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I have a regular language $a(a+b)^*$ to which i applied pumping lemma.

Let the pumping length be $'p'$

and the example string be $$w=a(a+b)^{p-1}$$.
The string satisfies the condition that it is at least length 'p'.

We now divide it into 3 parts $x y z$ with $$x=epsilon ,y= a(a+b)^{p-1} , z= epsilon$$.

This division also satisfies the condition that $|xy|<=p \land |y|\neq epsilon$.

Thus, $$w=xy^1z$$.
since , i can be zero in $y^i$, we pump down to get,

$$w=xy^0z = epsilon$$
which doesn't belong in the given language $a(a+b)^*$.

Thus,now that we have proved that $\exists w\in a(a+b)^*$ such that it can't be pumped.

I have already spent 2 hours trying to figure out what is wrong with this proof as the results doesn't make sense. Any help is appreciated.

Dhruv
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2 Answers2

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This isn't how the Pumping Lemma works. The Lemma states that "if $L$ is regular then for all $w \in L$ there exists factors $w = xyz$ satisfying ...", not "if $L$ is regular then for all $w \in L$ all factors $w = xyz$ satisfy ...". So just because you found one factorization of $w$ that can't be pumped doesn't mean that there is none.

There is a factorisation of $w$ that can be pumped, e.g.

$$x = a, y = (a + b)^{p - 1}, \text{ and } z = \varepsilon.$$

Knogger
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Every regular expression have an automata recognising it and the same holds for the converse too. Link to a previous post. Regular languages are defined as the language that can be recognized by a deterministic automaton.Thus every regular expression fall under the class of regular languages.

Dante
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