Before the proof starts note that $$ c$$ has $$ n+m$$ elements, and then $$ b_j$$ refers to the element at position $$ n+j$$ with respect to $$ c$$

We know that $$ b_j<<!--\; <\; -->a_i$$ because they form an inversion, we also know that $$ a$$ is sorted in ascending order, therefore we know that $$ b_j<<!--\; <\; -->a_i<<!--\; <\; -->a_{i+1}<<!--\; <\; -->\dots <!--\; \dots \; --><<!--\; <\; -->a_n$$, therefore for each $$ k\in <!--\; \in \; -->[i...n]$$ we know $$ (k,n+j)$$ forms an inversion in $$ c$$ as needed.

Since $$ a_i$$ and $$ b_j$$ do not form an inversion then we know that $$ a_i\le <!--\; \le \; -->b_j$$, since $$ b$$ is sorted in ascending order then we know that $$ a_i\le <!--\; \le \; -->b_j\le <!--\; \le \; -->b_{j+1}\le <!--\; \le \; -->\dots <!--\; \dots \; -->\le <!--\; \le \; -->b_m$$, and thus for every $$ k\in <!--\; \in \; -->[j\dots <!--\; \dots \; -->m]$$ we know $$ a_i$$ and $$ b_k$$ do not form an inversion.