Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 3

1. Show that the following quantified predicate R is equivalent to another 
quantified predicate that has no "not".

R = not (All x, (P(x) -> (Exist y, not Q(x,y))))

Sol:
  R 
= {Predicate R}
  not (All x, (P(x) -> (Exist y, not Q(x,y))))
= {Implication}
  not (All x, (not P(x) or (Exist y, not Q(x,y))))
= {De-Morgan's}
  (Exist x, not (not P(x) or (Exist y, not Q(x,y))))
= {De-Morgan's}
  (Exist x, (P(x) and (not (Exist y, not Q(x,y)))))
= {De-Morgan's}
  (Exist x, (P(x) and (All y Q(x,y))))

2. Let the domains of x and y be the set of all integers. Compute the Boolean 
values of the following quantified predicates:

  All x, Exist y, (x^2 < y)
  Exist y, All x, (x^2 < y)
  Exist x, All y, (x^2 >= y)
  All y, Exist x, (x^2 >= y)

Sol:
  All x, Exist y, (x^2 < y) = T
  Exist y, All x, (x^2 < y) = F
  Exist x, All y, (x^2 >= y) = F 
  All y, Exist x, (x^2 >= y) = T

3. Let the domain of x be the set of all objects and define the following 
predicates:

  B(x) = x is a book
  E(x) = x is expensive
  G(x) = x is good

Express the following English statement as quantified predicates:

  - Some books are not good
  - No book is expensive
  - All expensive books are good
  - Not all expensive books are good

Sol:
  - Some books are not good = Exist x, (B(x) and not G(x))
  - No book is expensive = All x, (B(x) -> not E(x))
  - All expensive books are good = All x, (B(x) and E(x) -> G(x))
  - Not all expensive books are good = Exist x, (B(x) and E(x) and not G(x))

4. Let the domain of x be {2, 3, 7} and the domain of y be {1, 7}. Express
the following quantified predicates as non-quantified predicates:

  Exist x, P(x)
  All x, P(x)
  Exist x, Q(x,1)
  All y, Q(2,y)

Sol:
  Exist x, P(x) = P(2) or P(3) or P(7)
  All x, P(x) = P(2) and P(3) and P(7)
  Exist x, Q(x,1) = Q(2,1) or Q(3,1) or Q(7,1)
  All y, Q(2,y) = Q(2,1) and Q(2,7)

5. Compute the Boolean value of predicate P:

  P = All x Exist z (x < z and not Exist y (x < y and y < z))

in the following two cases:
  Case 1. The domains of x, y, and z are the set of all integers
  Case 2. The domains of x, y, and z are the set of all rational numbers

Sol:
  1. P = T {choose for example z = x + 1, and so there is no y such that 
            x < y and y < z}
  2. P = F {choose for example y = (x + z)/2 and so x < y and y < z}

6. Use equivalence laws to prove that the following two formulas are equivalent
  x and (not z -> (x and y)) and y
  x and y

Sol:
  x and (not z -> (x and y)) and y
= {Associativity and Symmetry}
  (x and y) and (not z -> (x and y))
= {Implication}
  (x and y) and (z or (x and y))
= {Symmetry and Absorption}
  x and y

7. Use equivalence laws to prove that the following two formulas are equivalent:
  x -> y
  not y -> not x

Sol:
  x -> y
= {Implication}
  not x or y
= {Symmetry}
  y or not x
= {not (not y) = y}
  not (not y) or not x
= {Implication}
  not y -> not x
{Comment: If (x -> y) is called direct implication, then (not y -> not x) is 
called indirect implication. And if (not y -> not x) is called direct 
implication, then (x -> y) is called indirect implication.}

8. Let the domains of x, y, and z be the set of all positive integers. 
Give a guessing proof to prove
  Exist x, y, z, x^2 + y^2 = z^2

Sol:
  If x = 3, y = 4, and z = 5, then x^2 + y^2 = 9 + 16 = 25 = z^2.

9. Let the domain of x be the set of all integers. Give a case analysis proof
to prove (All x, x^2 != 5)

Sol:
  Case 1: (x >= -2 and x =< 2): x^2 =< 4 < 5
  Case 2: (x < -2 or x > 2): x^2 >= 9 > 5

10. Give a guessing proof to prove (Exist x, (2^x + 3) is not prime)
The domain of x is the set of all positive integers.

Sol:
  x = 1		2^1 + 3 = 5 is prime
  x = 2		2^2 + 3 = 7 is prime
  x = 3		2^3 + 3 = 11 is prime
  x = 4 	2^4 + 3 = 19 is prime
  x = 5		2^5 + 3 = 35 is not prime
{Comment: Poor man's induction does not always work.}

11. Give a direct inference proof to prove
  (All m,n, (m and n are odd) => (m+n is even))
where the domains of m and n are the set of all  non-negative integers.

Sol:
  m and n are odd
=> {definition of odd}
  m = 2k+1 and n = 2l+1, for some integers k and l
=> {add m + n}
  m+n = (2k+1) + (2l+1), for some integers k and l
=> {arithmetics}
  m+n = 2(k+l+1), for some integers k and l
=> {(k+l+1) is an integer}
  m+n = 2p, for some integer p
=> {definition of even}
  m+n is even