A. By using the truth table (Truth matrix), show that each statement is a tautology, contradictory or
a contingent statement.
1. ¬p ^ q
p
1
0
1
0
¬p ^ q is a contingency
¬p
0
1
0
1
q
0
1
1
0
¬p^q
0
1
0
0
2. ¬(p v q)q
p
q
1
0
0
1
0
0
1
1
¬(p v q)q is a contingency
pvq
1
1
0
1
¬(p v q)
0
0
1
0
¬(p v q)q
1
1
0
1
3. ¬(¬p ^ q) v q
p
q
1
0
0
1
1
1
0
0
¬(¬p^q) v q is a tautology
¬p
0
1
0
1
¬p^q
0
1
0
0
¬(¬p^q)
1
0
1
1
¬(¬p^q) v q
1
1
1
1
4. ¬p(pq)
p
q
1
0
0
1
1
1
0
0
¬p(pq) is a tautology
¬p
0
1
0
1
(pq)
0
1
1
1
¬p(pq)
1
1
1
1
pvq
1
1
1
0
¬q(p v q)
1
1
1
0
5. ¬q(p v q)
p
q
1
0
0
1
1
1
0
0
¬q(p v q) is a contingency
¬q
1
0
0
1
B. By using the truth table (Truth matrix), show that each statement is a tautology, contradictory or
contingent statement.
1. ¬(pq)¬q
p
1
q
0
¬q
1
pq
0
¬(pq)
1
¬(pq)¬q
1
0
1
1
1
0
0
¬(pq)¬q is a tautology
2. [p ^ (pq)]q
p
q
1
0
0
1
1
1
0
0
[p^(pq)]q is a tautology
0
0
1
1
1
1
pq
0
1
1
1
3. (p ^ ¬p)q
p
q
1
0
0
1
1
1
0
0
(p ^ ¬p)q is a tautology
4. (pq)(¬p v q)
p
q
¬p
1
0
0
0
1
1
1
1
0
0
0
1
(pq)(¬p v q) is a contingency
5. ((pq)^(qr))(pr)
p
q
r
pq qr
1
0
1
0
1
0
1
0
1
0
1
1
1
1
1
0
0
0
1
1
1
1
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
1
1
1
((pq)^(qr))(pr) is a tautology
0
0
0
[p ^(pq)]
0
0
1
0
¬p
0
1
0
1
[p^(pq)]q
1
1
1
1
p ^ ¬p
0
0
0
0
pq
0
1
1
1
pr
1
1
1
1
0
1
0
1
1
1
1
¬p v q
0
1
1
1
((pq)^(qr))
0
0
1
1
0
1
0
1
(p ^ ¬p)q
1
1
1
1
(pq)(¬p v q)
0
1
1
1
((pq)^(qr))(pr)
1
1
1
1
1
1
1
1
Exercise 1.10
A. Prove or disprove using the replacement rules that each of the following statement is a
tautology:
1. (p ^ q)(pq) ≡ (p^q)(¬p v q) by material implication
≡ ¬(p^q)v(¬p v q) by material implication
≡(¬q v ¬p)v(¬p v q) De Morgan’s Law
≡¬p v (¬q v q) associative law
≡¬p v T negation law
≡T
2. (pq)(¬p v q) ≡(¬p v q)(¬p v q)
3.
(p v q)(p ^ q) ≡¬(p v q) v (p ^ q) by material implication
≡(¬p ^ ¬q) v (p ^ q)by De Morgan’s Law
≡T by negation