Proposition: The exceptional scalar wavefront configuration

For the scalar kernels in (eq:triangle), the sufficient wavefront product criterion holds off $\mathcal{N}$. At a strictly ordered null-collinear triple $z\prec y\prec x$ it fails.

This failure does not prove nonexistence of a distributional extension, nor nonexistence of the fully contracted retarded coefficient.

Proof. The three pulled-back wavefront covectors have the forms

in the slot order $(x,z,y)$. Their sum vanishes precisely when

No two nonzero factors alone can give a zero sum because one endpoint would have an unpaired nonzero covector.

For distinct singular endpoints, all three differences must be null. Pairwise null-separated points in Minkowski space are collinear along a null line. Retarded support puts $z$ before $x$. The Feynman signs permit $l=-p$ only when $y$ lies between $z$ and $x$.

At such a triple, choose $p$ future null along the common line, $k=l=-p$, and use the two-sided null wavefront of the retarded kernel. The sum is zero, so the criterion fails. The partial-coincidence cases give the boundary parameters in (eq:nullset); $x=z\ne y$ does not allow cancellation of the two equal-orientation Feynman covectors.

Thus the product theorem certifies the complement, but not the exceptional configurations. A sufficient theorem's failed hypothesis is not a nonexistence proof.