In the example of language D, why does the pumping property not hold for the string S = 0^P 1^P 0^P 1^P?
In the example of language D, the pumping property does not hold for the string S = 0^P 1^P 0^P 1^P. To understand why, we need to examine the properties of context-sensitive languages and the pumping lemma for context-free languages.
Context-sensitive languages are a class of formal languages that can be described by context-sensitive grammars. These languages are more expressive than context-free languages, as they allow for rules that take into account the context in which symbols appear. The pumping lemma for context-free languages is a powerful tool used to prove that a language is not context-free.
The pumping lemma states that for any context-free language L, there exists a constant p (the pumping length) such that any string s in L with a length of at least p can be divided into five parts, s = uvwxy, satisfying the following conditions:
1. |vwx| ≤ p: The length of the substring vwx is at most p.
2. |vx| ≥ 1: The substring vx has a length of at least 1.
3. For all i ≥ 0, the string uv^iwx^iy is also in L.
In the case of language D and the string S = 0^P 1^P 0^P 1^P, let's assume that the pumping property holds. This means that for any decomposition s = uvwxy satisfying the conditions of the pumping lemma, the resulting strings uv^iwx^iy for i ≥ 0 should also be in language D.
Let's consider a possible decomposition of S: u = 0^a, v = 0^b, w = 0^c, x = 0^d, y = 1^P 0^P 1^P. Here, a, b, c, and d are non-negative integers, and a + b + c + d ≤ P.
According to the pumping lemma, we have three cases to consider:
1. If vwx contains only 0's: In this case, pumping up or down by changing the value of v and/or x would result in an imbalance of the number of 0's and 1's, violating the condition that the resulting string should be in language D.
2. If vwx contains both 0's and 1's: Pumping up or down by changing the value of v and/or x would break the order of the 0's and 1's in the string, again violating the condition that the resulting string should be in language D.
3. If vwx contains only 1's: In this case, pumping up or down by changing the value of v and/or x would result in an imbalance of the number of 0's and 1's, violating the condition that the resulting string should be in language D.
In all three cases, it is clear that pumping the string S = 0^P 1^P 0^P 1^P would lead to a violation of the conditions for language D. Therefore, the pumping property does not hold for this string.
To summarize, the string S = 0^P 1^P 0^P 1^P does not satisfy the conditions of the pumping lemma for context-free languages. This is because pumping the string would result in an imbalance of the number of 0's and 1's, or break the order of the symbols, violating the properties of language D.
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