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Mar 24, 2017 at 20:34 vote accept Alexander
Mar 24, 2017 at 19:10 comment added GoodDeeds @Alexander Yes, because $|y|\gt0$ would not be satisfied. If this is the case, we are done. Else, assume $k\gt0$, and the rest of the argument holds
Mar 24, 2017 at 19:00 comment added Alexander Thanks for the clarification regarding the random choosing of $x$, $y$ and $z$. If $k$ denotes the number of $a$s in $y$, then $y$ would have to be empty for $k = 0$ in the initial word, right? That would violate the restriction that $y$ must not be empty. Or am I wrong?
Mar 24, 2017 at 14:23 history answered GoodDeeds CC BY-SA 3.0