Question

negation form
~[(p∧q)→r]=​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\: \sim\bigg((p \land \: q) \to \: r \bigg)$

$\begin{gathered}\boxed{\begin{array}{c|c|c|c|c|c} \bf p & \bf q& \bf r & \bf p \land \: q& \bf \ \: (p \land \: q) \to \: r& \bf \sim \: (p \land \: q) \to \: r\\ \frac{\qquad}{} & \frac{\qquad}{}& \frac{\qquad}{} & \frac{\qquad}{}& \frac{\qquad}{}& \frac{\qquad}{}\\ \sf T & \sf T & \sf T& \sf T& \sf T& \sf F\\ \\\sf T & \sf T & \sf F& \sf T& \sf F& \sf T\\ \\\sf T & \sf F & \sf T& \sf F& \sf T& \sf F\\ \\\sf T & \sf F & \sf F& \sf F& \sf T& \sf F\\ \\\sf F & \sf T & \sf T& \sf F& \sf T& \sf F\\ \\\sf F & \sf T & \sf F& \sf F& \sf T& \sf F\\ \\\sf F & \sf F & \sf T& \sf F& \sf T& \sf F\\ \\\sf F & \sf F & \sf F& \sf F& \sf T& \sf F\\ \\ \sf & \sf \end{array}} \\ \end{gathered}$

Additional Information

Algebra of statements :-

$\red{\boxed{\sf \:p \lor \sim \: p \equiv \: t}}$

$\red{\boxed{\sf \:p \land\sim \: p \equiv \: f}}$

$\red{\boxed{\sf \:p \land\ \: t \equiv \: p}}$

$\red{\boxed{\sf \:p \land\ \: f \equiv \: f}}$

$\red{\boxed{\sf \:p \lor\ \: t \equiv \: t}}$

$\red{\boxed{\sf \:p \lor\ \: f \equiv \: f}}$

$\red{\boxed{\sf \: \sim \: (p \lor\ \: q) \equiv \: \sim \: p \: \land \: \sim \: q}}$

$\red{\boxed{\sf \: \sim \: (p \land\ \: q) \equiv \: \sim \: p \: \lor \: \sim \: q}}$

$\red{\boxed{\sf \: \sim \: f \equiv \: t}}$

$\red{\boxed{\sf \: \sim \: t \equiv \: f}}$