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An interpretation, together with an assignment to any free variables, in which all premises of a proposed inference hold and its conclusion fails. By soundness of the inference rules, a countermodel proves that the inference cannot be derived. For example, on $\{0,1\}$ with $P=Q=\{0\}$ and $x=0$, the formula $\forall y(P(y)\to Q(y))$ is true but $P(x)\to \forall yQ(y)$ is false. A pointwise assumption about $x$ cannot justify unrestricted universal quantification.

For a satisfiable equality-free universal logical conjunction of Horn clauses, intersecting the sets of true ground atoms of its Herbrand models gives another Herbrand model, contained in every one of them. A ground atom is true in this least Herbrand model exactly when it holds in every Herbrand model. Positive existential sentences are true in the least Herbrand model exactly when they follow from the Horn clauses; allowing negated atoms makes this equivalence fail. For example, the least Herbrand model of $\forall x(P(x)\to Q(x))$ over a single constant $a$ makes both predicates empty and satisfies $\exists x\neg P(x)$, whereas the first-order structure making both predicates universal is a countermodel to that consequence.

Solutions to the three questions of the University of Oxford computer science Finals examination held on 10 June 2015. Question 3(d) has a false assertion as printed; its solution gives a countermodel and explains the additional hypotheses under which the usual least Herbrand model argument works.^[ref]

The dot after $\forall x$ gives it scope over the following implication: read the premise as $H=\forall x(P(x)\to Q(x))$. Choose a fresh arbitrary eigenvariable $u$. A natural deduction proof isAt line 6, $u$ is free in no undischarged assumption, so the universal quantifier introduction rule is valid.

The proposed variant is not derivable in a sound first-order logic. For a countermodel, use the two-element domain $\{0,1\}$, interpret both predicates $P,Q$ as $\{0\}$, and assign the free variable $x$ the value 0. Then $\forall x(P(x)\to Q(x))$ is true, as is $P(x)$ under the assignment, but $\forall xQ(x)$ is false at element 1. The premise's bound $x$ and the conclusion's free $x$ have different scopes.Attempting to generalize from $Q(x)$ in the variant would violate the eigenvariable condition, because $x$ remains free in the undischarged assumption $P(x)$.