Pumping lemma for Context-Free Languages I have a question about a specific pumping lemma problem for Context-Free Languages.

Suppose we have the following Language:


  $L = \{a^{i}b^{j}c^{k}d^{l} \mid 0 &lt; i &lt; k \wedge j &gt; l &gt; 0 \}$


Here is my attemp to prove that the language is not context-free:

Assume $L$ is context-free. Let $n&gt;0$ be the pumping length given by the lemma.

Let $z = a^{n}b^{n+1}c^{n+1}d^{n}$, then $z \in L$. 

Than according to the lemma, $z$ can be written as $z = uvwxy$ where the following properties hold:


$|vx| \geq 1$
$|vwx| \leq n$
for every $i \geq 0$, $uv^{i}wx^{i}y \in L$.


We have 6 different possibilities for $vwx$:


$vwx = a^{i}$ where $i \leq n$
$vwx = a^{i}{b^j}$ where $i+j \leq n$
$vwx = b^i$ and $i \leq n$
$vwx = b^{i}c^{j}$ and $i+j \leq n$
$vwx = c^{i}$ with $i \leq n$
$vwx = c^{i}d^{j}$ and $i+j \leq n$ 


Is this right so far? The thing that I'm unsure of is if my different cases for $vwx$ are right. 

How do I choose the pumping length for case 2? If I choose $i$ = 2, what if $i$ is zero ? Then I don't have any contradiction.

Thanks in advance
Let me rephrase property 3 of the pumping lemma: for every $l \geq 0$, $uv^lwx^ly \in L$. Now let's consider the seven different cases:

Case 1: $vx = a^i$ where $i &gt; 0$. Choose $l = 2$ to get $a^{n+i} b^{n+1} c^{n+1} d^n \notin L$.

Case 2: $vx = a^ib^j$ where $i,j &gt; 0$. Like case 1 or case 3.

Case 3: $vx = b^i$ where $i &gt; 0$. Choose $l = 0$ to get $a^n b^{n+1-i} c^{n+1} d^n \notin L$.

Case 4: $vx = b^ic^j$ where $i,j &gt; 0$. Like case 3 or case 5.

Case 5: $vx = c^i$ where $i &gt; 0$. Choose $l = 0$ to get $a^n b^{n+1} c^{n+1-i} d^n \notin L$.

Case 6: $vx = c^id^j$ where $i,j &gt; 0$. Like case 5 or case 7.

Case 7: $vx = d^i$ where $i &gt; 0$. Choose $l = 2$ to get $a^n b^{n+1} c^{n+1} d^{n+i} \notin L$.