A tolerance below roundoff can pass with a zero embedded difference

Use the toy binary format whose nonzero representable numbers are $\pm m{2}^e$ with integer $e$ and $m\in \{4,5,6,7\}$ (three significant bits), rounded to nearest with ties to an even significand. The next larger representable value after $8=4\cdot 2$ is $10=5\cdot 2$, so their midpoint $9$ rounds to $8$ because $4$ is even and $5$ is odd. Apply a rounded Euler–Heun implementation to $y^{\mathrm{\prime }}=1$, $y(0)=8$ on $[0,1]$, with one step $h=1$, $h_{\min }=h_{\max }=h_{\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}}=1$, attempt budget $N=1$, $\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{l}=1\mathrm{/}4$, $\mathrm{r}\mathrm{t}\mathrm{o}\mathrm{l}=0$, safety $0.9$, and scale bounds $0.2$ and $5$. Both computed endpoints round to $8$; the stored difference is zero and the step is accepted, although the exact endpoint is $9$ and the actual error is $1$.

This example concerns the rounded realization described in Finite-budget adaptive Runge–Kutta controller. It does not contradict its exact-arithmetic specification and shows why that acceptance test is not a certificate.