The following question arose in the comments on this question, and it seems like a reasonable question to ask in its own right. I've added some additional details.
The word problem in any fixed finitely presented simple group $G$ is decidable; this is Kuznetsov's theorem. However, the procedure by which one solves it is not uniform. Indeed, given a word $w$ over the generators (of the fixed presentation for $G$), the procedure for deciding if $w=_G 1$ first begins by saying "if $G$ is trivial, then return 'yes'; if $G$ is not trivial, then choose a non-trivial element of $G$ such that... (etc.)". Of course, whether $G$ or not is trivial is a fixed property of $G$, so for a fixed $G$ there is a (trivial) Turing machine which returns whether or not $G$ is trivial (either the machine that always says "yes", or the machine which always says "no" will do).
Thus the procedure in Kuznetsov's theorem is not uniform. This raises the following interesting question.
($\ast$) Is the word problem uniformly decidable for finitely presented simple groups?
The procedure for non-trivial finitely presented simple groups (i.e. the continuation of the above procedure after one has chosen a non-trivial element) is, on the other hand, uniform (see below [1]). In particular, if one has an oracle which, on input a (presentation of a) finitely presented simple group, outputs whether or not the group is trivial, then the answer to ($\ast$) is yes: first decide whether the group is trivial or not by the oracle: if it is trivial, then the word problem is trivially solvable; if it is non-trivial, it can be uniformly decided by the above procedure.
Thus, the above question ($\ast$) is equivalent to the following:
($\ast'$) Is the triviality problem decidable for finitely presented simple groups?
Note that the triviality problem for finitely presented groups is, in general, undecidable, by the Adian-Rabin theorem, as being trivial is a Markov property of groups. However, one might suspect that this question is easier if one is promised that the groups given as input are always simple.
(Note: the above questions and statements can be identically considered for recursively presented groups.)
[1] To see that the word problem is uniformly decidable for non-trivial finitely presented simple groups, the following procedure can be used, as sketched by Ben Steinberg in the comments on this question. Let $\langle A \mid R \rangle$ be a presentation of a non-trivial finitely presented simple group $G$. Let $w$ be a word over the generators $A$ and their inverses. If $w=_G 1$, then we can detect this by enumerating the consequences of the relations $R$ until we find a finite sequence of derivations taking $w$ to $1$. On the other hand, if $w \neq_G 1$, then $\langle A | R \cup \{ w \} \rangle$ is a non-trivial quotient of $G$. But $G$ is simple, so the group $H$ presented by this presentation is necessarily trivial; now $H$ is trivial if and only if each generator $A$ is trivial in $H$, which we can detect by enumerating consequences of $R \cup \{ w \}$ until we have for every generator $a$ in $A$ found a finite derivation sequence taking $a$ to $1$. Thus, we can detect if we are in the case $w \neq_G 1$. By running the two algorithms above in parallel, we hence eventually end up detecting $w =_G 1$ or $w \neq_G 1$, so we have solved the word problem. Of course this also works for recursively presented groups.
