Models, Quantifiers, and Argument Evaluation Quiz

This classical syllogism uses two universal premises to derive a specific conclusion through universal instantiation and modus ponens. If all humans are mortal, and Socrates belongs to the class of humans, then Socrates must also be mortal because membership in that class guarantees the predicate “is mortal.” There is no model in which both premises are true and the conclusion false, so the argument is valid.

The universal quantifier distributes over conjunction. Saying “for all x, both P(x) and Q(x) are true” is the same as saying “for all x, P(x) is true” and “for all x, Q(x) is true.” The structure allows the conjunction to be split because both components apply universally and independently.

The existential quantifier distributes over disjunction. If there exists an x such that P(x) or Q(x) is true, then it must be that either there exists an x with P(x) true or there exists an x with Q(x) true. The witness for the existential makes at least one disjunct true, ensuring the disjunction of existentials holds.

The first premise says every P-object is also a Q-object. The second premise says every object is a P-object. Therefore every object must also be a Q-object, because substituting any element a satisfies P(a), and from P(a) → Q(a) we get Q(a). This holds for all elements in the domain, making ∀x Q(x) necessarily true.

The two conjuncts contradict: ∀x P(x) asserts every element satisfies P, while ∃x ¬P(x) asserts at least one element does not. No model can make both true simultaneously, so the conjunction is unsatisfiable.

A binary relation R is symmetric if whenever x is related to y, y is also related to x. The formula states exactly this: for any pair of elements x and y, if R(x,y) holds, then R(y,x) must hold. This matches the definition of symmetry and not transitivity, reflexivity, or antisymmetry.

∃x (P(x) → ∀y P(y)) is valid. Consider two cases: (1) If ∀y P(y) is true, then P(x) → ∀y P(y) is true for any x (true consequent makes any implication true), so the existential holds. (2) If ∀y P(y) is false, then ∃y ¬P(y), so there exists some element c where P(c) is false. For this c, the implication P(c) → ∀y P(y) is vacuously true (false antecedent), making c a witness for the existential. In either case, the formula is true, so it is valid.

If every element satisfies P, and every element that satisfies P also satisfies Q, then every element must satisfy Q. In every model fulfilling both premises, the conclusion must also be true. Since no countermodel exists, the semantic entailment holds universally.

For any x, P(x) ∨ ¬P(x) is always true because for any predicate P(x), either it holds or it does not hold. The law of excluded middle makes this disjunction true regardless of interpretation. Applying the universal quantifier preserves that truth across all domain elements and all possible models.

The two existential premises guarantee that at least one P-object exists and at least one Q-object exists, though not necessarily the same object. To validate ∃x(P(x) ∨ Q(x)), we only need one object that satisfies at least one of the predicates. The witness for P(x) already satisfies P(x) ∨ Q(x), so the conclusion follows.

If ∀x P(x) holds in a model, it asserts P(d) for every domain element d; therefore P(a) must also hold for the interpretation of a in that same model. There is no model in which ∀x P(x) is true and P(a) false, so no countermodel exists. Options A and D attempt impossible assignments (a must denote an element from the domain), and empty domains are usually normally disallowed.

∃x ∀y R(x,y) asserts that there is at least one element a such that for every y, R(a,y) holds. That is precisely "some element relates to all elements." Reflexivity or symmetry are unrelated requirements.

The premise says each element has at least one related element. The conclusion requires one element that works for all. These conditions are not logically connected. A classic countermodel uses a domain where each element relates to a different element, so each has a y but no single y works for all x. Thus the implication fails in some models and is not valid.

The universal implication fails exactly when the antecedent is true and the consequent false for some element. So to make ∀x(P(x) → Q(x)) false, the model must include at least one element a such that P(a) is true and Q(a) is false. No other configuration makes the implication false.

If empty domains are permitted, ∀x P(x) is vacuously true in the empty domain, while ∃x P(x) is false there, so the implication ∀x P(x) → ∃x P(x) can fail; hence it is not valid under empty-domain semantics.