Final answer:
To prove that these languages are not regular, we can use the pumping lemma. Using the pumping lemma, we can demonstrate that the languages a. m, n >0, b.{w w € {0,1}' is not a palindrome}, and c.{w w w | w,t e {0,1}*} are not regular.
Step-by-step explanation:
a. m, n >0
To prove that this language is not regular, we can use the pumping lemma. The pumping lemma states that if a language L is regular, then there exists a constant p (the pumping length) such that any string s in L with length greater than or equal to p can be divided into three parts, s = xyz, where y is non-empty and |xy| ≤ p, such that for any non-negative integer i, the string xy^iz is also in L.
Let's assume that the language a is regular. Then using the pumping lemma, we can choose a string s = 0^p1010. Since the length of s is greater than p, we can divide it into xyz such that |xy| ≤ p and |y| > 0. Considering all possible cases for the division, we will find that pumping up the string s will eventually create a string that is not in the language a, which contradicts the assumption that a is regular. Therefore, the language a is not regular.
b. {w w € {0,1}' is not a palindrome}
This language consists of all strings that are not palindromes. To prove that this language is not regular, we can again use the pumping lemma. Let's assume that the language b is regular. We can choose a string s = 0^p1^p, where p is the pumping length. Since s has a length greater than p, we can divide it into xyz such that |xy| ≤ p and |y| > 0. Pumping up the string s will eventually create a string that is a palindrome, which contradicts the assumption that b is regular. Therefore, the language b is not regular.
c. {w w w | w,t e {0,1}*}
Let's assume that the language c is regular. We can choose a string s = 0^p1^p0^p, where p is the pumping length. Since s has a length greater than p, we can divide it into xyz such that |xy| ≤ p and |y| > 0. Pumping up the string s will eventually create a string that does not have three equal parts, which contradicts the assumption that c is regular. Therefore, the language c is not regular.