Question
Question: \(\sim(p \Rightarrow q) \Leftrightarrow \sim p\mspace{6mu} \vee \sim q\) is...
∼(p⇒q)⇔∼p6mu∨∼q is
A
A tautology
B
A contradiction
C
Neither a tautology nor a contradiction
D
Cannot come to any conclusion
Answer
Neither a tautology nor a contradiction
Explanation
Solution
| 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 |
Last column shows that result is neither a tautology nor a contradiction.