Regular Pumping Examples

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<ul><li><p>More Applications </p><p>of</p><p>the Pumping Lemma</p><p>Costas Busch - RPI</p></li><li><p>The Pumping Lemma: Given a infinite regular language </p><p> there exists an integer (critical length)</p><p> for any string with length </p><p> we can write</p><p> with and</p><p> such that:</p><p>Costas Busch - RPI</p></li><li><p>Regular languagesNon-regular languages</p><p>Costas Busch - RPI</p></li><li><p>Theorem:The languageis not regularProof:Use the Pumping Lemma</p><p>Costas Busch - RPI</p></li><li><p>Assume for contradictionthat is a regular languageSince is infinitewe can apply the Pumping Lemma </p><p>Costas Busch - RPI</p></li><li><p> We pickLet be the critical length forPick a string such that: lengthand</p><p>Costas Busch - RPI</p></li><li><p>we can write:with lengths:From the Pumping Lemma: Thus:</p><p>Costas Busch - RPI</p></li><li><p>From the Pumping Lemma:Thus:</p><p>Costas Busch - RPI</p></li><li><p>From the Pumping Lemma:Thus:</p><p>Costas Busch - RPI</p></li><li><p>BUT:CONTRADICTION!!!</p><p>Costas Busch - RPI</p></li><li><p>Our assumption thatis a regular language is not trueConclusion:is not a regular languageTherefore:END OF PROOF</p><p>Costas Busch - RPI</p></li><li><p>Regular languagesNon-regular languages</p><p>Costas Busch - RPI</p></li><li><p>Theorem:The languageis not regularProof:Use the Pumping Lemma</p><p>Costas Busch - RPI</p></li><li><p>Assume for contradictionthat is a regular languageSince is infinitewe can apply the Pumping Lemma </p><p>Costas Busch - RPI</p></li><li><p>We pickLet be the critical length of Pick a string such that: lengthand</p><p>Costas Busch - RPI</p></li><li><p>We can writeWith lengthsFrom the Pumping Lemma: </p><p>Thus:</p><p>Costas Busch - RPI</p></li><li><p>From the Pumping Lemma:Thus:</p><p>Costas Busch - RPI</p></li><li><p>From the Pumping Lemma: Thus:</p><p>Costas Busch - RPI</p></li><li><p>BUT:CONTRADICTION!!!</p><p>Costas Busch - RPI</p></li><li><p>Our assumption thatis a regular language is not trueConclusion:is not a regular languageTherefore:END OF PROOF</p><p>Costas Busch - RPI</p></li><li><p>Regular languagesNon-regular languages</p><p>Costas Busch - RPI</p></li><li><p>Theorem:The languageis not regularProof:Use the Pumping Lemma</p><p>Costas Busch - RPI</p></li><li><p>Assume for contradictionthat is a regular languageSince is infinitewe can apply the Pumping Lemma </p><p>Costas Busch - RPI</p></li><li><p>We pickLet be the critical length ofPick a string such that: length</p><p>Costas Busch - RPI</p></li><li><p>We can writeWith lengthsFrom the Pumping Lemma: </p><p>Thus:</p><p>Costas Busch - RPI</p></li><li><p>From the Pumping Lemma:Thus:</p><p>Costas Busch - RPI</p></li><li><p>From the Pumping Lemma: Thus:</p><p>Costas Busch - RPI</p></li><li><p>Since:There must exist such that: </p><p>Costas Busch - RPI</p></li><li><p>However:forfor any </p><p>Costas Busch - RPI</p></li><li><p>BUT:CONTRADICTION!!!</p><p>Costas Busch - RPI</p></li><li><p>Our assumption thatis a regular language is not trueConclusion:is not a regular languageTherefore:END OF PROOF</p><p>Costas Busch - RPI</p></li></ul>