Pumping Lemma for CFL a^n b^m c^o for n

Question

Let be: L={aⁿ b^m c^o | n < m < o, n natural}
Using Pumping Lemma I have choosen: z = uvwxy = aⁿ bⁿ⁺¹ cⁿ⁺²
|uv|<=n and |v|>0
=> uv²wx²y
If vwx is of a's and / or b's it is okay and we would have more a's and/or b's than c's - but if vwx contains only c's it would be element of L.
As far as I understood all new words have to be not element of L to show that it is not a CFL. How would I do this?

Answer 1

If we have a mix of a's & b's use uv²wx²y
If we have a mix of b's & c's use uv⁰wx⁰y

Now all words that are created by pumping z are not element of L.