Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 2

1. Use equivalence laws to show that the following two formulas are equivalent.

f = (x and y)
g = (x and y) and ((not x or not z) or (y and x))

Sol:
  g
= {second formula} 
  (x and y) and ((not x or not z) or (y and x))
= {Symmetry}    
  (x and y) and ((not x or not z) or (x and y))
= {Symmetry}
  (x and y) and ((x and y) or (not x or not z))
= {Absorption}
  (x and y)
= {first formula}
  f

2. Use truth table to show that the following two formulas are equivalent.

f = (x and y)
g = (x and y) and ((not x or not z) or (y and x))

Sol:
  Consider two cases (x and y) = F and (x and y) = T.
  In the first case, f = F and g = F and (..) = F.
  In the second case, f = T and g = T and ((..) or T) = T and T = T.

3. Consider the following formula.
f = x and not y and not (x and not y or z)
How many value combinations of x, y, and z that make f = F?
How many value combinations of x, y, and z that make f = T?
Is f satisfiable or not?

Sol:
  If x = F, then f = F. 
  (These four value combinations make f = F)

  Otherwise, if x = T and y = T, then f = F. 
  (These two value combinations make f = F)

  Otherwise, x = T and y = F and in this case, f = T and T and not (T and T 
  or z) = F.
  (These two value combinations make f = F)

  Therefore, 8 value combinations of x, y, and z make f = F.

  Therefore, 0 value combinations of x, y, and z make f = T.

  Therefore, f is not satisfiable.

4. Use equivalence laws to show that the following two formulas are equivalent.
f = (x and (not y or not z)) or (z and y and x)
g = x

Sol:
  f 
= {first formula}
  (x and (not y or not z)) or (z and y and x)
= {Symmetry, Symmetry, Symmetry}
  (x and (not y or not z)) or (x and y and z)
= {De-Morgan's}
  (x and not (y and z)) or (x and y and z)
= {Distribution}
  (x and (not (y and z) or (y and z)))
= {not h(y,z) or h(y,z) = T}
  x and T
= {x and T = x}
  x
= {second formula}
  g

5. Use equivalence laws to show that the following two formulas are equivalent.
f = (x and y) -> (y or z)
g = T

Sol:
  f
= {first formula}
(x and y) -> (y or z)
= {Implication}
not (x and y) or (y or z)
= {De-Morgan's}
(not x or not y) or (y or z)
= {Associativity}
not x or (not y or y) or z
= {not y or y = T}
not x or T or z
= {Symmetry and g(x, z) or T = T}
T
= {second formula}
g

6. Use equivalence laws to show that the following two formulas are equivalent.
f = ((x and y) = x)
g = x -> y

Sol:
f 
= {first formula}
(x and y) = x
= {Equality}
((x and y) and x) or (not (x and y) and not x)
= {Symmetry, Associativity, x and x = x}
(x and y) or (not (x and y) and not x)
= {De-Morgan's, Absorption}
(x and y) or not x
= {Distribution}
(x or not x) and (y or not x)
= {x or not x = T}
T and (y or not x)
= {T and h(x,y) = h(x,y)}
y or not x
= {Implication}
x -> y
= {second formula}
g

7. Use Problems 5 and 6 and equivalence laws to show that the following two 
formulas are equivalent.
f = ((x and y) -> (y or z)) and ((x and y) = x)
g = (x -> y)

Sol:
f 
= {first formula}
((x and y) -> (y or z)) and ((x and y) = x)
= {Problem 5}
T and ((x and y) = x)
= {T and h(x,y) = h(x,y)}
(x and y) = x
= {Problem 6}
x -> y
= {second formula}
g

8. Let x be T iff "Alice attends today's lecture". Let y be T iff "Bob attends 
today's lecture". Write a boolean formula f, involving x and y, whose value is T
iff either Alice or Bob attends today's lecture.

Sol:
f = (not x and y) or (x and not y) = x xor y

9. Formula f(x,y,z) = not x and y and z is in Conjunctive Normal Form (CNF). 
Find an equivalent formula g(x,y,z) that is in Disjunctive Normal Form (DNF).

Sol:
  f(x,y,z)
= {CNF}
  not x and y and z
= {h(x,y,z) = h(x,y,z) or F}
  (not x and y and z) or F
= {DNF}
  g(x,y,z)
Note that the formula (not x and y and z) is in both CNF and DNF 

10. Formula f(x,y,z) = (not x and y) or z is in DNF. Find an equivalent 
formula g(x,y,z) that is in CNF.

Sol:
  f(x,y,z)
= {DNF}
  (not x and y) or z
= {Distribution}
  (not x or z) and (y or z)
= {CNF}
  g(x,y,z)

11. Let x be a parameter whose domain of values is the set {2, 3, 5}.
A. Define two predicates P(x) and Q(x) such that 
(Exist x, (P(x) and Q(x))) = (Exist x, P(x)) and (Exist x, Q(x))
B. Define two predicates P(x) and Q(x) such that 
(Exist x, (P(x) and Q(x))) != (Exist x, P(x)) and (Exist x, Q(x))

Sol:
A. P(x) = (x > 2), Q(x) = (x < 5)
B. P(x) = (x = 2), Q(x) = (x = 5)