315 Answers and Solutions to Selected Exercises ∀xP (x) ∧ ∀xQ(x) |= ∀x(P (x) ∧ Q(x)): ∀xP (x) ∧ ∀xQ(x) : T , ∀x(P (x) ∧ Q(x)) : F P (c) ∧ Q(c) : F P (c) : F Q(c) : F ∀xP (x) : T , ∀xQ(x) : T ∀xP (x) : T , ∀xQ(x) : T P (c) : T × Q(c) : T × Hence, ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x) (c) ∀xP (x) → ∀xQ(x) ≡ ∀x(P (x) → Q(x)) because ∀xP (x) → ∀xQ(x) ∀x(P (x) → Q(x)): ∀xP (x) → ∀xQ(x) : T , ∀x(P (x) → Q(x)) : F P (c1 ) → Q(c1 ) : F P (c1 ) : T , Q(c1 ) : F ∀xP (x) : F ∀xQ(x) : T P (c2 ) : F Q(c1 ) : T × A counter-model extracted from the tableau: a structure with domain {a1 , a2 } where c1 is interpreted as a1 , c2 is interpreted as a2 , P is interpreted as {a1 } and Q is interpreted as ∅ (d) ∃xP (x) ∧ ∃xQ(x) ≡ ∃x(P (x) ∧ Q(x)) because ∃xP (x) ∧ ∃xQ(x) ∃x(P (x) ∧ Q(x)):