Please scan the article on the smallest grammar problem. It is a simply-stated, major open problem because it directly relates to $\text{P vs NP}$ if an answer in either direction can be given.
Let $t \leqslant s$ mean that $t$ occurs somewhere in $s$ as a substring, or alternatively $utv = s$. Imagine you have a smallest grammar to look at. Call it $G$. It can be fully expanded to its input (for smallest grammar algorithm) $s \in \Sigma_s$. But what if you didn't expand any rules that are irreducible: any $T \to t$ such that for all $u \leqslant t$: $$|u| \lt 2$$ or $$\#_tu \lt 2,$$ where $\#_t u$ is the maximal number of non-overlapping occurences of the substring $u \leqslant t$. This number can be computed by leftmost packing $u$ into $t$.
What we should be left with is a starting rule $S' \to c_0T_{i_1} c_1 \cdots T_{i_n} c_n$ together with the rules $T_i, i=1\dots r$. Call this grammar $G'$.
Then, is it true that $S' \Rightarrow S$, or in other words that $G$ is a smallest grammar of start rule of $G'$ together with including $T_i$ rules from $G'$?
If true, then one possible recursive smallest grammar algorithm, while it still may be difficult to compute, is to first decompose fully into irreducibles and then run the algorithm again on the new alphabet $\Sigma_s \cup \{T_1, \dots, T_r\}$.
Therefore please help me prove or disprove the idea.
Let $\text{I}$ be the complete list of irreducible rules in $G$, and $S'= c_0 I_1 c_1 \cdots I_n c_n$, $ \ I_i \in \text{I}$ be expansion of $G$ up to not expanding any $I \in \text{I}$. Since $G'$ is a grammar for $s$ any grammar derived from $G'$ by taking a smallest grammar of $S'$ is also a grammar for $s$. This means that if $G$ is not in the set of smallest grammars derivable from $G'$ in this manner, then the smallest grammar size derivable $G'$ is strictly greater than $|G|$ (the cost of $G$: sum of lengths of RHS of rules).
But $G$ is in particular derivable from $G'$ by creating new rules and applying them to to grammar substrings (collapsing them into a variable symbol). This is true since we built $G'$ by doing the inverse operation of deleting rules and rule variables into strings.
Therfore, despite this inelegant proof, $G$ is a smallest grammar of $S'$ together with the terminal rules $I_i \in G'$.
Not sure, but we may need to indicate that this smallest grammar algorithm works by creating rules and collapsing into variables.