In the answer to this question, I'm not understanding how the string is derived for a given $l$.For example,Case 1: $vx = a^i$ where $i > 0$. Choose $l = 2$ to get $a^{n+i} b^{n+1} c^{n+1} d^n \notin L$.Why is $l = 2$ chosen and how is $a^{n+i}b^{n+1}c^{n+1}d^n$ derived from $l = 2$?Also, how can $vx$ be chosen instead of $vwx$ as the OP chose? What do we do about $w$? Is it the empty string?
Questions about an answer to a pumping lemma question for CFLs
context free;pumping lemma
The essential idea is that pumping lemma tells you about string $uv^lwx^ly$ with $l \geq 0$. That is, you can pump to $uwy$, if $l = 0$, that way shortening the initial string.The answer considers the string $a^n b^{n+1} c^{n+1} d^n$. Removing a single $b$ or inserting a single $a$ would move the string out of the language. Removal and insertion correspond to $l = 0$ and $l = 2$.The answer only considers $vx$, because $w$ does not matter - it will not be pumped.If $v$ or $x$ contains any $a$s, double them to get more $a$s then $c$s. If $v$ or $x$ contains any $b$s, remove them to get fewer $b$s than $d$s. $v$ and $x$ will never contain both $a$s and $c$s, because $|vwx| \leq n$. Same is true about the other pair of symbols.