Did I apply pumping lemma correctly?

Question

L = { w | w in {0,1}* and w has equal number of 0s and 1s }

Let n be the number of the pumping lemma.

I pick s = 0ⁿ 1ⁿ and y = 0^t where 1 <= t <= n.

Which gives xyz = 0^(n-t) 0^t 1ⁿ= 0ⁿ 1ⁿ which is in L.

But xz = 0^(n-t) 1ⁿ is not in L. Contradiction.

Did I apply it correct?

Answer 1

Hmmm ... You were almost ! there. Just in the last statement you are not pumping the string w = xyz at y.

Now we start by assuming that L is regular where L = { w | w in {0,1}* and w has equal number of 0s and 1s } and then we will go on to prove that for any i >= 0 the pumped string i.e w = xyⁱz does not contain the equal number of 0s and 1s ( a contradiction per se) therefore, the language is not regular :

L is given by :

L = {0ⁿ1ⁿ | n >= 0}

Iff y = 0^t => w = 0^n-t0^t1ⁿ

Now after pumping y for i >= 0 we get

xyⁱz = 0^n-t0^it1ⁿ

-> xyⁱz = 0^n+(i-1)t1ⁿ

Now since n+(i-1)t is not equal to n this contradicts our assumption that L = { w | w in {0,1}* and w has equal number of 0s and 1s } therefore xyⁱz does not belong to L

NOTE- You also need to consider other cases like y = 0^t1¹ , y = 1^t etc and later on prove that these do imply a contradiction.