Question

(.3. Show that (P^~p)^(pvq) is a contra diction.​

Answer 1

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

Given statement is

$\rm \: (p \: \land \sim \: p) \: \land \: (p \: \lor \: q)$

Now, Truth Table is as follow

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

$\rm\implies \:(p \: \land \sim \: p) \: \land \: (p \: \lor \: q) \: is \: contradiction$

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ADDITIONAL INFORMATION

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