For each formula below, determine whether or not it is 
satisfiable, and whether or not it is tautology:

a) ~(p \/ ~q) -> ~p

  answer: tautology (and therefore satisfiable), because:
       LEQV (p \/ ~q) \/ ~p   (by -> law)
       LEQV p \/ (~q \/ ~p)   (associativity)
       LEQV p \/ (~p \/ ~q)   (commutativity)
       LEQV (p \/ ~p) \/ ~q   (associativity)
       LEQV 1 \/ ~q
       LEQV 1

b) (p -> q) -> (q -> p)
   Satisfiable: p = q = 0, but not tautology: p = 0, q = 1

c) (p /\ (p -> q)) -> q
   Tautology (and therefore satisfiable), because
      LEQV ~(p /\ (p -> q)) \/ q             (by -> law)
      LEQV ~p \/ ~(p -> q) \/ q              (DeMorgan's law)
      LEQV ~p \/ ~(~p \/ q) \/ q             (by -> law)
      LEQV ~p \/ (p /\ ~q) \/ q              (DeMorgan's law)
      LEQV ((~p \/ p) /\ (~p \/ ~q)) \/ q    (distributivity of \/ over /\)
      LEQV (~p \/ ~q) \/ q                   (~p \/ p is tautology)
      LEQV ~p \/ 1
      LEQV 1

d) ~q /\ p /\ (p -> q)
  Unsatisfiable (and therefore, not tautology), because:
      LEQV ~q /\ p /\ (~p \/ q)              (-> law)
      LEQV ~q /\ ((p /\ ~p) \/ (p /\ q))     (distributivity of /\ over \/)
      LEQV ~q /\ (0 \/ (p /\ q))             
      LEQV ~q /\ p /\ q                   
      LEQV (~q /\ q) /\ p                    (commutativity and associativity)
      LEQV 0 /\ p
      LEQV 0

Prove or disprove each of the logical equivalences below, without 
using truth tables:

a) p -> (q /\ r) LEQV (p -> q) /\ (p -> r)

   answer: p -> (q /\ r) 
             LEQV ~p \/ (q /\ r)            (-> law)
             LEQV (~p \/ q) /\ (~p \/ r)    (distributivity of \/ over /\)
             LEQV (p -> q) /\ (p -> r)      (-> law)

b) p /\ (q \/ r) LEQV (p /\ q) \/ r
   False: because with p=0, r=1, the first formula gets 0 but the second one
          gets 1.

c) p <-> q LEQV ~p <-> ~q
   Answer:  p <-> q LEQV (p /\ q) \/ (~p /\ ~q)     (by <-> law)
                    LEQV (~p /\ ~q) \/ (p /\ q)     (commutativity)
                    LEQV (~p /\ ~q) \/ (~~p /\ ~~q) (double negation)
                    LEQV ~p <-> ~q                  (by <-> law)

d) (p /\ q /\ r) \/ (p /\ q /\ ~r) 
                     \/ (p /\ ~q /\ r) \/ (p /\ ~q /\ ~r) LEQV p

  Answer:     
   left-hand-side LEQV (p /\ ((q /\ r) \/ (q /\ ~r))) \/
                           (p /\ ((~q /\ r) \/ (~q /\ ~r)))   (by distributivity)
                  LEQV p /\ ((q /\ r) \/ (q /\ ~r) \/ (~q /\ r) \/ (~q /\ ~r))
                                                              (by distrivutivity)
                  LEQV p /\ ((q /\ (r \/ ~r)) \/ (~q /\ (r \/ ~r)))
                  LEQV p /\ (q \/ ~q)
                  LEQV p

Find a CNF formula (using logical equivalence laws) for the following 
formula:
  (~x -> (y /\ x)) /\ (~y -> (x /\ z))

  Answer:   formula is  LEQV (x \/ (y /\ x)) /\ (y \/ (x /\ z)) (by -> law)
                        LEQV (x \/ y) /\ x /\ (y \/ x) /\ (y \/ z) 
                                          (by distributivity of \/ over /\)