Question

using the truth table prove that the following is a logical equivalence.
1)~p^q=(p v q)^~p​

Answer 1

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

Consider,

$\red{\rm :\longmapsto\: \sim \: p \: \land \: q}$

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

Now, Consider,

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

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

Hence, from above, we concluded that

$\red{\bf :\longmapsto\: \sim \: p \: \land \: q \: = \: (p \: \lor \: q) \land \: \sim \: p}$

Additional Information :-

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