Problem- 26
Show that each of these conditional statements is a tautology by completing the truth tables.
We conclude that each of these conditional statements is a tautology because the entries in the last column contain ………..?
a) (p∧q)→p
b) p→(p∨q)
c) ¬p→(p→q)
d) (p∧q)→(p→q
e) ¬(p→q)→p
f) ¬(p→q)→¬q
Solution
a) (p∧q)→p
|
p |
q |
p∧q |
(p∧q)→p |
|
T |
T |
T |
T |
|
T |
F |
F |
T |
|
F |
T |
F |
T |
|
F |
F |
F |
T |
b) p→(p∨q)
|
p |
q |
p∨q |
p→(p∨q) |
|
T |
T |
T |
T |
|
T |
F |
T |
T |
|
F |
T |
T |
T |
|
F |
F |
F |
T |
c) ¬p→(p→q)
|
p |
q |
¬p |
p→q |
¬p→(p→q) |
|
T |
T |
F |
T |
T |
|
T |
F |
F |
F |
T |
|
F |
T |
T |
T |
T |
|
F |
F |
T |
T |
T |
d) (p∧q)→(p→q)
|
p |
q |
p∧q |
p→q |
(p∧q)→(p→q) |
|
T |
T |
T |
T |
T |
|
T |
F |
F |
F |
T |
|
F |
T |
F |
T |
T |
|
F |
F |
F |
T |
T |
e) ¬(p→q)→p
|
p |
q |
p→q |
¬(p→q) |
¬(p→q)→p |
|
T |
T |
T |
F |
T |
|
T |
F |
F |
T |
T |
|
F |
T |
T |
F |
T |
|
F |
F |
T |
F |
T |
f) ¬(p→q)→¬q
|
p |
q |
p→q |
¬(p→q) |
¬q |
¬(p→q)→¬q |
|
T |
T |
T |
F |
F |
T |
|
T |
F |
F |
T |
T |
T |
|
F |
T |
T |
F |
F |
T |
|
F |
F |
T |
F |
T |
T |
We conclude that each of these conditional statements is a tautology because the entries in the last column contain all Ts.