First, $\zerovector\in\nsp{B}$ trivially. Now suppose that $\vect{x}\in\nsp{B}$. Then
\begin{align*}
AB\vect{x}
&=A(B\vect{x})&&\knowl{./knowls/theorem.MMA.knowl}{\text{Theorem MMA}}\\
&=A\zerovector&&\vect{x}\in\nsp{B}\\
&=\zerovector&&\knowl{./knowls/theorem.MMZM.knowl}{\text{Theorem MMZM}}
\end{align*}
Since we have assumed $AB$ is nonsingular, Definition NM implies that $\vect{x}=\zerovector$.
Second, $\zerovector\in\csp{B}$ and $\zerovector\in\nsp{A}$ trivially, and so the zero vector is in the intersection as well (Definition SI). Now suppose that $\vect{y}\in\csp{B}\cap\nsp{A}$. Because $\vect{y}\in\csp{B}$, Theorem CSCS says the system $\linearsystem{B}{\vect{y}}$ is consistent. Let $\vect{x}\in\complex{n}$ be one solution to this system. Then
\begin{align*}
AB\vect{x}
&=A(B\vect{x})&&\knowl{./knowls/theorem.MMA.knowl}{\text{Theorem MMA}}\\
&=A\vect{y}&&\text{$\vect{x}$ solution to $\linearsystem{B}{\vect{y}}$}\\
&=\zerovector&&\vect{y}\in\nsp{A}
\end{align*}
Since we have assumed $AB$ is nonsingular, Definition NM implies that $\vect{x}=\zerovector$. Then $\vect{y}=B\vect{x}=B\zerovector=\zerovector$.
When $AB$ is nonsingular and $m=n$ we know that the first condition, $\nsp{B}=\set{\zerovector}$, means that $B$ is nonsingular (Theorem NMTNS). Because $B$ is nonsingular Theorem CSNM implies that $\csp{B}=\complex{m}$. In order to have the second condition fulfilled, $\csp{B}\cap\nsp{A}=\set{\zerovector}$, we must realize that $\nsp{A}=\set{\zerovector}$. However, a second application of Theorem NMTNS shows that $A$ must be nonsingular. This reproduces Theorem NPNT.