We have a string $s$ containing $n \leq 100$ bits. The move we can make on it is erasing from $s$ some substring $x$, but only if $x$ is directly preceded by $x^R$, where $x^R$ means string $x$ reversed. In other words, we're choosing some even-length palindrome and erasing it's "right half".
The goal is to minimize the length of the word left when no further can moves be done. In example, from a bit word $100110$ we can erase $01$ from the middle, and be left with $1010$ where no moves can be done. However, starting differently, we can reduce the word to a single $10$, which is optimal. I want to compute this minimal length for a given string, for $100110$ it would be $2$.
Now, I have a dynamic-programming algorithm which I know produces right answers (it has been tested a lot), but don't know how to prove it. It computes the best result for every substring of $s$ in increasing length. For those of length $k$, the answer is surely at most $k$. Also, it's obvious that the only strings in which no moves can be done are of form $1010...$ or $0101...$. For some word we're playing on, as it's first bit can never be erased, we know the exact form of a word that will be left from it, the only thing that we lack is it's length. Now, I assumed that for a word $t$, there exist some words $x$, $y$ that $t = xy$, and if we do the optimal operations just on $x$, producing a word $x'$, then just on $y$ producing $y'$, and in the end on the word $x' y'$, we will acquire the optimal result for $t$. The last step is very easy, because we know how words $x'$ and $y'$ look like only knowing it's lengths, so there are only a couple of cases to consider. The complexity of this process is $O(n^3)$ as it spends linear time on each substring of $s$.
If $DP[i][j]$ is the answer for a substring $[i..j]$, the complete recurrence is as follows:
$$DP[i][j] = \min \{DP[i][k] + F(i,k,j) \cdot DP[k+1][j] : i \leq k < j \}.$$
where
$$ F(i,k,j) = \left\{ \begin{array}{ll} 1 & \textrm{for $s[i] = s[k+1]$ and $DP[i][k] = 1$}\\ 1 - DP[i][k] \mod 2 & \textrm{for $s[i] = s[k+1]$ and $DP[i][k] > 1$}\\ DP[i][k] \mod 2 & \textrm{otherwise} \end{array} \right. $$
what corresponds to the fact that in some cases $y'$ can be erased using $x'$ (possibly in multiple moves) - in those cases $F(i,k,j)$ is $0$, and it's $1$ otherwise. Also note, that if $s[i] = s[i+1]$ then when dealing with $DP[i][j]$ for $j > i$ at first we initialize $DP[i][j] = DP[i+1][j]$.
For me, the assumption that $t$ can be split into $x$ and $y$ in a such a way that the mentioned procedure is optimal, is non-trivial. I'm asking for a proof that this assumption can be made, or some different solutions for this problem, either would be nice.
