I want to prove that for strings $s$ over a singleton alphabet $\Sigma = \{a\}$, if $s = a^{x\cdot y}, \ x,y \neq 1$, then there exists a smallest grammar derived from the grammar: $$ S \to A^x \\ A \to a^y $$
By derived, I mean you can then compute a smallest grammar solely by creating new rules and collapsing variables. My intuition tells me that this is the case. This is interesting because now the problem has been effectively divided-and-conquered, as $\{a\} \cap \{A\} = \varnothing$ are disjoint alphabets and so you proceed by computing the smallest grammar of $A^x$ alone and of $a^y$ alone, and simply union the resultant rule sets.
Of course, we would need to handle the case where $x$ or $y$ is prime, and for that, we would make another lemma, saying that a smallest grammar is any smallest grammar of $a^{x-1}$ with $a$ appended to the start rule. My intuition also tells me that is the case.
For example $s = a^8$, then $S \to A^4; A \to a^2$ is clearly a smallest grammar.
My idea is to look at all possible substiututions of repeated substrings. For the case $s = a^8$, we have:
$$ A = a^2, \ B = aA, \ C = aB = A^2 $$
Those are the only possibilities. Now define $\varphi(C) = aB; \ \varphi(x) = x$ elsewhere; then we can get from the reduced grammar $S \to C^2$ to $S \to (aB)^2$ simply by applying $\varphi$ to the rule string $C^2$.
Can we somehow define a finite semigroup of $\varphi$'s and magically say that we've tried all possible combinations?
What's nice about the singleton alphabet is that all variables commute with each other and $a$! If the above can be proved, then we've proven that at least over a singleton alphabet, the smallest grammar problem is easy. This comes from the fact that factoring the exponent $n = x\cdot y$ can be done naively in time $\sqrt n $ and $n = |s|$ is the length of the input so, the algorithmn, because it divides and conquers will turn out to be polynomial in $|s|$.