376 j;J, Solutions (or Hints) to Chapter-end Exercises
1.6 -,p=pJ--p
p v Q = (P J-- Q) J-- (P J-- Q)
P 1\ Q = (P J-- P) J-- (Q J-- Q)
1.7 (a) The truth table is given in Table Al.2.
TABLE A1.2 Truth Table for Exercise 1.7(a)
P Q
R
PvQ
PvR
RvQ PvR=:,RvQ P v Q =:, ((P v R) =:, (R v Q))
T T
T
T
T
T
T
T
T T
F
T
T
T
T
T
T F
T
T
T
T
T
T
T F
F
T
T
F
F
F
F
T
T
T
T
T
T
T
F
T
F
T
F
T
T
T
F F
T
F
T
T
T
T
F F
F
F
F
F
T
T
1.8 (a) -, P ~ (-, P 1\ Q) == -, (-, P) v (-, P 1\ Q)
by 1 12
== P V (-, P 1\ Q)
by 1 7
== (P V -, P) 1\ (P V Q)
by 1 4
== T 1\ (P V Q)
by Is
==PvQ
by 1 9
-,P~(-,P~ (-,P 1\ Q))== -,(-,P) V (P V Q) by 1 12
== P V (P V Q)
by 1 7
==PvQ
by h and Ii
1.9 We prove Is and 1 6 using the truth table.
TABLE A1.3 Truth Table for Exercise 1.9
P
Q
PI\Q
-,P
-,Q
P v (P 1\ Q)
-, (P 1\ Q) -,Pv-,Q
T
T
T
F
F
T
F
F
T
F
F
F
T
T
T
T
F
T
F
T
F
F
T
T
F
F
F
T
T
F
T
T
P v (P 1\ Q) == P since the columns corresponding to P and
P v (P 1\ Q) are identical; -, (P 1\ Q) =-, P v -, Q is true since the
columns corresponding to -, (P 1\ Q) and -, P v -, Q are identical.
1.11 We construct the truth table for (P ~ -, P) ~ -, P.
TABLE A1.4 Truth Table for Exercise 1.11
P
T
F
-,P
F
T
P =:, -,P
F
T
(P =:, -,P) =:, -,P
T
T
As the column corresponding to (P ~ -, P) ~ -, P has T for all
combinations, (P ~ -, P) ~ -, P is a tautology.