Question

Verify that the proposition p^(q ^ ¬p) is a Contradiction, use truth table.

Answer 1

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

The given preposition is

$\rm :\longmapsto\:p \wedge \: (q \: \wedge \: \sim \: p)$

Truth table is as follow :-

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

Hence,

The given preposition

$\rm :\longmapsto\:p \wedge \: (q \: \wedge \: \sim \: p) \: is \: contradiction.$

Additional Information :-

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