Example: A non-Born scalar candidate and a compensated table

For $0<\eta \le 1/2$, set

It has anchors at $0,1/3,1/2,2/3,1$ and obeys ${\phi }_{\eta }(1-x)=1-{\phi }_{\eta }(x)$. The derivative of the added polynomial has absolute value at most one on $[0,1]$: the product rule bounds its terms by $4/9$, $2/9$, and $1/3$. Thus ${\phi }_{\eta }^{\prime }\ge 1-\eta >0$, and it maps $[0,1]$ increasingly onto itself.

At $\eta =1/2$,

However, the cocycle instance

has defect

The associated Born table is

One can nevertheless satisfy both sequential orders on this one strictly supported input by setting the proposed joint table to

After the first-qubit readout, assign a one-qubit residual with computational zero-weight $m$; after the second-qubit readout, assign one with zero-weight $a$. If one-qubit readout uses ${\phi }_{\eta }$, both sequential products equal $J_{ij}$.

The first assigned residual has weight $m=1/4$, whereas the projective residual after first outcome zero has weight $u=1/2$. The compensation is therefore diagonally unfaithful.

This is an exact partial assignment at the displayed state and continuations. It is not a strategy on all rays. It does not prove continuity across support boundaries, all record equations, or no-signaling for arbitrary local programs.