Then $v_2=u\cdot v_3$ for some $v_3$. Furthermore $v_1=u^2\cdot v_3$. Arguing in this way we obtain that $v_1=u^{d}\cdot
v_{{d}+1}$ since $|u|<x/ {d}$, $|v_2|\geqslant ({d}-1)x/ {d}$. A contradiction.

\[c:lem1.2\] If a word $V$ of length $k\cdot t$ includes not more than $k$ different subwords of length $k$ then $V$ includes a period of length $t$.

We use induction in $k$. The base $k = 1$ is obvious. If there are not more than $(k - 1)$ different subwords of length $(k - 1)$ then we apply induction assumption. If there exist $k$ different subwords of length $(k - 1)$ then every subword of length $k$ is uniquely determined by its first $(k - 1)$ letters. Thus $V=v^t$ where $v$ is a $k$-tail of $V$.

a\) A word $W$ is *$n$-divisible in ordinary sense* if there exist $u_1, u_2,\ldots,u_n$ such that $W=v\cdot u_1\cdots
u_n$ and $u_1\succ\ldots \succ u_n$.

b\) In our proof we will call a word $W$ *$n$-divisible in tail sense* if there exist tails $u_1,\ldots,u_n$ such that $u_1\succ u_2\succ \ldots \succ u_n$ and for any $i=1, 2,\ldots, n - 1$ the beginning of $u_i$ is to the left from the beginning of $u_{i+1}$. If the contrary is not specified, an *$n$-divisible* 