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

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  • $\begingroup$ 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 $|xyz| = |y| \leq |xy| \leq p < 2^p = |a^{2^p}|$ ↯ $\endgroup$
    – Knogger
    Feb 12 at 8:38
  • $\begingroup$ But if $|y| = p$, $|xy| <= p$ would be satisfied, no? $\endgroup$
    – Robert
    Feb 12 at 14:01
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    $\begingroup$ Okay I get it, $|y|$ can't consist of $2^p$ because of $|xy| <= p$. $\endgroup$
    – Robert
    Feb 12 at 14:03

1 Answer 1

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In the question you've linked, we have $|xy| = |x| + |y| \leq p$ (property b) so $|y| \leq p$ since $|x| \geq 0$. Therefore $$|xy^2z| = |xyz| + |y| = 2^p + |y| \leq 2^p + p < 2^{p + 1}.$$

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  • $\begingroup$ 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... $\endgroup$
    – Robert
    Feb 12 at 14:09

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