Question

Let alphabet Σ = {a, b, c}, and consider L1 = {w ∈ Σ ∗ | more than half the symbols in w are c’s}. Prove that L1 is not FS using the pumping lemma.

Answer 1

The language is . To prove that the language is not FS, use the pumping lemma as follows.

L1 = {we {a,b,c}* more than half the symbols in w are c}

Let be the pumping length. Then consider the word
w = qPP+1
. In this word, more than half the symbols are 'c', hence
w = apcP+1E L1
.

Now consider any breakup of the word such that:

• $\lvert xy\rvert \leq p$
• $\lvert y\rvert \geq 1$.

Then as the first letter in are 'a's, the entire string can only consist of 'a's. Hence $y = a^k, k \geq 1$ .

Now if the language were regular, then for each $i \geq 0$ , it must be that $xy^iz \in L1$ .

Consider . Then , as an additional will contribute additional 'a's to the word.

But in , there are c's, while the length of the word is $2p+k+1 \geq 2p+2$ as $k \geq 1$ .

This means that at most half of the symbols are 'c's, thus violating the property for the language L1, hence $w' \notin L1$ .

This contradics the pumping lemma, hence proving that L1 is not FS.

Comment in case of any doubts.