Two guard crossings between endpoints defeat sign-only screening

The incoming-mode continuation is $x(t)=t$, but the guard is independent of $x$, so $h(t)=(t-1\mathrm{/}3)(t-2\mathrm{/}3)$. Its endpoint values are $h(0)=h(1)=2\mathrm{/}9>0$. A rule that checks only endpoint signs therefore reports no crossing. The roots $1\mathrm{/}3$ and $2\mathrm{/}3$ are not dyadic, so neither lies on a midpoint boundary of the isolator's dyadic bisection of $[0,1]$.

The two roots are $1\mathrm{/}3$ and $2\mathrm{/}3$, and $h^{\mathrm{\prime }}(t)=2t-1$. Thus $h^{\mathrm{\prime }}(1\mathrm{/}3)=-1\mathrm{/}3$ is a downward crossing while $h^{\mathrm{\prime }}(2\mathrm{/}3)=1\mathrm{/}3$ is an upward crossing. Under the declared positive orientation, the first root is nontriggering and the second is enabled.

With exact polynomial range and derivative enclosures and a finite box budget sufficient for a complete result, the interval isolator in [F2] finds both roots and classifies their derivative signs. It cannot certify no event, because the root at $2\mathrm{/}3$ is enabled and is the earliest enabled event. Thus equal endpoint signs do not justify an event-absence conclusion; a certified absence report requires complete root isolation and orientation classification with no enabled root.