Advanced Models, Countermodels, and Logical Consequence Quiz

A unary predicate is interpreted as a subset of the domain, meaning any element may or may not belong to the extension of P. For a domain of three elements, each of the three elements can independently be “in P” or “not in P.” Since each element has two choices and there are three elements, the total number of subsets is 2 × 2 × 2 = 8. These represent all possible interpretations of P.

The negation of an existential is a universal negation, based on quantifier duality. Saying “there does not exist an x such that P(x) holds” means that every x fails to satisfy P(x). This is exactly what ∀x ¬P(x) states. The two are logically interchangeable in every model.

The existential quantifier requires the existence of at least one witness a such that both P(a) and Q(a) hold. It does not require all elements to satisfy the conjunction or that P and Q be distributed in any particular way across the domain. One such element is enough to satisfy the formula.

The formula P(a) ∧ P(b) asserts that at least two specific elements satisfy P. From this, we can conclude that at least one element satisfies P, which is exactly ∃x P(x). We cannot conclude universality, nor can we deduce conditional relationships that are not directly supported.

The first part says every P-element must also be a Q-element. The second part says at least one P-element is not a Q-element. These two conditions contradict each other directly: one demands no exceptions, the other requires exactly such an exception. No model can satisfy both simultaneously, so the combined formula is unsatisfiable.

∃x P(x) supplies a witness a with P(a) true; ∀x ¬Q(x) ensures Q(a) is false for that same a, so a satisfies P ∧ ¬Q; hence ∃x (P(x) ∧ ¬Q(x)) holds. Disjointness is true extensionally but the phrasing of Option A is ambiguous; Option C is false.

From ∀x(P(x) ∨ Q(x)), applying universal instantiation yields P(a) ∨ Q(a). Combined with ¬P(a), one must conclude Q(a) by disjunctive syllogism. There is no model where the premise and negation of P(a) are true but Q(a) is false, so the argument is valid.

To break the inference from an existential to a universal claim, we need a model where the existential is true but the universal false. This requires at least one element satisfying P and at least one not satisfying P. Such a model makes the premise true and the conclusion false, proving invalidity.

The formula ∀x P(x) → ∀x Q(x) states that if all elements satisfy P, then all satisfy Q. The formula ∀x(P(x) → Q(x)) states that every element that satisfies P also satisfies Q. These differ in strength: the second allows some elements to satisfy neither P nor Q, while the first requires Q(x) for all x once P(x) holds universally. Therefore, they are not equivalent.

The correct entailment argument uses an arbitrary element a: from ∀x P(x) we get P(a); from ∀x (P(x) → Q(x)) we get P(a) → Q(a); modus ponens gives Q(a); since a was arbitrary, ∀x Q(x) follows. Options A, C, D are irrelevant or false.

In any nonempty domain, from ∀x(P(x) ∨ Q(x)) we can conclude ∃x P(x) ∨ ∃x Q(x). For any element a, either P(a) or Q(a) holds, which guarantees at least one of the existentials is true. However, we cannot conclude which predicate applies universally (option A) or that both apply universally (option B), since different elements might satisfy different disjuncts.

For any element x, we can choose y to be that same element. Since x = x is always true in standard first-order logic with equality, the existential is automatically satisfied. This works for every x in every model, making the universally quantified statement valid.

If there exists a single element that satisfies both P and Q, then certainly there exists at least one element that satisfies P, and there exists at least one element that satisfies Q. This allows us to conclude the conjunction of the two existential statements. Nothing stronger, such as universality, follows.

∃x ∀y (y = x) requires a single element x equal to every y, which holds precisely when the domain has one element. Hence the formula is satisfiable (in singleton domains) but not valid in general. It is not equivalent to ∀y ∃x (y = x), which is valid (choose x = y).

If empty domains are permitted, ∀x P(x) is vacuously true in the empty domain, but ∃x P(x) is false there, so the implication ∀x P(x) → ∃x P(x) fails in that model; thus it is not valid under empty-domain semantics. Other listed formulas remain valid or are contingent in ways that do not exhibit this particular empty-domain failure.