= Solution
For a finite generating set $A$, use the symmetric alphabet $\Sigma=A\cup A^{-1}$. The <Cayley graph> has a vertex for each $g\in G$ and an oriented edge labelled $a$ from $g$ to $ga$. Give every edge length one. Its vertex <word metric> is
$$
\boxed{d_A(g,h)=\min\{|w|:w\text{ over }\Sigma\text{ represents }g^{-1}h\}}.
$$
Two <finitely presented groups> are quasi-isometric when their <Cayley graphs>, with these <path> metrics, are quasi-isometric. The choice of finite generating set is immaterial, as the following proof shows; the <relators> identify vertices but do not otherwise enter the metric definition.
Given finite generating sets $A,B$ of the same group, put $L=\max(1,\max_{a\in A^{\pm1}}|a|_B)$ and $M=\max(1,\max_{b\in B^{\pm1}}|b|_A)$. Replace each letter in a shortest $A$-word by a fixed $B$-word to obtain $|g|_B\le L|g|_A$. Reversing the roles gives $|g|_A\le M|g|_B$. Therefore the vertex identity map satisfies
$$
\boxed{M^{-1}d_A(g,h)\le d_B(g,h)\le Ld_A(g,h)}.
$$
It is onto, and hence a <quasi-isometry>. Map each edge to the fixed word <path> for its label, or pass through the uniformly dense vertex sets, to obtain the corresponding graph <quasi-isometry> with a bounded additive error. This proves <equivalence of finite word metrics> and independence of the two <finite group presentations>.
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