Prime avoidance and height of an ideal 
How can I use prime avoidance to prove that if $R$ is a Noetherian ring and $J\subset R$ an ideal with $\mathrm{height}(J)=n$ then there exists $x_1,\ldots,x_n\in J$ such that $\mathrm{height}((x_1,\ldots,x_i))=i$ for all $i=1,\ldots,n$?

I know that if $R$ is local and Noetherian of dimension $d$ then there exist $x_1,\ldots,x_d$ such that $\mathrm{height}(x_1,\ldots,x_i)\geq i$ for all $i=1\ldots,d$.
 A: We can assume that $R$ is an integral domain as modding out a prime minimal over $0$ does not change the length of chains of primes. Below is a proof of the theorem the way I learned it, which does not use PAL (although that is used to derive earlier results). I'm not sure how to give one that does use it without doing a great deal of work proving theorems I assume below.
Step 1: There exists $x_1,\ldots,x_n\in J$ such that $\mathrm{height}((x_1,\ldots,x_i))\geq i,\forall i=1,\ldots,n$.
Proof: It suffices to prove that there exists an $x_1\in J$ such that $\mathrm{height}((x_1)) \geq 1$, as then $\mathrm{height}(J/(x_1)) \leq n-1$ since the chain $(0)\subset P_1\subset \cdots \subset P_n = P$ loses at least $P_1$ when $(x_1)$ is modded at as $P_1$ is out most minimal over $(x_1)$. We can then make $R/(x_1)$ an integral domain and repeat the argument, so $\mathrm{height}(J/(x_1,\ldots,x_i)) \leq n-i$ meaning $\mathrm{height}((x_1,\ldots,x_i))\geq i$ as otherwise we would not be able to remove $i$ primes from all prime chains. Assume that $\mathrm{height}((x_1)) = 0,\forall x\in J$, thus any prime containing $(x_1)$ has height $0$. But since $R$ is an integral domain this is only true of $(0)$, so we get $J = (0)$ hence $\mathrm{height}(J) = 0$, and so our claim still holds.
Step 2: For any $i=1,\ldots,n$, $\mathrm{height}((x_1,\ldots,x_i))\leq i$ (this is known as the Generalized Principle Ideal Theorem or GPIT).
Results I Assume: Nakayama's lemma (if $M\subset R$ is a finitely generated ideal, $I\subset R$ an ideal, $N\subseteq M$ an ideal and $M = N + IM$, then $M = N$), that every ideal has a unique irreducible primary decomposition, and that a Noetherian ring of dimension $0$ is Artinian. 
Proof: We begin with the base case $(x_1)$. If $\mathrm{height}((x_1))>1$, then we have some $P$ minimal over $(x_1)$ which contains a nonzero prime $Q$. We let $Q^{(n)}$ denote the unique $Q$-primary component of $Q^n$. Note that $Q^{(n+1)}\subseteq Q^{(n)}$ as $Q = Q^{(n+1)}\cap K_1\cap \cdots\cap K_j$ where each $K_m$ is a primary ideal that is not $Q$-primary, and if we take $t\in K_1\cap\cdots\cap K_j\setminus Q,x\in Q^{(n+1)}$ then $tx\in Q^{n+1}\subseteq Q^n\subseteq Q^{(n)}$, yet $t^m\notin Q^{(n)}$ for any $m\in\mathbb{N}$ so $x\in Q^{(n+1)}$. Also, the intersection of all $Q^{(n)}$ is $(0)$, as for any element $x$ in this intersection we have some $t\notin Q$ such that $tx\in Q^n$ for all $n$ hence $x\in Q^nR_Q$ for all $n$ so $x\in \bigcap\limits_{n\geq 1} Q^n R_Q = 0$ hence $x=0$ in $R_Q$ so $x=0$ in $R$. Finally, since $R_P/(x_1)$ has dimension $0$ it must be Artinian so the chain $(Q+(x_1))/(x_1)\supseteq (Q^{(2)}+(x_1))/(x_1)\supseteq \cdots$ stabilizes thus $Q^{(n)}+(x_1)=Q^{(n+1)}+(x_1)$ for some $n$. This means $Q^{(n)}=Q^{(n+1)}+Q^{(n)}(x_1)$ as subtracting a multiple of $x_1$ from each side gives $a=b+rx_1$ for $a\in Q^{(n)},b\in Q^{(n+1)}\subseteq Q^{(n)}$ hence $rx_1=a-b\in Q^{(n)}$ and since $x_1\notin Q$ we have $r\in Q^{(n)}$. Nakayama's lemma gives us that $Q^{(n)}=Q^{(n+1)}$ and the same holds for all $m>n$ by induction, so the intersection is $Q^{(n)}$. But we already established that the intersection is $(0)$, hence $Q$ must be $(0)$ as $Q^{(n)}$ is $Q$-primary and $R$ is an integral domain, contradicting our assumption. The proof for $i>1$ relies on much the same methods, so I shall not give it here.
