Consider the NAND operator $\uparrows, which is defined as follows $p \uparrow q \equiv eg (p \wedge q)$. (1) (3 points) Using truth tables show that $\wedge$ is associative, i.e., $p \wedge (9 \wedge r \equiv(p \wedge q) \wedge r$. \begin{tabular}{ccc||CCCC) $\mathrm{p} $ & $\mathrm{q} $ & $\mathrm{r}$ & $(q\wedge r$ & $(p \wedge (9 \wedge r))$ & $(p \wedge 9)$ & $((p \wedge q) \wedge r$ & $(p wedge (9 \wedger)) Leftrightarrow(p \wedge q \wedge r$ \hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & & & & & 1 $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & & & & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ & & & & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & & & & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & & & & W $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{F}$ & & & & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & & & & W $\mathrm{F}$ & $\mathrm{F} $ & $\mathrm{F} $ & & & & \end{tabular) (2) (4 points) Using truth tables show that $\uparrow is not associative, i.e., $p \uparrow(q \uparrow r ot \equiv(p \uparrow q) \uparrow r$. CS.PB. 007