Question
Without using truth table prove that : $\sim p \wedge q=(p \vee q) \wedge \sim p$
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$\left|\begin{array}{lll}1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12\end{array}\right|+4\left|\begin{array}{lll}2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6\end{array}\right|=10\left|\begin{array}{lll}1 & 2 & 1 \\ 3 & 1 & 7 \\ 3 & 2 & 6\end{array}\right|$