By hypothesis, is not a code. Then there exists an equation over with a nontrivial solution: it is not restrictive to suppose that . Since however the factorization over is unique, the such that is the same as the such that : then with , so is not injective.
Now suppose, for the sake of contradiction, that exists. Let . Then any equation
can be rewritten as
with , : this is an equation over , and as such, has only the trivial solution , , for all . This implies that (1) only has trivial solutions over : by the characterization of free submonoids, is a code. However, because no element of “starts with ”, and is a proper subset of because the former does not contain and the latter does. Then is a free submonoid of which contains and is properly contained in , against the definition of as the free hull of .
- 1 M. Lothaire. Combinatorics on words. Cambridge University Press 1997.
|Date of creation||2013-03-22 18:21:43|
|Last modified on||2013-03-22 18:21:43|
|Last modified by||Ziosilvio (18733)|
|Defines||has only trivial solutions over|