1.46 Prove that the following languages are not regular. You may use the pumping lemma and the closure of the class of regular languages under union, intersection, and complement. a. {0^n1^m0^n | m, n ? 0} b. {0^m1^n | m ? n}
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The pumping lemma is a tool used to prove that a language is not regular. It states that for any regular language L, there exists a constant K such that any string in L of length at least K can be divided into three parts, u, v, and w, such that v is not empty, Show more…
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Using the Pumping Lemma, show why the following language cannot be a regular language: L = {x ∈ {0,1}* | ∃i ∈ I : x = 10^i10^i1 ∧ i > 0} Example: 10101, 1001001, 100010001, etc.
Madhur L.
Use the Pumping Lemma (with the adversary argument) to prove that the language {ww|w ∈ (0+1+2)*} is not regular.
One important technique used to prove that certain sets are not regular is the pumping lemma. The pumping lemma states that if $M=\left(S, I, f, s_{0}, F\right)$ is a deterministic finite-state automaton and if $x$ is a string in $L(M),$ the language recognized by $M,$ with $l(x) \geq|S|,$ then there are strings $u, v,$ and $w$ in $I^{*}$ such that $x=u v w, l(u v) \leq|S|$ and $l(v) \geq 1,$ and $u v^{i} w \in L(M)$ for $i=0,1,2, \ldots$ Prove the pumping lemma. [Hint: Use the same idea as was used in Example $5 . ]$
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