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## Here's the Solution to this Question

a. $(p\wedge q)\vee (p\vee r)=p\vee (q\wedge r)$

Solution.

$\begin{matrix} p & q & r & p\wedge q & p\vee r & (p\wedge q)\vee (p\vee r) & q\wedge r & p\vee (q\wedge r) \\ F & F & T & F & T & T & F & F \end{matrix}$

There is a row in the truth table above where the truth values of $(p\wedge q)\vee (p\vee r)$  and $p\vee (q\wedge r)$  are different. Hence the given statement is false. (In this case, other rows of the truth table are not important.)

b. $(p\vee q)\wedge (p\vee r)=p\vee (q\wedge r)$

Solution.

$\begin{matrix} p & q & r & p\vee q & p\vee r & (p\vee q)\wedge (p\vee r) & q\wedge r & p\vee (q\wedge r) \\ F & F & F & F & F & F & F & F \\ F & F & T & F & T & F & F & F \\ F & T & F & T & F & F & F & F \\ F & T & T & T & T & T & T & T \\ T & F & F & T & T & T & F & T \\ T & F & T & T & T & T & F & T \\ T & T & F & T & T & T & F & T \\ T & T & T & T & T & T & T & T \end{matrix}$

In all rows, the truth values of $(p\vee q)\wedge (p\vee r)$  and $p\vee (q\wedge r)$  are equal. Hence the given statement is true.

c. $\sim (p\vee q) = \sim (p\wedge \sim q)$

Solution.

$\begin{matrix} p & q & p\vee q & \sim (p\vee q) & \sim q & p\wedge \sim q & \sim (p\wedge \sim q) \\ F & T & T & F & F & F & T \end{matrix}$

There is a row in the truth table above where the truth values of $\sim (p\vee q)$  and $\sim (p\wedge \sim q)$  are different. Hence the given statement is false. (In this case, other rows of the truth table are not important.)

d. $(p\wedge q)\wedge (p\vee r)=p\vee (q\wedge r)$
$\begin{matrix} p & q & r & p\wedge q & p\vee r & (p\wedge q)\wedge (p\vee r) & q\wedge r & p\vee (q\wedge r) \\ T & F & F & F & T & F & F & T \end{matrix}$
There is a row in the truth table above where the truth values of $(p\wedge q)\wedge (p\vee r)$  and $p\vee (q\wedge r)$  are different. Hence the given statement is false. (In this case, other rows of the truth table are not important.)