I have a few questions about Pumping Lemma Contrapositive. First of all, how do I choose pumping length $n$? Is it just any constant from the language definition? i.e. I have $L=\{a^kb^gc^hd^j\}$ so I can choose any constant from set $k, g, h ,j$ since those are 'pumping' a length of word?
Secondly, lets say I have language defined as follows:
$L=\{a^ib^jc^k : i,j,k >0 \land i+k\leq j\}$
What I understand, I have to make word definition which length will be greater or equal to chosen parameter $n$ so: $z=a^nb^{n+k}c^k$
Then I have to do a split into $uvw$ which satisfies conditions: $|uv|\leq n$ and $|v|\geq 1$ so:
$uv=a^n$
$u=a^s \land v=a^{n-s}$ so my word looks as follows $a^sa^{n-s}b^{n+k}c^k$ and I have to find such $i$ so word $uv^iw$ is not in the language. For $i = 2$ I have equation:
$s+2n-2s+k \leq n+k$
$2n-s+k \leq n+k$ and since $s\lt n$ then this word will not be in the language -> L is not regular.
Is it that easy? Or is there something I misunderstood ?