1. **State the problem:** We want to prove that the biconditional operation is associative, i.e., $$(p \leftrightarrow q) \leftrightarrow r = p \leftrightarrow (q \leftrightarrow r).$$ 2. **Recall the biconditional definition:** The biconditional $p \leftrightarrow q$ is true exactly when $p$ and $q$ have the same truth value. 3. **Construct the truth table:** We list all possible truth values for $p$, $q$, and $r$, then compute $p \leftrightarrow q$, $(p \leftrightarrow q) \leftrightarrow r$, $q \leftrightarrow r$, and $p \leftrightarrow (q \leftrightarrow r)$. | $p$ | $q$ | $r$ | $p \leftrightarrow q$ | $(p \leftrightarrow q) \leftrightarrow r$ | $q \leftrightarrow r$ | $p \leftrightarrow (q \leftrightarrow r)$ | |-----|-----|-----|-----------------------|------------------------------------------|-----------------------|------------------------------------------| | T | T | T | T | T | T | T | | T | T | F | T | F | F | F | | T | F | T | F | F | F | F | | T | F | F | F | T | T | T | | F | T | T | F | F | F | F | | F | T | F | F | T | T | T | | F | F | T | T | F | F | F | | F | F | F | T | T | T | T | 4. **Compare columns:** The columns for $(p \leftrightarrow q) \leftrightarrow r$ and $p \leftrightarrow (q \leftrightarrow r)$ are identical for all truth assignments. 5. **Conclusion:** Since the truth values match for all cases, the biconditional operation is associative: $$ (p \leftrightarrow q) \leftrightarrow r = p \leftrightarrow (q \leftrightarrow r). $$