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Question

Answers

A. $p \wedge q$

B. \[p \leftrightarrow q\]

C. $\sim p \vee \sim q$

D. $\sim p \wedge \sim q$

Answer

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We will see another method of solving these types of questions using the Venn diagram concept.

Method-I:

The given expression is $p \vee \left( { \sim p \wedge q} \right)$.

Thus, the negation of this expression can be written as:

$ \sim \left( {p \vee \left( { \sim p \wedge q} \right)} \right)$

$ = \sim p \wedge \sim \left( { \sim p \wedge q} \right)$ $\left[ {\because \sim \left( {a \vee b} \right) \equiv \sim a \wedge \sim b} \right]$

$ = \sim p \wedge \left( {p \vee \sim q} \right)$ $\left[ {\because \sim \left( {a \wedge b} \right) \equiv \sim a \vee \sim b} \right]$

$ = \left( { \sim p \wedge p} \right) \vee \left( { \sim p \wedge \sim q} \right)$ $\left[ {\because c \wedge \left( {a \vee b} \right) \equiv \left( {c \wedge a} \right) \vee \left( {c \wedge b} \right)} \right]$

$ = F \vee \left( { \sim p \wedge \sim q} \right)$ $\left[ {\because \sim a \wedge a \equiv F} \right]$

$ = \left( { \sim p \wedge \sim q} \right)$ $\left[ {\because F \vee a \equiv a} \right]$

Hence option (D) is the correct answer.

Method-II:

Let us draw the Venn diagram for the two compound statements $p$ and $q$.

Venn Diagram for two statements p and q

In the Venn diagram,

$p$ has its truth value as $T$ in regions $1$ and $3$.

$q$ has its truth value as $T$ in regions $2$ and $3$.

$ \sim p$ has its truth value as $T$ in regions $2$ and $4$.

$ \sim q$ has its truth value as $T$ in regions $1$ and $4$.

The given expression is $p \vee \left( { \sim p \wedge q} \right)$.

Thus, the negation of this expression can be written as $ \sim \left( {p \vee \left( { \sim p \wedge q} \right)} \right)$. We will solve this expression to find region(s) in the Venn diagram in the following steps:

1) $ \sim p \wedge q$ has its truth value as $T$ in region(s) common to regions of $ \sim p$ and $q$.

2) The common region is region $2$ by observation. Thus, $ \sim p \wedge q$ has its truth value as $T$ in region $2$.

3) Now, $p \vee \left( { \sim p \wedge q} \right)$ has its truth value as $T$ in $p \vee region2$, i.e., regions $1,2,3$.

4) The last step is to find the region of Truth Values of $ \sim \left( {p \vee \left( { \sim p \wedge q} \right)} \right)$as $T$. $ \sim \left( {p \vee \left( { \sim p \wedge q} \right)} \right)$ comes by taking negation of ($p \vee region2$). i.e., region $4$.

Let us consider the options now. The one that has region 4 as the $T$ value is the required answer:

A. $p \wedge q$ has Truth value as $T$ in region(s) $2$ [By observing the Venn Diagram]

B. \[p \leftrightarrow q\] has Truth value as $T$ in region(s) $3,4$ [when both $F$ or both $T$]

C. $ \sim p \vee \sim q$ has Truth value as $T$ in region(s) $1,2,4$ [By observing the Venn Diagram]

D. $ \sim p \wedge \sim q$ has Truth value as $T$ in region(s) $4$ [By observing the Venn Diagram]

Hence, $ \sim \left( {p \vee \left( { \sim p \wedge q} \right)} \right) = \sim p \wedge \sim q$