Lets suppose that there exists a comparison-based algorithm that turns an arbitrary array to a state $A$ in $o(n\log k)$, and there is another comparison-based algorithm that turns an array in state $A$ to completely sorted in $O(n\log (n/k))$. $n$ is the size of the array, while $k$ is another non-constant parameter given. I want to get a contradiction. Here is my reasoning:
Running the first algorithm followed by the second completely sorts the array, and takes
$$o(n\log k)\ +\ O(n\log(n/k)) = o(n\log k\ +\ n\log(n/k)) =$$ $$o(n(\log k\ +\ \log(n/k))) = o(n\log (k \times n/k)) = o(n\log n)$$
which is a contradiction to the known lower bound to comparison-based sorting algorithms.
My question is: Is my reasoning correct? Why or why not? I've tried to prove it using the definitions, but I'm stuck.