Two layers cannot see three hops

After $k$ applications node $1$ depends on the degrees and features within $k$ hops. In both paths nodes $1$, $2$ and $3$ have the same degrees; node $4$ does not.

The values agree for two layers and differ at the third. Node $1$‘s value after $k$ layers uses the entries of $\hat{\mat A}$ along walks of length at most $k$ from it. Those entries depend on the degrees at both ends of each edge walked. Within two hops of node $1$ the walks touch only nodes $1$, $2$ and $3$, whose degrees are $1$, $2$ and $2$ in both paths. The third layer walks the edge from node $3$ to node $4$, and node $4$ has degree $1$ in the shorter path and $2$ in the longer.

So a two-layer network cannot tell whether node $1$ sits on a path of four or five, whatever its width or weights. That is I.14.T2 at work: a task that depends on something three hops away needs three layers, and the averaging of proposition 3 makes each extra layer cost something.