The statement (sim( p Leftrightarrow sim q )) | vedclass

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(C) We are given the expression $(\sim( p \Leftrightarrow \sim q )) \wedge q$.First,recall that $\sim( p \Leftrightarrow \sim q )$ is equivalent to $p

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equivalent to $( p \Rightarrow q ) \wedge q$

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(C) We are given the expression $(\sim( p \Leftrightarrow \sim q )) \wedge q$.First,recall that $\sim( p \Leftrightarrow \sim q )$ is equivalent to $p \Leftrightarrow q$ because $\sim( p \Leftrightarrow \sim q ) \equiv p \Leftrightarrow \sim(\sim q) \equiv p \Leftrightarrow q$.Now,the expression becomes $( p \Leftrightarrow q ) \wedge q$.We know that $p \Leftrightarrow q$ is $( p$ $\Rightarrow q ) \wedge ( q$ $\Rightarrow p )$.So,$( p \Leftrightarrow q ) \wedge q \equiv ( p$ $\Rightarrow q ) \wedge ( q$ $\Rightarrow p ) \wedge q$.Since $( q \Rightarrow p ) \wedge q \equiv q \wedge p$,the expression simplifies to $( p \Rightarrow q ) \wedge ( p \wedge q ) \equiv p \wedge q$.Alternatively,checking the truth table for $( p \Leftrightarrow q ) \wedge q$:If $p=T, q=T$,then $(T \Leftrightarrow T) \wedge T = T \wedge T = T$.If $p=F, q=T$,then $(F \Leftrightarrow T) \wedge T = F \wedge T = F$.If $p=T, q=F$,then $(T \Leftrightarrow F) \wedge F = F \wedge F = F$.If $p=F, q=F$,then $(F \Leftrightarrow F) \wedge F = T \wedge F = F$.Comparing this with option $(C)$: $( p \Rightarrow q ) \wedge q$.If $p=T, q=T$,then $(T \Rightarrow T) \wedge T = T \wedge T = T$.If $p=F, q=T$,then $(F \Rightarrow T) \wedge T = T \wedge T = T$.This does not match.Wait,let's re-evaluate the original expression: $(\sim( p \Leftrightarrow \sim q )) \wedge q$.Since $\sim( p \Leftrightarrow \sim q ) \equiv p \Leftrightarrow q$,the expression is $( p \Leftrightarrow q ) \wedge q$.This is equivalent to $p \wedge q$.Looking at the options,none of the provided options are $p \wedge q$. Let us re-check the logic.Actually,$( p \Rightarrow q ) \wedge q$ is equivalent to $q \wedge p$ if $p$ is true,but generally $( p \Rightarrow q ) \wedge q \equiv q \wedge p$ is not true. Wait,$( p \Rightarrow q ) \wedge q \equiv (\sim p \vee q) \wedge q \equiv q$.Let's re-examine the expression: $(\sim( p \Leftrightarrow \sim q )) \wedge q \equiv (p \Leftrightarrow q) \wedge q$. This is $p \wedge q$.Given the options,there might be a typo in the question or options. However,based on standard logic,$(p \Leftrightarrow q) \wedge q \equiv p \wedge q$.

The statement $(\sim( p \Leftrightarrow \sim q )) \wedge q$ is :

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