Splitting strings in pumping lemma for regular language

I was recently reading the book Introduction to the Theory of Computation, Second Edition by Michael Sipser, and encountered the following example:

Let $$F=\{ww\ |\ w\in \{0, 1\}^*\}$$. We show that $$F$$ is nonregular, using the pumping lemma.

Assume to the contrary that $$F$$ is regular. Let $$p$$ be the pumping length given by the pumping lemma. Let $$s$$ be the string $$0^p10^p1$$. Because $$s$$ is a member of $$F$$ and $$s$$ has length more than $$p$$, the pumping lemma guarantees that $$s$$ can be split into three pieces, $$s = xyz$$, satisfying the three conditions of the lemma.

The lemma contains the condition "$$|xy|\leq p$$", so I thought the string $$s$$ can be split into

$$x=\epsilon, y=0^p,z=10^p1$$

which does not satisfy the condition of the lemma, since $$xy^2z$$ ($$0^{2p}10^p1$$) is not in $$F$$. Also tried splitting the string into

$$x=0^{p-n},y=0^n,z=10^p1$$

where $$n$$ is an integer such that $$0\leq n\leq p$$. This also does not satisfy the lemma.

I would like to know where I am making a mistake.

Edit:

I was in a hurry and forgot the fact that one has to prove that $$s$$ is not in $$F$$. The following paragraph of the quoted paragraph above mentions that there cannot be a split that satisfies the condition of the lemma, thus proving that $$F$$ is non-regular.

For ones who are interested, this is the entire example given by Michael Sipser:

Let $$F=\{ww\ |\ w\in \{0, 1\}^*\}$$. We show that $$F$$ is nonregular, using the pumping lemma.

Assume to the contrary that $$F$$ is regular. Let $$p$$ be the pumping length given by the pumping lemma. Let $$s$$ be the string $$0^p10^p1$$. Because $$s$$ is a member of $$F$$ and $$s$$ has length more than $$p$$, the pumping lemma guarantees that $$s$$ can be split into three pieces, $$s = xyz$$, satisfying the three conditions of the lemma.

Condition 3 ($$|xy|\leq p$$) is once again crucial, because without it we could pump $$s$$ if we let $$x$$ and $$z$$ be the empty string. With condition 3 the proof follows because $$y$$ must consist only of 0s, so $$xyyz\notin F$$

Observe that we chose $$s=0^p10^p1$$ to be a string that exhibits the "essence" of the nonregularity of $$F$$, as opposed to, say, the string $$0^p0^p$$. Even though $$0^p0^p$$ is a member of $$F$$, it fails to demonstrate a contradiction because it can be pumped.

• To prove the language $F$ is non-regular, the string $xy^2z$ is not supposed to be in $F$. This is how you draw the contradiction and conclude your proof. Commented Jul 7 at 8:16
• Also, think of the pumping lemma as a two player game. Since you choose the string, the opponent splits it. So your argument should be based on a general split (the second case) and not on any particular fixed split (like in the first case). Commented Jul 7 at 8:19
• @codeR I reread the paragraph and understood my mistake. It kinda flew over my head when I first read it. Thanks for the answer. I'll close the question later. Commented Jul 7 at 8:56

When using the pumping lemma to prove that a language is not regular, you cannot just find ONE decomposition $$xyz$$ that does not satisfy the three properties: you must prove that ANY decomposition does not satisfy the three properties.