Why Does the Pumping Lemma Require |xy| ≤ p?

[colt]

It says that |xy| < p. But I don't understand why even after reading the proof. If I have a four state DFA, whose last state is the one that is going to repeat for a given input string of length p, |x| is already going to be four, since it represents the states necessary to reach the repetition state. So in this worst case scenery, |x| is already equal to p, so with |y|>0 we already have |xy|>p. SO what's wrong with my line of thought

 

[Robert1986]

If I understand correctly, the pumping lemma gives you a number, p, such that a bunch of stuff happens. Can you reference the exact pumping lemma you are talking about?

 

[verty]

And show us the proof so we can understand.

 

[colt]

I am sorry, I forgot that there is more than one pumping lemma so I aborted the copy that I was doing. Very clever from my part. Anyway I attached a screenshot of the proof:

 

[CellCoree]

?

The third condition of the pumping lemma states that for any regular language L, there exists a constant p such that for any string w in L with length greater than or equal to p, there exists a partition of w into three substrings x, y, and z, where |xy| ≤ p, |y| > 0, and xy^kz is also in L for all k ≥ 0.

Your line of thought is incorrect because it assumes that |x| is always equal to the number of states in the DFA, which is not necessarily true. The number of states in the DFA may vary depending on the input string, and therefore, the length of x may also vary. In fact, the length of x can be much smaller than the number of states in the DFA.

The pumping lemma is a proof technique used to show that a language is not regular. It is based on the idea that if a language is regular, then it can be recognized by a finite automaton (such as a DFA). The pumping lemma conditions ensure that if a language is regular, it must satisfy certain properties that can be exploited to show that it is not regular.

In the third condition, the length of x is limited to p, which means that even if it is smaller than the number of states in the DFA, it cannot be larger than p. This allows for the creation of a cycle within the DFA that can be "pumped" to generate an infinite number of strings that are still in the language. This is the key to proving that a language is not regular.

In summary, the third condition of the pumping lemma is necessary to ensure that the language can be recognized by a finite automaton and to show that it is not regular. Your line of thought, while logical, does not take into account the varying lengths of x and therefore does not accurately apply to the pumping lemma.