Section 1.3

• 10. a) Forall x F(x,Fred); b)Forall y F(Evelyn,y); c) Forall x Exists y F(x,y); d) ~Exists x Forall y F(x,y); e) Forall y Exists x F(x,y)
• 12.
  • a) Exists s Exists m P(s,junior,m). True.
  • b) Forall S Exists c P(x,c,computer science). False.
  • c) Exists s Exists c Exists m(P(s,c,m) ^ (c ~= junior) ^ (m ~= mathematics)). True.
  • d) Forall s ( Exists c P(s,c,computer science) V Exists m P(s,sophomore,m)). False.
  • e) Exists m Forall c Exists s P(s,c,m). False.
• 20. a) Forall x (P(x) -> ~S(x)); b) Forall x (R(x) -> S(x)); c) Forall x (Q(x) -> P(x)); d) Forall x (Q(x) -> ~R(x)); e) Yes.
• 24. a) Two cases. If A is true, then (Forall x P(x)) V A is true, and since P(x) V A is true for all x, Forall x (P(x)VA) is also true. Thus both sides are true. Now suppose A is false. If P(x) is true for all x, then the left-hand side is true. Furthermore, the right-hand side is also true. On the other hand, if P(x) is false for some x, then both sides are false. Therefore again the two sides are equivalent. A similar argument works for part b.
• 28. We must show that each implies the other. Suppose the first expression is true. Now either P(x) is universally true or Q(x) is universally true. In either case, the second expression (in the exercise statement) is true. Now we must prove the converse. Suppose the second expression is true. If P(x) is universally true, then we are done. Otherwise, P(x0) is not true for some x0. The hypothesis tells us that P(x0) V Q(y) is true, no matter what y is, so Q(y) must be universally true (for all y). We can also say that Q(x) is universally true for all x. Now the first expression is true.
• 30. a) false; b) false; c) true; d) false.
• 32. (P(1) ^ ~P(2) ^ ~P(3)) V (~P(1) V P(2) V ~P(3)) V (~P(1) V ~P(2) V P(3)).
• 34.
  • a) Exists x (P(x) V Q(x) V A)
  • b) Exists x Exists y ~(P(x)V Q(y))
  • c) Forall x Exists ~y (~P(x) V Q(y))