Past exam of the computer science course of the University of Oxford / 2015 / Knowledge representation and reasoning / 1

The condition about taking the course restricts which students the first-order sentence talks about. Within that condition we need both an inclusive logical disjunction and a prohibition on satisfying both alternatives:The logical conjunction on the right expresses exactly one of the two categories. It does not prohibit unrelated individuals from belonging to both categories: the restriction applies to the students in the antecedent.

The supervisor can depend on the student, so the existential quantification belongs inside the universal quantification:The first-order sentence permits different students to share a supervisor. It requires neither that the supervisor be different from the student nor that the supervisor be a faculty member, since neither additional condition was given.

Using the supplied counting function, an appropriate first-order sentence isHere the counting comparison expresses the threshold and the existential quantification expresses teaching at least one course. The word “if” supplies this direction of logical implication; it does not assert that these are the only reasons a faculty member can be busy. With the intended meanings of the nonlogical symbols, no further counting axioms are needed for this translation. If those meanings were unspecified, the displayed first-order sentence alone would not define arithmetic or connect the counting function to the supervision relation.

Write $S,T,U,M,F,Q,J$ for the atomic formulas $Student(x)$, $Takes(x,KRR,2015)$, $Undergrad(x)$, $Masters(x)$, $FacultyMember(x)$, $GreaterOrEqual(CountSupervise(x),2)$, and $Teach(x,y)$, respectively. These are abbreviations for formulas, not new symbols in the vocabulary. In clausal normal form, with all remaining variables implicitly universally quantified, the three translations become:For (ii), Skolemization introduces the fresh Skolem function $f$. One supervisor suffices even when both categories hold, so the same $f$ can be used in both first-order clauses. For (iii), moving the existentially quantified course out of an antecedent produces a universally quantified variable in the resulting first-order clause:

The resolution with free selection calculus for first-order Horn clauses is sound and refutationally complete for the clause sets from $\boxed{\text{(ii) and (iii)}}$. For example, select the unique positive first-order literal in every definite first-order Horn clause, and select one negative first-order literal in every nonempty clause without a positive one. On the displayed definite first-order Horn clauses, this selects $SupervisedBy(x,f(x))$ or $Busy(x)$. A fair saturation must still consider all permitted inferences. Functions do not invalidate this completeness theorem, although they can prevent termination. ^[ref]

The first first-order clause of (i) has two positive first-order literals, so the Horn resolution with free selection theorem does not apply to that sentence. General first-order resolution, with its usual completeness requirements, can handle it. Soundness of an individual resolution inference is not the issue; the issue is applicability and completeness of this particular restricted calculus.

With existential rules understood as function-free tuple-generating dependencies, $\boxed{\text{only (ii)}}$ belongs to the fragment. Split its antecedent to obtainThese two existential rules have exactly the same meaning as (ii): either category requires a supervisor, and when both hold the same supervisor may witness both requirements.

Sentence (i) requires a genuine choice between two positive alternatives and also an exclusion condition. Positive existential rules cannot enforce that exclusion: interpreting all their relations as full relations on a one-element domain always satisfies them, whereas that interpretation violates (i). Sentence (iii), in the supplied vocabulary, uses the nonconstant function $CountSupervise$ and is therefore outside the function-free fragment. Introducing a new relation for the counting comparison would change the representation and require additional axioms; it is not an equivalent sentence in the stated vocabulary.

For ordinary function-free Datalog, $\boxed{\text{none of the three sentences}}$ has the required form. Sentence (i) is not a definite first-order Horn clause: the category choice gives two positive alternatives, and its second clause is a negative constraint. Even allowing constraints would not remove the positive choice. Semantically, the category-choice condition fails closure under intersection of interpretations with fixed domain and functions: one model can choose only the undergraduate category, another only the masters category, and their intersection chooses neither.

Sentence (ii) needs an existentially supplied supervisor. Function-free Datalog cannot demand such a witness without naming or otherwise fixing it in the rule. Its Skolemized form is a first-order Horn clause with a function, rather than a Datalog rule. Sentence (iii) also contains a nonconstant function. Being Horn is therefore insufficient for membership in Datalog.

The standard relational guarded fragment contains $\boxed{\text{(i) and (ii)}}$. In (i), $Student(x)$ guards the only quantified variable:The constants in the course-taking relation do not introduce additional free variables that need guarding.

For (ii), use the two implications from part (c)(i). Each outer universal quantification is guarded by its unary category predicate. The inner existential quantification can be writtenwhose guard contains both $x$ and $y$. Sentence (iii) uses a nonconstant function and so is outside the standard relational guarded fragment. This conclusion uses the original vocabulary; it does not silently replace the counting term by a fresh relation.

In ALC, $\boxed{\text{only (ii)}}$ is representable over the given symbols. It is the general concept inclusionEquivalently, use one general concept inclusion for each category. Unary predicates become description logic concepts and the supervision predicate becomes a description logic role.

Sentence (i) depends on a ternary predicate evaluated at two named constants. Basic ALC has unary description logic concepts and binary description logic roles, with no constructor for that course-and-year slice of a ternary predicate. Sentence (iii) uses a counting function and a comparison relation that basic ALC cannot express. Replacing these conditions by freshly named description logic concepts would give an informal encoding, but would not define their meanings or establish equivalence in the supplied vocabulary.

