If one of the case obeys all rules of Pumping Lemma, can we conclude there is no contradiction?
This is a common question regarding pumping lemmas, and it is completely answered by carefully understanding the quantifiers in the lemma.
The lemma itself says the following:

For every CFL $L$, there exists a constant $p$ such that for every word $w\in L$, if $|w|&gt;p$, then there exist words $u,v,x,y,z$ such that $w=uvxyz$, and (1) $|vxy|\le p$, (2) $|vy|\ge 1$, (3) for every $i\ge 0$, $uv^ixy^iz\in L$.

Now, the typical usage of the lemma is by contradiction, so we have a language $L$ which we suspect to be not a CFL, and we show the negation of the lemma (read this carefully, and see how we negate the formulation of the lemma):

For every constant $p$, there exists a word $w\in L$ with $|w|&gt;p$, such that for every decomposition $w=uvxyz$, if (1) and (2) hold, then there exists $i\ge 0$ such that $uv^ixy^iz\notin L$.

As you can see, in order to apply the lemma, we can choose the word $w$, but we are given $u,v,x,y,z$. Then, we assume that cases (1) and (2) hold, and deduce that (3) doesn't hold.
It is really important to understand which elements we can choose, and which we need to be given without choice. These correspond to existential/universal quantification.