Can any presentation of a finitely presented group be reduced to a finite one? Suppose $G = \langle x_1, \ldots, x_n \mid p_1, \ldots, p_m \rangle$ is a finitely presented group, and let $\langle A \mid R \rangle$ be another presentation of $G$, with $A$ and $R$ possibly infinite.  Do there always exist finite subsets $A' \subset A$, $R' \subset R$ such that $G = \langle A' \mid R' \rangle$?
I feel like the answer should be "yes."  Here's my idea: we can write each $x_i$ as a product of finitely many $a \in A$ (and their inverses); denote by $A_i$ this finite set of $a$'s.  Then the finite set $A_1 \cup \cdots \cup A_n$ generates $G$.
Similarly, each relator $p_i$ can be derived from a finite set of relators $R_i \subset R$.  Here's my problem, though: how do I know that any relator $w$ in the letters $A_1 \cup \cdots \cup A_n$ can be reduced using these $R_i$?  Not every such $w$ blocks off into $x_i$-chunks.

What about a group $G$ with generators $x_i$ for $i \ge 1$, and relations $x_i^6=1$ and $x_{i+1}^{-1}x_i^2x_{i+1}=x_i^3$ for all $i \ge 1$.
The conjugation relation implies that $x_i$ is trivial for each $i$, so $|G|=1$.
But in the group $G_i$ generated by $x_1,\ldots,x_i$ with relations involving only these generators, the generator $x_i$ does not get killed off, so $|G_i|=6$.
Added later: As suitangi has pointed out, this does not work, because we can take $A'$ and $R'$ to be empty sets to get a presentation of the trivial group. So here is  asecond try.
The generators are still $x_i$ with $i \ge 1$. The relations are now $x_i^6=1$ for $i > 1$, and $x_{i+1}^{-1}x_i^2x_{i+1}=x_i^{-3}$ for all $i \ge 1$.
Now we still have $x_i=1$ for all $i > 1$, and then the relation $x_2^{-1}x_1^2x_2=x_1^{-3}$ reduces to $x_1^5=1$, so $|G|=5$.
But the group $G_i$ is now infinite cyclic for $i=1$, of order $3990$ for $i=2$, and the free product $C_5*C_6$ for $i>2$.  More generally, for any nonempty subset $A'$ of $A$, if $i$ is maximal with $x_i \in A'$, then $x_i$ has order $6$ in the group defined by taking only those relations involving generators in $A'$, so no subset of those relations can define a group on $A'$ that is isomorphic to $G$.
