An ANM noise is independent of every set of nondescendants

For each $k\in S$, define $A_k$ to be $k$ together with all vertices having a directed path to $k$, and put $A={\bigcup }_{k\in S}{A}_k$. In topological order, a root variable is a measurable function of its own noise; if this representation has been obtained for each parent of $k$, substituting those representations into $X_k=f_k(X_{\mathrm{pa}(k)})+N_k$ shows that $X_k$ is a measurable function of $(N_i{)}_{i\in A_k}$. Finite induction therefore makes $X_S$ a measurable function of $(N_i{)}_{i\in A}$. Since $S$ contains only nondescendants of $j$, no directed path goes from $j$ to a vertex of $S$, so $j\mathrm{\notin }A$.

The finite noise product law [F2] implies that $N_j$ is independent of the vector $(N_i{)}_{i\in A}$ because the two coordinate blocks are disjoint. Applying [F3] to the measurable map from that vector to $X_S$ from step 1.1 shows $N_j\perp X_S$. If $S=\mathrm{\varnothing }$, $X_S$ is the constant empty tuple and the same independence statement holds.