Let $\Sigma$ be an alphabet and $L\subseteq\Sigma^*$. We define $\verb+lmult+(L)=\left\{x^iu\;|\;x\in\Sigma,u\in\Sigma^*,i>0,xu\in L\right\}\cup\{\epsilon\}.$ [...]
Show the following: Iff for all $x\in\Sigma,u\in\Sigma^*$ the implication $xu\in L\Rightarrow u\in L$ is true, then 1 is a regular pumping lemma number for $\verb+lmult+(L)$.
My thoughts so far:
Assume that the implication is valid for all combinations of $x\in\Sigma,u\in\Sigma^*$ then we can pick an arbitrary $y\in \verb+lmult+(L)$ which can be split up into $y=vxu$. Under the condition of the pumping lemma we know that $x\neq\epsilon$ and $|vx|\leq1$ implies $v\overset{!}{=}\epsilon$. We now have to proove that $\forall i:x^iu\in\verb+lmult+(L)$.
Here is the point I am not really sure about which needs some "polishing":
By definition of $\verb+lmult+(L)$ we know that $x^iu\in\verb+lmult+(L)$ (and therefore that $\verb+lmult+(L)$ has the regular pumping lemma number 1) iff $xu\in L$. We know nothing about $u\in\Sigma^*$ but neither we have any restrictions for that. The next step is to decompose $y=x^iu$ so that we can check whether $xu\in L$. With the very first implication we can write $x^iu=x^{i-1}xu\in\verb+lmult+(L)\Rightarrow x^{i-1}u\in\verb+lmult+(L)\Rightarrow\ldots\Rightarrow xu\in\verb+lmult+(L)\Rightarrow xu\in L.$
I would like to know whether my approach is correct and how you would solve/rewrite it.