Question

Let
$
  \;\left( i \right)\;(p \vee q) \vee (p \vee \sim q),\;{\text{ }}\;{\text{ }}\; \\
  \left( {ii} \right)\;(p \wedge q) \wedge (p \vee \sim q),\;{\text{ }}\; \\
  \left( {iii} \right)\;(p \vee q) \wedge (p \vee \sim q),\;{\text{ }}\; \\
  \left( {iv} \right)\;(p \vee q) \vee (p \wedge \sim q) \\
$
Which one is tautology?

$
  (a){\text{ (i)}} \\
  (b){\text{ (ii)}} \\
  (c){\text{ (iii)}} \\
  (d){\text{ (iv)}} \\
$

Answer 1

Hint – A tautology is a statement that is always true, no matter what. If you construct a truth table for a statement and all of the column values for the statement are true (T), then the statement is a tautology because it's always true!

Complete step-by-step answer:
$
  \left( i \right)\;(p \vee q) \vee (p \vee \sim q) = p \vee (q \vee \sim q) = p \vee t = t \\
  \left( {ii} \right)\;(p \wedge q) \wedge (p \vee \sim q) \\
  \left( {iii} \right)\;(p \vee q) \wedge (p \vee \sim q) = p \vee (q \wedge \sim q) - p \vee f = p \\
  \left( {iv} \right)\;(p \vee q) \vee (p \wedge \sim q) \\
$
\[\begin{array}{*{20}{l}}
  p&q&{ \sim q}&{p \vee q}&{p \wedge \sim q}&{(p \vee q) \vee (p \wedge \sim q)} \\
  T&T&F&T&F&T \\
  T&F&T&T&T&T \\
  F&T&F&T&F&T \\
  F&F&T&F&F&F
\end{array}\]
$
  (p \wedge q) \wedge (p \vee \sim q) \\
  \begin{array}{*{20}{l}}
  p&q&{ \sim q}&{p \wedge q}&{p \wedge \sim q}&{(p \wedge q) \wedge (p \vee \sim q)} \\
  T&T&F&T&T&T \\
  T&F&T&F&T&F \\
  F&T&F&F&F&F \\
  F&F&T&F&T&F
\end{array} \\
$
Therefore, you can see now option (a) is correct.

Note – In this problem, first we try to construct a truth table for all the given statements. After that we will see whether these statements are a tautology or not. Hence, we get the required answer.