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Raphael
Raphael
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I need to show that

S = {(10^p)^m | p >= 0, m >= 0}

$\qquad \displaystyle S = \{(10^p)^m \mid p \geq 0, m \geq 0\}$

is not a regular language using pumping lemma.

Can I multiply the product of the powers and express it to: S= (1^m 0^pm)$S = \{ 1^m 0^{pm} \mid \dots \}$ and apply the pumping lemma where I pump 1's then say that the language doesn't accept the new string?

I need to show that

S = {(10^p)^m | p >= 0, m >= 0}

is not a regular language using pumping lemma.

Can I multiply the product of the powers and express it to: S= (1^m 0^pm) and apply the pumping lemma where I pump 1's then say that the language doesn't accept the new string?

I need to show that

$\qquad \displaystyle S = \{(10^p)^m \mid p \geq 0, m \geq 0\}$

is not a regular language using pumping lemma.

Can I multiply the product of the powers and express it to: $S = \{ 1^m 0^{pm} \mid \dots \}$ and apply the pumping lemma where I pump 1's then say that the language doesn't accept the new string?

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Kaveh
Kaveh
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Post Migrated Here from stackoverflow.com (revisions)
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thokthak
thokthak

Pumping Lemma: is it valid to "multiply the product of powers" in this case?

I need to show that

S = {(10^p)^m | p >= 0, m >= 0}

is not a regular language using pumping lemma.

Can I multiply the product of the powers and express it to: S= (1^m 0^pm) and apply the pumping lemma where I pump 1's then say that the language doesn't accept the new string?