how can I prove that if I have a regular language L
and I create a new language L'
where L' = (L but with last letter repeated)
(i.e. if ab is in language L then abb is in language L')
that L' is a regular language
I tried solving the question but I have no way of being sure of
my solutions so I'll post it here:
L is a regular language therefor there is a regular expression for it,
we create a new language x where x =L*($\Sigma$),
$\Sigma$ is all the letters in language L
,x is a regular language because it is regular language chained with a letter
,and letters are a regular expression
the language L' is a subset of x,
there for x can be written as x=L' $\cup$ t (t=x/L'),
and therefor L' has to be a regular expression
because if it wasn't then x would not be a a regular language
Best Answer
Let $A$ be the alphabet. For each letter $a \in A$, let $$ La^{-1} = \{u \in A^* \mid ua \in L\} $$ It is a well known fact that if $L$ is regular, then so is $La^{-1}$. Now your language $L'$ can be written as $$ L' = \bigcup_{a \in A}\ (La^{-1})aa $$ For each $a \in A$, $(La^{-1})aa$ is the product of the two regular languages $La^{-1}$ and $aa$ and this is regular. It follows that $L'$ is a finite union of regular languages and this is regular.