# How is $|xy^{2}z| < 2^{p+1}$ (Pumping Lemma application)

In the Question here it is said that

$$|xy^2z|<2^{p+1}$$

Considering that $$|x| = 0$$ and $$|z| = 0$$, y consists of $$2^{p}$$. It's probably trivial, but how do I see, that $$|xy^2z| < 2^{p+1}$$?

• It can't be the case that both $|x| = 0$ and $|z| = 0$. By property b) you have $|xy| \leq p$, so if $|x|, |z| = 0$ then $|xyz| = |y| \leq |xy| \leq p < 2^p = |a^{2^p}|$ ↯ Commented Feb 12 at 8:38
• But if $|y| = p$, $|xy| <= p$ would be satisfied, no? Commented Feb 12 at 14:01
• Okay I get it, $|y|$ can't consist of $2^p$ because of $|xy| <= p$. Commented Feb 12 at 14:03

In the question you've linked, we have $$|xy| = |x| + |y| \leq p$$ (property b) so $$|y| \leq p$$ since $$|x| \geq 0$$. Therefore $$|xy^2z| = |xyz| + |y| = 2^p + |y| \leq 2^p + p < 2^{p + 1}.$$
• what if I pump up i? $|xyz| + |y| + |y| + ... \geq 2^p+p$ EDIT: Then it's the same game again, it's bigger than $2^{p+1}$ but smaller than $2^{p+2}$ etc... Commented Feb 12 at 14:09