Let $A,B,C$ be $n\times n$ complex matrices. Denote by $(A \mid B)$ the block matrix derived from matrices $A$ and $B$.
Does the following inequality hold? $$ \operatorname{rank}(A \mid B \mid C) \leq \operatorname{rank}(A \mid B)+\operatorname{rank}(B \mid C)-\operatorname{rank}(B) $$
the rank of a $n\times kn$ matrix is $n$ minus the dimension of the left kernel (called Null). One can thus rewrite the claim as $$Null(A|B) - Null(A|B|C) \le Null(B) - Null(B|C)$$ Notice that if $v$ is in the left kernel of $(A|B|C)$ in particular is also in the left kernel of $(A|B)$, and the same for $(B|C)$ and $(B)$.
If now we take $v\in Ker(A|B)$ but $v\not\in Ker(A|B|C)$, then it will surely be in $Ker(B)$ but not in $Ker(B|C)$, proving the result.