Complete the following truth table for $f ( \boldsymbol { A } , \boldsymbol { B } ) = \sim \boldsymbol { A } \wedge \sim \boldsymbol { B }$, then use the truth table to evaluate $f ( 1,1 )$.
$\begin{array}{|c|c|c|c|c|}
\hline \boldsymbol{A} & \boldsymbol{B} & \sim \boldsymbol{A} & \sim \boldsymbol{B} & f(\boldsymbol{A}, \boldsymbol{B}) \
\hline 1 & 1 & & & \
\hline 1 & 0 & & & \
\hline 0 & 1 & & & \
\hline 0 & 0 & & & \
\hline
\end{array}$

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The Answer of Complete the following truth table for $f ( \boldsymbol { A } , \boldsymbol { B } ) = \sim \boldsymbol { A } \wedge \sim \boldsymbol { B }$, then use the truth table to evaluate $f ( 1,1 )$.
$\begin{array}{|c|c|c|c|c|}
\hline \boldsymbol{A} & \boldsymbol{B} & \sim \boldsymbol{A} & \sim \boldsymbol{B} & f(\boldsymbol{A}, \boldsymbol{B}) \
\hline 1 & 1 & & & \
\hline 1 & 0 & & & \
\hline 0 & 1 & & & \
\hline 0 & 0 & & & \
\hline
\end{array}$

Complete the Following Truth Table For f(A,B)=∼A∧∼Bf ( \boldsymbol { A } , \boldsymbol { B } ) = \sim \boldsymbol { A } \wedge \sim \boldsymbol { B }f(A,B)=∼A∧∼B

Complete the Following Truth Table For $f ( \boldsymbol { A } , \boldsymbol { B } ) = \sim \boldsymbol { A } \wedge \sim \boldsymbol { B }$