Problem 1092

This property is known as a "confluence" property. To demonstrate it, we will consider a sequence of operations $p_{1}, \cdots p_{n}$ that cannot be extended, and a sequence (possibly infinite) $q_{1}, q_{2}, \cdots$ of operations. We will show that we can modify the first sequence of operations without changing its result, so that $q_{1}=p_{1}$. By induction, we will be done: the sequence of operations $\left(q_{k}\right)$ will be finite and will yield the same result.

If we draw a bar between an $a$ on the left and a $b$ on the right of the transformed word by the operation $q_{1}$, then as long as this bar is not used, that is, at the center of a word $a^{k} b^{\ell}$ that we transform, we leave it and the letters surrounding it remain unchanged. Since after $p_{n}$ no more operations can be performed, there exists $k$ such that $p_{k}$ uses this bar.

Thus, we can perform $p_{k}$, then $p_{1}, p_{2}, \cdots, p_{k-1}, p_{k+1}, \cdots p_{n}$, which gives the same result. Hence the result.