Answers and Solutions to Selected Exercises 303 ¬∃x¬A(x) S , v |= ∀xA(x) implies S , v [x := s] |= A(x) for all s ∈ S From S , v ¬∃x¬A(x), we have S , v |= ∃x¬A(x), which means that there is an s ∈ S such that S , v [x := s ] |= ¬A(x) Hence, S , v [x := s ] A(x), contradicting the fact that S , v [x := s] |= A(x) for all s ∈ S Likewise, ¬∃x¬A(x) |= ∀xA(x) (c) Again, we show that each side logically implies the other ¬∀x¬A(x) Then there is a structure S and Suppose ∃xA(x) ¬∀x¬A(x) an assignment v such that S , v |= ∃xA(x) but S , v From S , v |= ∃xA(x) it follows that there is s ∈ S such that ¬∀x¬A(x) implies that S , v [x := s ] |= A(x) Besides, S , v S , v |= ∀x¬A(x) Hence, S , v [x := s] |= ¬A(x) for all s ∈ S, and so S , v [x := s] A(x) for all s ∈ S, contradicting the fact that S , v [x := s ] |= A(x) Likewise, ¬∀x¬A(x) |= ∃xA(x) (h) Again, we show that each side logically implies the other Suppose first that ∃x∃yA(x, y ) ∃y ∃xA(x, y ) Then there is a structure S and an assignment v such that S , v |= ∃x∃yA(x, y ) but S , v ∃y ∃xA(x, y ) S , v |= ∃x∃yA(x, y ) implies that there are s , s ∈ S ∃y ∃xA(x, y ), such that S , v [x := s ][y := s ] |= A(x, y ) From S , v we have S , v [x := s][y := t] A(x, y ) for all s, t ∈ S This means that S , v [x := s ][y := s ] A(x, y ), which is a contradiction Likewise, ∃y ∃xA(x, y ) |= ∃x∃yA(x, y ) 3.4.13 All equivalences follow from the fact that if x does not occur free in Q, then S, v |= Q iff S, v |= Q for every x-variant v of v iff S, v |= ∀xQ 3.4.14 (a) No Consider for example the structure H with P and Q interpreted as the predicates M and W, respectively (c) Yes (e) No Consider for example the structure N with Q interpreted as the predicate < (g) No Consider for example the structure Z with Q(x, y ) interpreted as “x ≤ y ” (i) Yes The two formulae are essentially contrapositives to each other 3.4.15 (b) ¬ ∀x((x = x2 ∧ x > 1) → x2 < 1) ≡ ¬∀x(¬(x = x2 ∧ x > 1) ∨ x2 < 1) ≡ ∃x(x = x2 ∧ x > ∧¬x2 < 1) (d) ¬ ∀x(x = ∨∃y ¬(xy = x)) ≡ ∃x(¬x = ∧¬∃y ¬(xy = x)) ≡ ∃x(¬x = ∧∀y (xy = x)) (f) ¬∃x∃y (x > y ∨ −x > y ) ≡ ∀x∀y (¬x > y ∧ ¬(−x > y ))