word means $n$-divisible in tail sense.

c\) A word $W$ is *$n$-cancellable* if either it is $n$-divisible in ordinary sense or there exists a word of the form $u^{d}\subseteq W$..

Now we describe a sufficient condition for $n$-cancellability and its connection with $n$-divisibility.

\[c:lem1.3\] If a word $W$ includes $n$ identical disjoint subwords $u$ of length $n\cdot{d}$ then $W$ is $n$-cancellable.

Suppose the contrary. Consider the tails $u_1,u_2,\ldots,u_n$ of the word $u$ which begin from each of the first $n$ letters of $u$. Renumerate the tails to provide the inequalities $u_1\succ\ldots\succ u_n.$ By lemma 1 the tails are incomparable. Consider the subword $u_1$ in the left-most copy of $u,$ the subword $u_2$ in the second copy from the left, $\ldots, u_n$ in the $n$th copy from the left. We get an $n$-division of $W$. A contradiction.

\[c:lem1.4\] If a word $W$ is $4nd$-divisible then it is $n$-cancellable.

Suppose the contrary. Consider the numbers of positions of letters $a_i$, $a_1<a_2<\ldots<a_{4nd}$, that begin the tails $u_i$ dividing $W$. Set $a_{4nd+1} = |W|$. If $W$ is not $n$-cancellable then there exists $i$, $1\leqslant i\leqslant 4(n-1)d + 1$, such that for any $i\leqslant b<c\leqslant d<e\leqslant i+4d$ the $(a_c - a_b)$-tail $u_b$ is incomparable with 