\z \in C(k)$. Then, $$\begin{gathered}
A(x_i,\y)-A(x_i,\z)=j-k\\
A'(x_i,\y)-A'(x_i,\z)=p(j-k).\end{gathered}$$

Lemma 3.5 of [@Cabling] shows that, by making the cabling parameter $n$ sufficiently large, we can ensure that exterior intersection points generate the highest $A$ and $A'$ gradings. More precisely, we have the following restatement of Lemma 3.5 of [@Cabling]:

\[lemma:isolate\] Pick $l\in \Z$, and let $$g=\mathrm{max}\{ A(\x) \ | \ \x \text{\ is\ any\ intersection\ point,\ exterior\ or\ not} \}.$$ Then there exists a constant $N > 0$ such that for all $n$ with $n> N$, the only intersection points with $A(\x)\ge g-l$ are exterior.

The lemmas are summarized by the table and caption in Figure \[fig:chaincomplex\].

![\[fig:chaincomplex\] Table of $A$ and $A'$ gradings of exterior points for the Heegaard diagram, $H(p,n)$. For each ordered pair, the number on the left is the $A$ grading (i.e. the companion knot’s grading). The number on the right is $A'$ (the cabled knot’s grading). According to Lemma \[lemma:pindown\], we have $g'=pg+\frac{(pn)(p-1)}{2}$. The area below the top (solid) line represents the chain complex $\FiltY(K_{p,pn+1},g'-p-1)=\FiltY(K,g-2)$. The area below the bottom (dashed) line represents the chain complex $\FiltY(K_{p,pn+1},g'-2p-1)=\FiltY(K,g-3)$. ](chaincomplextau.eps)

Theorem \[thm:filt\] follows quickly from the table. The key point is that $H(p,n)$ is a diagram for $Y$, regardless 