Deductive Reasoning in First-order Logic 175 4.2.2 Suppose x is not free in Q Prove the following logical consequences (a) ∀x(P (x) ∨ Q) |= ∀xP (x) ∨ Q (b) ∀xP (x) ∨ Q |= ∀x(P (x) ∨ Q) (c) ∃x(P (x) ∧ Q) |= ∃xP (x) ∧ Q (d) ∃xP (x) ∧ Q |= ∃x(P (x) ∧ Q) (e) ∀x(Q → P (x)), Q |= ∀xP (x) (f) ∃xP (x) → Q |= ∀x(P (x) → Q) (g) ∃x(P (x) → Q), ∀xP (x) |= Q (h) ∃x(Q → P (x)), Q |= ∃xP (x) 4.2.3 Check which of the following first-order formulae are logically valid For each of those which are not, construct a counter-model, that is, a structure which falsifies it (a) ∃x(P (x) → ∀yP (y )) (b) ∀x(P (x) → ∃yP (y )) (c) ∀x∃yQ(x, y ) → ∀y ∃xQ(x, y ) (d) ∀x∃yQ(x, y ) → ∀y ∃xQ(y, x) (e) ∀x∃yQ(x, y ) → ∃y ∀xQ(x, y ) (f) ∀x∃yQ(x, y ) → ∃xQ(x, x) (g) ∃y ∀xQ(x, y ) → ∃xQ(x, x) (h) ∀x(∃yQ(x, y ) → ∃yQ(y, x)) (i) ∃x¬∃yP (x, y ) → ∀y ¬∀xP (x, y ) (j) (∀x∃yP (x, y ) ∧ ∀x∀y (P (x, y ) → P (y, x))) → ∃xP (x, x) 4.2.4 Check which of the following logical consequences hold For those which not, use the tableau to construct a counter-model, that is, a structure and assignment in which all premises are true, while the conclusion is false (a) ∀xA(x), ∀xB (x) |= ∀x(A(x) ∧ B (x)) (b) ∀x(A(x) ∧ B (x)) |= ∀xA(x) ∧ ∀xB (x) (c) ∀xA(x) ∨ ∀xB (x) |= ∀x(A(x) ∨ B (x)) (d) ∀x(A(x) ∨ B (x)) |= ∀xA(x) ∨ ∀xB (x) (e) ∀xA(x) → ∀xB (x) |= ∀x(A(x) → B (x)) (f) ∀x(A(x) → B (x)) |= ∀xA(x) → ∀xB (x) (g) ∀x(A(x) → B (x)) |= ∃xA(x) → ∃xB (x) (h) ∃xA(x), ∃xB (x) |= ∃x(A(x) ∧ B (x)) (i) ∃x(A(x) ∧ B (x)) |= ∃xA(x) ∧ ∃xB (x) (j) ∃x(A(x) ∨ B (x)) |= ∃xA(x) ∨ ∃xB (x) (k) ∃xA(x) ∨ ∃xB (x) |= ∃x(A(x) ∨ B (x)) (l) ∃x(A(x) → B (x)) |= ∃xA(x) → ∃xB (x) (m) ∃xA(x) → ∃xB (x) |= ∃x(A(x) → B (x)) 4.2.5 Determine which of the following logical equivalences hold For each of those that not hold, construct a counter-model, that is, a structure in which one formula is true while the other is false (a) ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x) (b) ∀x(P (x) ∨ Q(x)) ≡ ∀xP (x) ∨ ∀xQ(x) (c) ∀x(P (x) → Q(x)) ≡ ∀xP (x) → ∀xQ(x) (d) ∃x(P (x) ∧ Q(x)) ≡ ∃xP (x) ∧ ∃xQ(x)