184 Logic as a Tool Theorem 138 (Soundness and completeness of ND) Each of the systems ND of Natural Deduction for first-order logic, with and without equality, is sound and complete, that is, for all first-order formulae A1 , , An , C of the respective language (respectively, with or without equality): A1 , , A n , ND C iff A1 , , An , |= C References for further reading For more details on the theory and examples of derivations in Natural Deduction, see van Dalen (1983), Jeffrey (1994) (who referred to “synthetic trees”), Smullyan (1995), Fitting (1996), Huth and Ryan (2004), Nederpelt and Kamareddine (2004), Prawitz (2006, the original development of the modern version of Natural Deduction), Chiswell and Hodges (2007), and van Benthem et al (2014), as well as Kalish and Montague (1980) and Bornat (2005), where Natural Deduction derivations are presented in a boxed form rather than in tree-like shape Exercises Use Natural Deduction for all exercises below In all derivations, indicate the rules applied and the succession of all steps by numbering them Check that the provisos for all applications of rules are satisfied 4.3.1 Prove the following logical validities and consequences (a) |= ∀xA(x) → ∀yA(y ) (b) |= ∃xA(x) → ∃yA(y ) (c) ∀xA(x) |= ¬∃x¬A(x) (d) ∃xA(x) |= ¬∀x¬A(x) 4.3.2 Suppose x is not free in Q Prove the validity of the following logical consequences by deriving them with Natural Deduction (a) (b) (c) (d) (e) ∀x(P (x) ∨ Q) |= ∀xP (x) ∨ Q ∀xP (x) ∨ Q |= ∀x(P (x) ∨ Q) ∃x(P (x) ∧ Q) |= ∃xP (x) ∧ Q ∃xP (x) ∧ Q |= ∃x(P (x) ∧ Q) ∃xP (x) → Q |= ∀x(P (x) → Q) (f) (g) (h) (i) (j) ∀x(Q → P (x)), Q |= ∀xP (x) ∃x(Q → P (x)), Q |= ∃xP (x) ∃x(P (x) → Q), ∀xP (x) |= Q ∃x(¬P (x) ∨ Q) |= ∀xP (x) → Q ∃x(P (x) ∨ ¬Q) |= Q → ∃xP (x) 4.3.3 Determine which of the following logical consequences hold by searching for a derivation in Natural Deduction For those that you find not derivable in ND, consider A and B as unary predicates and look for 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) ∀xA(x) ∨ ∀xB (x) |= ∀x(A(x) ∨ B (x)) (c) ∀x(A(x) ∧ B (x)) |= ∀xA(x) ∧ ∀xB (x)