1. **State the problem:** We need to determine if the expression $$(((P \oplus Q) \wedge (Q \lor R)) \to (P \lor R)) \leftrightarrow ((P \to R) \lor (Q \leftrightarrow R))$$ is a tautology, meaning it is true for all truth values of $P$, $Q$, and $R$. 2. **Recall definitions:** - $P \oplus Q$ (exclusive or) is true if exactly one of $P$, $Q$ is true. - $P \to R$ is false only if $P$ is true and $R$ is false; otherwise true. - $Q \leftrightarrow R$ is true if $Q$ and $R$ have the same truth value. 3. **Analyze left side:** - $(P \oplus Q) \wedge (Q \lor R)$ is true if exactly one of $P$, $Q$ is true and at least one of $Q$, $R$ is true. - The implication $((P \oplus Q) \wedge (Q \lor R)) \to (P \lor R)$ is false only if the antecedent is true and the consequent $P \lor R$ is false. 4. **Analyze right side:** - $(P \to R) \lor (Q \leftrightarrow R)$ is true if either $P \to R$ is true or $Q \leftrightarrow R$ is true. 5. **Check all truth assignments:** We test all $2^3=8$ combinations of $P$, $Q$, $R$: | P | Q | R | $P \oplus Q$ | $Q \lor R$ | Left antecedent | $P \lor R$ | Left implication | $P \to R$ | $Q \leftrightarrow R$ | Right side | Biconditional | |---|---|---|-------------|------------|-----------------|------------|-----------------|-----------|---------------------|------------|--------------| | F | F | F | F | F | F | F | T | T | T | T | T | | F | F | T | F | T | F | T | T | T | F | T | T | | F | T | F | T | T | T | F | F | T | F | F | F | | F | T | T | T | T | T | T | T | T | T | T | T | | T | F | F | T | F | F | T | T | F | F | F | F | | T | F | T | T | T | T | T | T | T | T | T | T | | T | T | F | F | T | F | T | T | F | F | F | F | | T | T | T | F | T | F | T | T | T | T | T | T | 6. **Interpretation:** The biconditional is false in some cases (rows 3, 5, 7), so the expression is **not** a tautology. **Final answer:** The expression is not a tautology because it is false for some truth assignments of $P$, $Q$, and $R$.