Pumping lemma for Context-Free Languages
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 $a^n b^{n+1} c^{n+1} d^{n+i} \notin L$.