Let $V_1, V_2, V_3$ be linear subspaces of $\mathbb R^n$. One says they have 'normal intersection' if $V_i\pitchfork(V_j\cap V_k)$ whenever $i\ne j$ and $i\ne k$. Prove that this holds iff $$\operatorname{codim}(V_1\cap V_2\cap V_3)=\operatorname{codim} V_1+\operatorname{codim} V_2+\operatorname{codim} V_3.$$
I guess I proved one direction. If the spaces intersect normally, then in particular we have $V_1 \pitchfork (V_2\cap V_3)$ and $V_2 \pitchfork V_3$ (if $j=k=3$). Now $$\operatorname{codim}(V_1\cap V_2\cap V_3)=\operatorname{codim}(V_1\cap (V_2\cap V_3))=\operatorname{codim}V_1+\operatorname{codim}(V_2\cap V_3)=\operatorname{codim} V_1+\operatorname{codim} V_2+\operatorname{codim} V_3,$$ where I have used that if two submanifolds $X, Z$ of $Y$ intersect transversally then $$\operatorname{codim}(X\cap Z)=\operatorname{codim} X + \operatorname{codim} Z$$ (Theorem on p.30 of the book).
Is this direction correct? How do I prove the other direction?