Question

prove that:
p^ (~qV p) = p​

Answer 1

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

Consider the given expression is

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

Now,

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

Hence,

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

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

Answer 2

Step-by-step explanation:

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

Consider the given expression is

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

Now,

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

Hence,

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Learn More :-

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