Answer
The provided statement is neither a tautology nor a self-contradiction.
Work Step by Step
A conjunction \[\wedge \] is true only when both simple statements are true.
The provided compound statement is\[\left[ \left( q\to \sim r \right)\wedge \left( \sim r\to p \right) \right]\leftrightarrow \left( q\wedge \sim p \right)\].It can be first deduced by finding negation of \[p,r\] as\[\sim p,\sim r\].
Now, find\[\left( q\to \sim r \right)\]. It is a conditional statement, which would mean that it is false only when the former statement is true, and the latter statement is false. In all other cases, it is true.
Similarly, find\[\left( \sim r\to p \right)\].
Then, find\[\left[ \left( q\to \sim r \right)\wedge \left( \sim r\to p \right) \right]\]by taking AND combination or conjunction of two ingredient variables. Similarly find\[\left( q\wedge \sim p \right)\].
Then at last, find\[\left[ \left( q\to \sim r \right)\wedge \left( \sim r\to p \right) \right]\leftrightarrow \left( q\wedge \sim p \right)\]. It is a biconditional statement, which would mean that it is true only when both the simple statements\[\left[ \left( q\to \sim r \right)\wedge \left( \sim r\to p \right) \right]\]and \[\left( q\wedge \sim p \right)\]assumes the same truth value.
To construct the truth table, draw the table having 5 rows and 9 columns in which the elements of the first row are:
\[\begin{align}
& p,q,r,\sim r,\left( q\to \sim r \right),\left( \sim r\to p \right),\left[ \left( q\to \sim r \right)\wedge \left( \sim r\to p \right) \right], \\
& \left( q\wedge \sim p \right),\left[ \left( q\to \sim r \right)\wedge \left( \sim r\to p \right) \right]\leftrightarrow \left( q\wedge \sim p \right) \\
\end{align}\]
The final output is obtained as follows: