as in Equation .

\[prop:diag\] Let $B({\ensuremath{\mathbf{z}}},t)$ be an element of $\mathbb{Q}[z_1,{\ensuremath{\overline{z}}}_1,\dots,z_d,{\ensuremath{\overline{z}}}_d][\![t]\!]$. Then $$\label{eq:postodiag}
[z_1^{\geq}]\cdots[z_d^{\geq}]B({\ensuremath{\mathbf{z}}},t) \bigg|_{z_1=1,\dots,z_d=1} = \Delta \left(\frac{B\left({\ensuremath{\overline{z}}}_1,\dots,{\ensuremath{\overline{z}}}_d,z_1\cdots z_d\cdot t\right)}{(1-z_1)\cdots(1-z_d)}\right).$$

Suppose that $B$ has the expansion $$B({\ensuremath{\mathbf{z}}},t) = \sum_{n \geq 0} \left( \sum_{{\ensuremath{\mathbf{i}}}\in \mathbb{Z}^d} b_{{\ensuremath{\mathbf{i}}}}(n)z_1^{i_1}\cdots z_d^{i_d} \right)t^n.$$ Then the right hand side of Equation  is given by $$\Delta \left(\sum_{k \geq 0} z_1^k \right)\cdots\left(\sum_{k \geq 0} z_d^k \right)
\left(\sum_{n \geq 0} \left( \sum_{{\ensuremath{\mathbf{i}}}\in \mathbb{Z}^d} b_{{\ensuremath{\mathbf{i}}}}(n) z_1^{n-i_1}\cdots z_d^{n-i_d} \right)t^n\right)$$ so that the coefficient of $t^n$ in the diagonal is the sum of all terms $b_{{\ensuremath{\mathbf{i}}}}(n)$ with $i_1,\dots,i_d\geq 0$ (by assumption there are only finitely many which are non-zero). But this is exactly the coefficient of $t^n$ on the left hand side.

We note also that in the context of lattice path models with step set ${\ensuremath{\mathcal{S}}}\subseteq \{\pm1,0\}^d \setminus \{\mathbf{0}\}$, the modified generating function $F\left({\ensuremath{\overline{z}}}_1,\dots,{\ensuremath{\overline{z}}}_d, z_1\cdots z_d\cdot t\right)$ is actually a power series in the variables $z_1,\dots,z_d,t$ (as a walk cannot move farther on the integer lattice than its number of steps). Combining Lemma \[thm:pospart\] and Proposition \[prop:diag\] implies that the generating function for the number of walks can be represented as $F({\ensuremath{\mathbf{1}}},t) = \Delta
\left(\frac{G({\ensuremath{\mathbf{z}}},t)}{H({\ensuremath{\mathbf{z}}},t)}\right)$, where $$\begin{aligned}
\frac{G({\ensuremath{\mathbf{z}}},t)}{H({\ensuremath{\mathbf{z}}},t)} &= \frac{(1-z_1^2)\cdots(1-z_d^2)}{1 - t(z_1\cdots z_d)S({\ensuremath{\mathbf{z}}})}\cdot \frac{1}{(1-z_1)\cdots(1-z_d)}\notag \\
&= \frac{(1+z_1)\cdots(1+z_d)}{1 - t(z_1\cdots z_d)S({\ensuremath{\mathbf{z}}})}. \label{eq:rat}\end{aligned}$$ 