= Solution
Combine a <minimum> with an <adjacent pair in a linear order>:
$$
\boxed{\xi=\bigl(\exists m\,\forall x\,\neg(x<m)\bigr)\land\exists u\,\exists v\,\bigl(u<v\land\neg\exists w\,(u<w\land w<v)\bigr).}
$$
In $[0,1]\cup[2,3]$, choose $m=0,u=1,v=2$. The <minimum> and the <adjacent pair in a linear order> both exist. In $[0,1]$, the <dense order> property rules out such an adjacent pair. In $\mathbb Z$, adjacent pairs exist, but no <minimum> exists: for any $m$, the <integer> $m-1$ is smaller. Thus this <first-order sentence> has precisely the required three <truth values>. Neither the endpoints nor the numerical witnesses occur as constant symbols in the <first-order sentence>.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/digital-systems-linear-algebra-introduction-to-formal-proof-ordered-structures.png]
{title=A minimum and an adjacent pair distinguish the two-interval order from a closed interval and the integers}
{description=Orange highlights the consecutive domain members $1$ and $2$ in the two-interval order. The dotted arrow crosses points absent from that domain. A closed real <interval> has intervening domain members; the <integers> have adjacent pairs but no <minimum>.}
Back to article page
