Consider the following formula:
 (~x -> (y /\ x)) /\ (~y -> (x /\ z))

- Give the truth table of this formula.
  (to do this use the method described on pages 146 and 147 of the book)

- Using the truth table, give a DNF and CNF formula that is logically 
  equivalent to it
  Answer:
    DNF: (x /\ ~y /\ z) \/ (x /\ y /\ ~z) \/ (x /\ y /\ z)
    CNF: (x \/ y \/ z) /\ (x \/ y \/ ~z) /\ (x \/ ~y \/ z) /\
         (x \/ ~y \/ ~z) /\ (~x \/ y \/ z)

- Using only the laws for LEQV (and without using the truth table) 
  derive a CNF formula that is LEQV to the above one.

  (~x -> (y /\ x)) /\ (~y -> (x /\ z))
        LEQV (~~x \/ (y /\ x)) /\ (~~y \/ (x /\ z)) (by -> law, twice)
        LEQV (x \/ (y /\ x)) /\ (y \/ (x /\ z))     (double negation, twice)
        LEQV ((x \/ y) /\ (x \/ x)) /\ ((y \/ x) /\ (y \/ z)) (distrib. twice)
        LEQV ((x \/ y) /\ x)) /\ ((y \/ x) /\ (y \/ z)) (Idempotency)
        LEQV x /\ (x \/ y) /\ (y \/ x) /\ (y \/ z)    (Commutativity)

-  XOR (drawn as a O with a "+" inside) is a new connective which has
   the following truth table:

    x  y  x XOR y
   -----|--------
    0  0|    0
    0  1|    1
    1  0|    1
    1  1|    0
 
   Basically, x XOR y is 1 if and only if exactly one of x or y is 1.
   Prove that { XOR, ->} is a complete set of connectives.
  
   Answer: we show that we can make ~ and \/ using XOR and ->, and
   we know that { ~, \/} is a complete set.
   (explain each step in details).

   . By truth table: x XOR x LEQV 0 
   . So, (x XOR x) -> x LEQV 0 -> x 
                        LEQV 1
   . By truth table: x XOR 1 LEQV ~x 
   . Therefore, x XOR ((x XOR x) -> x) LEQV ~x
   . To implement x \/ y we use the fact that x -> y LEQV ~x \/ y
     So x \/ y LEQV ~x -> y.
     We know already how to make ~x using only XOR and ->. Therefore
     x \/ y LEQV (x XOR ((x XOR x) -> x)) -> y


If you have extra time, go over Theorem 5.26 (page 166) of the textbook