Every nonregular $k$-part partition has a refinement with energy gain greater than ${ϵ}^5$ and at most $k{2}^{k+1}$ parts

For each ordered irregular pair $(X,Y)$ choose witness sets $A_{XY}\subseteq X$ and $B_{XY}\subseteq Y$.

For each $X\in \mathcal{P}$, refine $X$ by all witness subsets that occur in $X$, whether as a first or a second coordinate. There are at most $2k$ such subsets, and the common refinement they generate has at most $2^{2k}$ cells. By choosing one witness orientation for each unordered pair and retaining the two diagonal witnesses separately, the same construction uses at most $k+1$ subsets per part and hence at most $2^{k+1}$ cells. Thus the resulting partition $\mathcal{Q}$ has at most $k{2}^{k+1}$ parts.

For every selected irregular ordered pair, the restriction of $\mathcal{Q}$ to its two old parts refines the witness split. By [L2] and [L3], its contribution gains more than ${ϵ}^4\mathrm{\mid }X\mathrm{\mid }\mathrm{\mid }Y\mathrm{\mid }\mathrm{/}{n}^2$; all other old-pair contributions are nondecreasing.

Although only one orientation of each off-diagonal irregular pair supplied witness sets in step 2.1, that single split refines both old parts, so step 3.1 applies to both ordered pairs $(X,Y)$ and $(Y,X)$, which carry equal weight $\mathrm{\mid }X\mathrm{\mid }\mathrm{\mid }Y\mathrm{\mid }\mathrm{/}{n}^2$ and equal regularity status. Summing the gains over all ordered irregular pairs therefore recovers the full normalized irregular weight, and by [L1] the total gain is greater than ${ϵ}^4ϵ={ϵ}^5$.