1. **Stating the problem:** Construct truth tables for statements 6, 17, 14, and 21 and replace T/F with 1/0. 2. **Step 1: Statement 6a:** "Stocks are increasing but interest rates are steady." - Symbolic form: $s \wedge i$ - Truth table for $s \wedge i$: $$\begin{array}{cc|c} s & i & s \wedge i \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}$$ 3. **Step 2: Statement 17:** $\sim(p \wedge q)$ vs $\sim p \wedge \sim q$ - Truth table: $$\begin{array}{cc|c|c} p & q & \sim(p \wedge q) & \sim p \wedge \sim q \\ \hline 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \end{array}$$ 4. **Step 3: Statement 14:** $p \wedge (q \wedge r)$ - Truth table for $p \wedge (q \wedge r)$: $$\begin{array}{ccc|c} p & q & r & p \wedge (q \wedge r) \\ \hline 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{array}$$ 5. **Step 4: Statement 21:** $(p \wedge q) \wedge r$ vs $p \wedge (q \wedge r)$ - Truth table: $$\begin{array}{ccc|c|c} p & q & r & (p \wedge q) \wedge r & p \wedge (q \wedge r) \\ \hline 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \end{array}$$ 6. **Explanation:** - For 6a, the conjunction $s \wedge i$ is true only when both stocks are increasing and interest rates steady. - For 17, $\sim (p \wedge q)$ is true except when both p and q are true, and $\sim p \wedge \sim q$ is true only when both are false, so the tables show they differ. - For 14 and 21, associativity of $\wedge$ makes the tables for $p \wedge (q \wedge r)$ and $(p \wedge q) \wedge r$ identical. 7. **Final:** All truth values are displayed with 0 (false) and 1 (true) as requested.