302 Logic as a Tool (g) Yes For the sake of a contradiction, suppose ∃x(A(x) ∧ B (x)) ∃xA(x) ∧ ∃xB (x) Then there exist a structure S and an assignment v in S such that S , v |= ∃x(A(x) ∧ B (x)), while S , v ∃xA(x) ∧ ∃xB (x) Hence, S , v ∃xA(x) or S , v ∃xB (x) This means S , v [x := s] A(x) for all s or S , v [x := s] B (x) for all s, while S , v [x := c] |= A(x) and S , v [x := c] |= B (x) for some c ∈ S, which is a contradiction (i) Yes For the sake of a contradiction, suppose ∃x(A(x) ∨ B (x)) ∃xA(x) ∨ ∃xB (x) Then there exist a structure S and an assignment v in S such that S , v |= ∃x(A(x) ∨ B (x)), while S , v ∃xA(x) ∨ ∃xB (x) Hence, S , v ∃xA(x) and S , v ∃xB (x) This means that there is some c ∈ S such that S , v [x := c] |= A(x) or S , v [x := c] |= B (x), while S , v [x := s] A(x) and S , v [x := s] B (x) for all s ∈ S, which is a contradiction (k) No We will show that ∃x(A(x) → B (x)) ∃xA(x) → ∃xB (x) It is sufficient to find any counter-model, that is, a structure S and a variable assignment v such that: S , v |= ∃x(A(x) → B (x)) and S , v ∃xA(x) → ∃xB (x) Take for example the structure S with domain the set Z of all integers, with an interpretation of the unary predicate A to be {0} and interpretation of the unary predicate B to be ∅ (We know that the variable assignment is irrelevant here because all formulae involved in the logical consequence are sentences, but we will nevertheless need it in order to process the truth definitions of the quantifiers.) Now let v1 be a variable assignment obtained from v by redefining it on x as follows: v (x) := Then: (1) S , v1 A(x) (2) S , v1 B (x) According to the truth definition of → it follows from (1) and (2) that: (3) S , v1 |= A(x) → B (x) According to the truth definition of ∃ it therefore follows from (3) that: (4) S , v |= ∃x(A(x) → B (x)) Now let v2 be a variable assignment obtained from v by redefining it on x as follows: v2 (x) := Then: (5) S , v2 |= A(x) From (5) it follows according to the truth definition of ∃ that: (6) S , v |= ∃xA(x) It follows directly from the definition of B and the truth definition of ∃ that: (7) S , v ∃xB (x) According to the truth definition of → it follows that: (8) S , v ∃xA(x) → ∃xB (x) S is therefore a counter-model, falsifying the logical consequence ∃x(A(x) → B (x)) |= ∃xA(x) → ∃xB (x) 3.4.12 (a) We show that each side logically implies the other First, to show ∀xA(x) |= ¬∃x¬A(x), suppose the contrary Then there is a structure S and an assignment v such that S , v |= ∀xA(x) but S , v