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 = 0t where 1 <= t <= n.
Which gives xyz = 0(n-t) 0t 1n= 0n 1n which is in L.
But xz = 0(n-t) 1n is not in L. Contradiction.
Did I apply it correct?
嗯......你几乎！那里。就在最后一条语句中，您没有将字符串抽到w = xyz.y
现在我们首先假设 L 是规则的 whereL = { w | w in {0,1}* and w has equal number of 0s and 1s }然后我们将继续证明对于任何泵送字符串i >= 0，即w = xy i z不包含相等数量的 0 和 1（本身是矛盾的）因此，语言不规则：
L 由 给出：
L = {0 n 1 n | n >= 0}
如果 y = 0 t => w = 0 n-t 0 t 1 n
现在在为 i >= 0 抽 y 之后，我们得到
xy我z = 0 n-t 0它1 n
-> xy i z = 0 n+(i-1)t 1 n
现在因为 n+(i-1)t 不等于 n 这与我们的假设相矛盾， L = { w | w in {0,1}* and w has equal number of 0s and 1s }因此xy i z不属于L
注意-您还需要考虑其他情况，例如y = 0 t 1 1、 y = 1 t等，稍后证明这些确实意味着矛盾。