How "bad" can presentation of the trivial group get? These questions are sort of preliminary questions and reference requests for a project I am doing.
Lets say, for concreteness, that $R$ is a set of words in the free group of rank two and that $\langle a,b \mid R \rangle$ is the trivial group. How "bad" can $R$ be? I guess my ideal "bad" is that $R$ is infinite and if $\varnothing \neq T \subseteq R $, then $\langle a,b \mid R \setminus T \rangle$ is not the trivial group. Also how would one go about finding/constructing these "bad" $R$, maybe for finitely generated group in general. 
I guess a more general question: Let $R_{\text {fam}}$ be a countable family of disjoint sets of words in the free group generated by the set $S$ such that $\langle S \mid \cup R_{\text{fam}} \rangle $ is the trivial group; How "bad" can $R_{\text{fam}}$ be? The "bad" here is essentially the same except looking at $\cup (R_{\text{fam}}\setminus T)$ where $T$ is some non empty subset of $R_\text{fam}$.
I am mostly looking for an answer to the specific example and references for these sorts of questions. There are plenty of variations, maybe looking at  finite $\langle S|R \rangle$ and looking at how bad that $R$ can get, also looking at "preloaded" $R$, that is $R$ has to have certain relations. 
 A: Expanding on the comments, there is a very strong sense in which every "infinitary" description of the trivial group can be boiled down to a "finitary" one - namely, the compactness theorem for first-order logic. This asserts roughly that any set of axioms which proves some single sentence has a finite subset which proves that sentence. 
Finitely generated group presentation fall into this setting: the sentence we're interested in  is $$(a_1=e)\,\,\wedge\,\, ...\,\,\wedge\,\, (a_n=e)$$ (where "$\wedge$" is "and" and the $a_i$s are the generators of our group), and our set of axioms consists of the axioms of group theory together with axioms corresponding to each relation. By the compactness theorem, any set of relations which makes the finitely many generators each trivial has a finite subset which does the same thing.
Note that this argument breaks down for infinitely-generated groups - and indeed the claim itself is false, the simplest counterexample having infinitely many generators, each of which is trivialized directly - the point being that to say that each of infinitely many generators is trivial is no longer first-order, since we need an infinite conjunction.
The compactness theorem similarly applies to your more general $R_{fam}$ situation: some finitely many sets in $R_{fam}$ must enforce triviality, as long as $S$ is infinite.
