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Which one of the following, statements is not a tautology
$\left( {p \vee q} \right) \to \left( {p \vee \left( { \sim q} \right)} \right)$
$\left( {p \vee q} \right) \to p$
$p \to \left( {p \vee q} \right)$
$\left( {p \wedge q} \right) \to \left( { \sim p} \right) \vee q$
Solution
$\left( A \right)\,\,\,\,\left( {p \vee q} \right) \to \left( {p \vee \left( { \sim q} \right)} \right)$
$ = \sim \left( {p \vee q} \right) \vee \left( {p \vee \sim q} \right)$
$ = \left( { \sim p \wedge \sim q} \right) \wedge \left( {p \vee \sim q} \right)$
$ \ne T$
$\left( B \right)\,\,\,\,\left( {p \wedge q} \right) \to p$
$ = \sim \left( {p \wedge q} \right) \vee p = \left( { \sim p \vee \sim q} \right) \vee p$
$ = \left( { \sim p \wedge p} \right) \vee \sim q$
$ = T$
$\left( C \right)\,\,\,\, \sim p \vee \left( {p \vee q} \right)$
$ = \left( { \sim p \vee p} \right) \vee q = T$
$\left( D \right)\,\,\,\, \sim \left( {p \vee q} \right) \vee \left( { \sim p \vee q} \right)$
$ = \left( { \sim p \vee \sim q} \right) \vee \left( { \sim p \vee q} \right)$
$ = \sim p \vee T = T$