Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Digital systems, linear algebra and introduction to formal proof / 8 / c / Solution

Read the initial universal quantifiers as governing the conjunction of irreflexivity and transitivity. Explicitly, the intended first-order sentence assertsThus $Q$ is an irreflexive relation, a transitive relation, and a serial relation.

Suppose a first-order model had a finite nonempty domain of cardinality $N$. Choose $a_0$. Seriality supplies successive elements $a_1,\dots ,a_N$ with $Q(a_i,a_{i+1})$. By the pigeonhole principle, $a_i=a_j$ for some $i<j$. Repeated transitivity along that part of the chain gives $Q(a_i,a_j)=Q(a_i,a_i)$, contradicting irreflexivity. Hence there is no finite nonempty model. This is the infinite-model property of a serial strict partial order.

An infinite model exists: take domain $\mathbb{N}$ and $Q(x,y)$ to mean $x<y$. It is irreflexive and transitive, and $y=x+1$ witnesses seriality for every $x$.

The quantifier-scope convention matters: giving each printed universal quantifier only the immediately following atomic scope would leave the transitivity implication with free variables and would not express the claimed theorem. The explicit parenthesisation above records the reading under which the source's requested assertion is valid.