# How do I prove that a language is deletion closed?

For example, how could I prove that the following language is deletion closed:

{$a^k$$b^j$ : $j$, $k$ $\geqslant$ 0}

The reason seems obvious to me, I just can't see a way to prove it.

• If you can't prove it it's not obvious. What have you tried and where did you get stuck? (What is "deletion closed"?) – Raphael Nov 21 '15 at 20:02
• I encourage you to edit the question to include the definition of "deletion closed" in the question. – D.W. Nov 21 '15 at 22:29

You'd prove that a language $L$ is deletion-closed as follows.
Consider a word $w\in L$ and let $w = w_1\dots w_\ell$. Now, for any $r$ and $s$ with $1\leq r\leq s\leq \ell$, consider the word $w' = w_1\dots w_{r-1}w_{s+1}\dots w_\ell$, i.e., the word that results from deleting the substring $w_r\dots w_s$ from $w$. We have $w'\in L$ because [some argument that will be specific to the language $L$ you're working with].