Can any presentation of a finitely presented group be reduced to a finite one?

Not every presentation of a finitely presented group can be reduced to a finite one.

In his highly advisable notes on Geometric Group theory, Charles F. Miller III provides the presentation: \begin{equation} \langle a,b,c_o,c_1,c_2,\ldots \mid a^4=1, b^3 =1, c_0^{-1} b c_0 = a^2, c_1^{-1} c_0 c_1 =b, c_2^{-1} c_1 c_2 = c_0 , c_3^{-1}c_2 c_3 = c_1,\ldots\rangle \end{equation} which defines a cyclic group of order $2$, but any of its finite subpresentations define a group of order $3$, a group of order $4$, or an infinite group.

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 $G$.