i$.

Steps 1.1, 1.2, and 2.1 build the required list $a_1,\dots ,a_r\in \mathfrak{a}$.