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The statement $p\rightarrow (q\rightarrow p)$ is equivalent to:
Option: 1
$p\rightarrow q$

Option: 2
$p\rightarrow (p\vee q)$

Option: 3
$p\rightarrow (p\rightarrow q)$

Option: 4
$p\rightarrow (p\wedge q)$

Answers (1)

best_answer

$\begin{array}{|c|c|c|c|c|c|} \hline \mathrm{q} & \mathrm{p} & q \rightarrow p & p \rightarrow(q \rightarrow p) & p \vee q & p \rightarrow(p \vee q) \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\ \hline \end{array}$

since the truth value of $p\rightarrow(q\rightarrow p )\;\;\text{and }\;p\rightarrow(p\vee q)$

are same.

Posted by

vinayak

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