Deductive Reasoning in First-order Logic 185 (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) ∃xA(x) ∧ ∃xB (x) |= ∃x(A(x) ∧ B (x)) (h) ∃x(A(x) ∧ B (x)) |= ∃xA(x) ∧ ∃xB (x) (i) ∃x(A(x) ∨ B (x)) |= ∃xA(x) ∨ ∃xB (x) (j) ∃xA(x) ∨ ∃xB (x) |= ∃x(A(x) ∨ B (x)) (k) ∃xA(x) → ∃xB (x) |= ∃x(A(x) → B (x)) (l) ∃x(A(x) → B (x)) |= ∃xA(x) → ∃xB (x) (m) ∃x∃yA(x, y ) |= ∃y ∃xA(x, y ) (n) ∀y ∃x(P (x) → Q(y )) |= ∀xP (x) → ∀zQ(z ) (o) ∃y ∀x(P (x) → Q(y )) |= ∃xP (x) → ∃zQ(z ) (p) ∀y (P (y ) → Q(y )) |= ∃yP (y ) → ∃yQ(y ) (q) ∀y (P (y ) → Q(y )) |= ∃x¬Q(x) → ∃x¬P (x) (r) |= ∃x(P (x) → ∀yP (y )) 4.3.4 Formalize the following arguments in first-order logic and try to prove their logical correctness by deriving them in Natural Deduction For those that you find not derivable in ND, look for a counter-model (a) No yellow and dangerous reptiles are lizards Some reptile is dangerous or is not a lizard Therefore, not every reptile is a yellow lizard (b) All penguins are birds No penguin can fly Every white bird can fly Therefore, no penguin is white (c) All penguins are birds Some penguins are white All penguins eat some fish Therefore, there is a white bird that eats some fish (d) No lion eats birds Every penguin is a bird Simba is a lion Therefore some lion eats no penguins