1) A tautology
2) A contradiction
3) Neither a tautology nor a contradiction
4) Cannot come to any conclusion
Solution: (3) Neither a tautology nor a contradiction
|
p |
q |
p ⇒ q |
~ (p ⇒ q) |
~ p |
~ q |
~ p ˅ ~ q |
~ (p ⇒ q) ⇔ (~ p ˅ ~ q) |
|
T |
T |
T |
F |
F |
F |
F |
T |
|
T |
F |
F |
T |
F |
T |
T |
T |
|
F |
T |
T |
F |
T |
F |
T |
F |
|
F |
F |
T |
F |
T |
T |
T |
F |
The last column shows that the result is neither a tautology nor a contradiction.
