Question
Statement - 1 : $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is equivalent to $${p \leftrightarrow q}.$$
Statement - 2 : $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is a tautology
A.
Statement - 1 is true, Statement - 2 is true;
Statement - 2 is not a correct explanation for Statement - 1.
B.
Statement - 1 is true, Statement - 2 is false.
C.
Statement - 1 is false, Statement - 2 is true.
D.
Statement - 1 is true, Statement - 2 is true,
Statement - 2 is a correct explanation for statement - 1
Answer :
Statement - 1 is true, Statement - 2 is false.
Solution :
The truth table for the logical statements, involved in statement - 1, is as follows :
| $$p$$
| $$q$$ |
$$ \sim q$$ |
$$p \leftrightarrow \, \sim q$$ |
$$ \sim \left( {p \leftrightarrow \, \sim q} \right)$$ |
$$p \leftrightarrow q$$ |
| T |
T |
F |
F |
T |
T |
| T |
F |
T |
T |
F |
F |
| F |
T |
F |
T |
F |
F |
| F |
F |
T |
F |
T |
T |
We observe the columns for $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ and $${p \leftrightarrow q}$$ are identical, therefore
$$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is equivalent to $${p \leftrightarrow q}$$
But $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is not a tautology as all entries in its column are not $$T.$$