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Let us construct a truth table for each of these compound propositions.

(i) $(p → q) ↔ (¬q → ¬p)$

$\begin{array}{||c|c||c|c|c|c|c||} \hline\hline p & q & p → q & \neg q & \neg p & ¬q → ¬p & (p → q) ↔ (¬q → ¬p) \\ \hline\hline 0 & 0 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 1 & 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 1 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 0 & 0 & 1 & 1\\ \hline\hline \end{array}$

ii) $p ⊕ (p ∨ q)$

$\begin{array}{||c|c||c|c|c|c|c||} \hline\hline p & q & p ∨ q & p ⊕ (p ∨ q) \\ \hline\hline 0 & 0 & 0 & 0\\ \hline 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 1 & 0\\ \hline 1 & 1 & 1 & 0 \\ \hline\hline \end{array}$

(i) Let us determine by using truth tables if $(p ∧ q) → p$ is a tautology, contradiction or a contingency.

$\begin{array}{||c|c||c|c||} \hline\hline p & q &p ∧ q & (p ∧ q) → p \\ \hline\hline 0 & 0 & 0 & 1\\ \hline 0 & 1 & 0 & 1 \\ \hline 1 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 1\\ \hline\hline \end{array}$

Since the last column contains only 1, we conclude that this formula is a tautology. Therefore, this formula neither a contradiction, nor a contingency.

(ii) Let us show that $¬(p ⊕ q)$ and $p ↔ q$ are logically equivalent using the truth table.

$\begin{array}{||c|c||c|c|c||} \hline\hline p & q & p ⊕ q & ¬(p ⊕ q) & p ↔ q \\ \hline\hline 0 & 0 & 0 & 1 & 1\\ \hline 0 & 1 & 1 & 0 & 0 \\ \hline 1 & 0 & 1 & 0 & 0\\ \hline 1 & 1 & 0 & 1 &1\\ \hline\hline \end{array}$

Since the last two columns are coinside, the formulas $¬(p ⊕ q)$ and $p ↔ q$ are logically equivalent.