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Prove that the following language holds the pumping lemma for context-free languages: (Although it is not context-free)

L is a language under alphabet {a,b,c,d}

L={$a^ib^ic^j$ : i,j $\ge$ 0}∪{$a^kb^kd^{2n}c^k$ : k,n $\ge$ 0}

Point out what is the smallest pumping length m to which the pumping lemma holds.

I don't quite understand why the minimal pumping length is 2, the only one character word in L is c and the pumping lemma holds for it.

and there's no other one character word in L to my understanding. so shouldn't the minimal pumping lemma be 1?

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  • $\begingroup$ The problem is not well-stated. If you mean "Prove that the following lemma fulfillss the conditions of the pumping lemma" then this cannot be done. Int first component, for $j=0$, any pumping in $a^ib^i$ will lead out of the language. The same in the second component for $n=0$. 2 is actually the smallest length where you can find words that cannot be pumped. For 0 and 1 you cannot. $\endgroup$ – Peter Leupold Feb 20 '18 at 17:25
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tl;dr: try pumping $z=abddc$ with $n=1$.

This is a little tricky. You should remember that the pumping length $n$ is used twice in the lemma. First, it is used to claim that any word such that $|z|\ge n$ can be pumped. This is not relevant here; indeed, $c$ can be pumped. But it has a second use, limiting the size of the "pumpable" segment, which fails here for $n=1$.

Let's assume $n=1$ and observe the word $z=abddc\in L$. We need to find $z=uvwxy$ such that $|vwx|\le n$ and $uv^iwx^iy\in L$.

Since $n=1$ then $vwx$ must contain of exactly one letter (at least one, since $|vx|>0$ and at most one because $|vwx| \le 1$). If we choose $v=a$ then $i=0$ would yield $bddc\notin L$ and similarly we can't have $v=b$ or $v=c$. Choosing $v=d$ seems the obvious choice, but then for $i=0$ we'll end up with $abdc\notin L$ becuase $d$ should appear an even number of times. So you need at least $n\ge 2$ in order to be able to pump an even number of d's.

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