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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?

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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.

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