A cocircular four-site tie selects the stable-id diagonal

The unperturbed heights are $0,1,2,1$. Each is $x+y$ at its site, so the four lifted points lie in the plane $z=x+y$.

All four original sites lie on the circle centered at $(1\mathrm{/}2,1\mathrm{/}2)$ with radius squared $1\mathrm{/}2$. No other sites are present, so neither selected triangle has a site strictly inside its circumdisk. They are weakly empty by [F4].

With the perturbation from [F2], the plane through $A,B,C$ has value $1+\epsilon -{\epsilon }^2+{\epsilon }^4$ at $D$, while $D$'s lifted height is $1+{\epsilon }^8$. Their difference is ${\epsilon }^8-\epsilon +{\epsilon }^2-{\epsilon }^4$, whose least nonzero exponent is $1$ with coefficient $-1$. Thus $D$ lies formally below the plane through $A,B,C$, and diagonal $AC$ is illegal. The flip selects $BD$ by [F3].

The plane through $A,B,D$ has value $2+{\epsilon }^2+{\epsilon }^8-\epsilon$ at $C$; subtracting from $C$'s height $2+{\epsilon }^4$ gives $\epsilon -{\epsilon }^2+{\epsilon }^4-{\epsilon }^8$, which is formally positive. The plane through $B,C,D$ has value ${\epsilon }^8-{\epsilon }^4+{\epsilon }^2$ at $A$; subtracting from $A$'s height $\epsilon$ gives $\epsilon -{\epsilon }^2+{\epsilon }^4-{\epsilon }^8$, also formally positive. Hence both $ABD$ and $BCD$ are lower triangles for the common lift.

When three synthetic identifiers are inserted below $1,2,3,4$, each input rank increases by three and each exponent $2^{r_i}$ is multiplied by $8$. Thus every input-only determinant polynomial becomes the four-site polynomial with $\epsilon$ replaced by ${\epsilon }^8$, which preserves the first nonzero coefficient sign. The augmented run therefore retains diagonal $BD$, and [F5] returns the stated weakly empty triangulation.