For resolution, a first-order sentence is in clausal normal form when it is a universally quantified logical conjunction of first-order clauses:where every $L_{ij}$ is a first-order literal, meaning an atomic formula or the logical negation of an atomic formula. The quantifiers are often left implicit and the first-order clauses treated as a set. The empty first-order clause represents falsity; an empty set of first-order clauses represents truth.

A convention that allows an arbitrary quantifier prefix followed by a conjunctive matrix is weaker: that is a prenex normal form with a conjunctive matrix. The elimination of existential quantifiers by Skolemization is the extra step relevant to resolution and to part (ii).

The usual transformation to clausal normal form preserves $\boxed{\text{equisatisfiability, but not necessarily logical equivalence}}$. Removing implications and distributing logical conjunction and logical disjunction preserve logical equivalence. Skolemization introduces new symbols and chooses witnesses as functions of preceding universally quantified variables.

For example,On the domain $\{0,1\}$, let $R$ be equality. Then $\phi$ is true. Expand this same first-order structure by interpreting $f$ as the constant function with value $0$. The resulting first-order structure falsifies $\psi$ at $x=1$. Interpreting $f$ as the identity instead satisfies $\psi$.

Thus every model of $\psi$ has a reduct satisfying $\phi$, and a model of $\phi$ can be expanded with suitable witnesses to satisfy $\psi$. That model-expansion property gives equisatisfiability, not logical equivalence for every interpretation of the added function. If “CNF” means a conjunctive matrix with existential quantifiers retained, Skolemization has not yet happened and an equivalent form is possible; this is why the convention must be specified.

An exponential increase is unavoidable for an equivalent conjunctive normal form over the original variables. Consider the linear-size Boolean formulaDistributing gives the $2^n$ Boolean clausesTo prove that the growth is necessary, fix one choice $r_i\in \{p_i,q_i\}$. Let ${\alpha }_r$ make every $r_i$ false and its partner true. Each pair then contains a false variable, so $F_n$ is false. Any equivalent conjunctive normal form must contain a Boolean clause $C$ falsified by ${\alpha }_r$ and entailed by $F_n$.

Every individual term $p_i\wedge q_i$ entails $C$, because that term implies $F_n$. After tautological Boolean clauses are discarded, this forces $C$ to contain a positive $p_i$ or positive $q_i$ for every $i$: otherwise set both variables in that pair true and independently falsify all Boolean literals of $C$, contradicting the entailment. Since $C$ is false at ${\alpha }_r$, it must contain the false member $r_i$, never its true partner. Consequently the same $C$ cannot be false at any different ${\alpha }_s$. The $2^n$ different assignments therefore require at least $2^n$ different Boolean clauses.

Writing the answer already takes $\boxed{\Omega (2^n)\text{ time in the worst case}}$. This obstruction applies to first-order logic as well, since the example can be expressed using ground unary atomic formulas. It concerns exact equivalence without auxiliary symbols. A Tseytin transformation can produce a polynomial-size equisatisfiable conjunctive normal form with auxiliary variables, so the statement must not be extended to that different task.

The algorithm is $\boxed{\text{incorrect: it can return true for a non-entailed ground atom}}$. Its recursive calls return only Booleans. They do not return and propagate the first-order substitutions obtained while satisfying earlier rule premises.

Take distinct constants $a,b,c$, factsand the single Datalog ruleAsk for $H(c)$. The head matches the goal without binding $x$. In line 15, the recursive call for $P(x)$ succeeds by first-order unification with $P(a)$, but returns only true. The next recursive call still receives $Q(x)$ with $x$ unbound, and succeeds by first-order unification with $Q(b)$. Neither subgoal is the visited head $H(c)$, so the visited test does not stop these calls. Both recursive results are true, and the algorithm returns true for $H(c)$.

However, the least Herbrand model is just $\{P(a),Q(b)\}$. No single constant makes both $P(x)$ and $Q(x)$ true, so the rule cannot derive $H(c)$. The two individually successful premises need one consistent first-order substitution. A correct backward chaining procedure must thread answer first-order substitutions through the remaining premises and backtrack when they cannot be extended; forward chaining on ground Datalog instances is another way to avoid this error.

A fragment has the finite model property if $\boxed{\text{every satisfiable sentence in the fragment has a finite model}}$. This does not mean that every model is finite, nor that there is one uniform bound on model size for the entire fragment.

ALC has the finite model property, including satisfiability with respect to finite TBoxes. A tableau algorithm for ALC uses finitely many possible labels and direct blocking in description logic to obtain a finite representation of a model. The resulting finite-model claim is a property of the logic; termination alone, for an arbitrary procedure, would not establish it.

Full first-order logic does not have the finite model property. For example, the first-order sentencerequires an irreflexive transitive relation with a successor for every element. It is satisfied by the usual strict order on the natural numbers. In a nonempty finite domain, repeatedly taking successors eventually repeats an element; transitivity would then force $R(x,x)$, contradicting irreflexivity. Hence this satisfiable first-order sentence has no finite model.