Question

If the Boolean expression $\left( p\oplus q \right)\wedge \left( \sim P\odot q \right)$ is equivalent to $p\wedge q$ where $\oplus ,\odot \in \\{\wedge ,\vee \\}$ then the ordered pair $\left( \oplus ,\odot \right)$ is
$\begin{aligned} & \text{a) }\left( \wedge ,\vee \right) \\\ & \text{b) }\left( \vee ,\vee \right) \\\ & \text{c) }\left( \wedge ,\wedge \right) \\\ & \text{d) }\left( \vee ,\wedge \right) \\\ \end{aligned}$

Answer 1

Now there are 4 possibilities we will create a truth table with values $\left( p\vee q \right)$ , $\left( p\wedge q \right)$ , $(\sim p\vee q)$ , $(\sim p\wedge q)$
Now from the truth table we will check when the expression $\left( p\oplus q \right)\wedge \left( \sim P\odot q \right)$ will become equivalent to $p\wedge q$

Complete step-by-step solution:
Let us prepare the truth table for $\left( p\vee q \right)$ , $\left( p\wedge q \right)$ , $(\sim p\vee q)$ , $(\sim p\wedge q)$
Now let us understand the meaning of these first
Now $\left( p\vee q \right)$ means P or Q hence as the name suggests this expression is true when either of P or Q is true
Now consider $\left( p\wedge q \right)$ this means P and Q. the expression for P and Q is true when Both P and Q are true.
Now note that the symbol ~ is used to express negation
Hence $\sim p$ means the negation of P
Now consider $(\sim p\vee q)$
This expression is true when either of $\sim p$ or q is true
Similarly $(\sim p\wedge q)$ is true when both $\sim p$ and q are true.
Hence now we will prepare the truth table for the following

p & q & \sim p & p\wedge q & p\vee q & \sim p\vee q & \sim p\wedge q & {} \\\ T & T & F & T & T & T & F & {} \\\ T & F & F & F & T & F & F & {} \\\ F & T & T & F & T & T & T & {} \\\ F & F & T & F & F & T & F & {} \\\ \end{matrix}............(1)$$ Now from equation (1) we prepare the truth table for$\left( p\wedge q \right)\wedge \left( \sim p\vee q \right)$ $$\begin{aligned} & \left( p\wedge q \right)\wedge \left( \sim p\vee q \right) \\\ & T \\\ & F \\\ & F \\\ & F \\\ \end{aligned}$$ Similarly from (1) let us prepare the table for $\left( p\vee q \right)\wedge \left( \sim p\vee q \right)$ $$\begin{aligned} & \left( p\vee q \right)\wedge \left( \sim p\vee q \right) \\\ & T \\\ & F \\\ & T \\\ & F \\\ \end{aligned}$$ Now again from equation (1) we will prepare the truth table for $$\left( p\wedge q \right)\wedge \left( \sim p\wedge q \right)$$ $$\begin{aligned} & \left( p\wedge q \right)\wedge \left( \sim p\wedge q \right) \\\ & F \\\ & F \\\ & F \\\ & F \\\ \end{aligned}$$ And finally we prepare the table for $\left( p\vee q \right)\wedge \left( \sim p\wedge q \right)$ from equation (1) $\begin{aligned} & \left( p\vee q \right)\wedge \left( \sim p\wedge q \right) \\\ & F \\\ & F \\\ & T \\\ & F \\\ \end{aligned}$ Hence we get that the truth tables for $\left( p\wedge q \right)\wedge \left( \sim P\vee q \right)$ match with the truth table $\left( p\wedge q \right)$ Hence the equation we want is $\left( p\wedge q \right)\wedge \left( \sim P\vee q \right)$ Comparing the equation with the given equation we get $\left( \wedge ,\vee \right)$ as the correct ordered pair for $\left( \oplus ,\odot \right)$ **Hence option a is the correct option.** ** Note:** While doing such problems we can also solve by the algebraic method we can use properties of Boolean algebra and further solve the equations accordingly. Also while writing truth tables always note that $p\wedge q$ is true only if both p and q are true at once and false if any one of p or q is false. Also, $p\vee q$ is false if both p and q are false and true if any one of p and q is true.