GATE CSE 2009 | Question: 24

Question

The binary operation $\Box$ is defined as follows

$$\begin{array}{|c|c|c|} \hline \textbf{P} & \textbf{Q} & \textbf{P} \Box \textbf{Q}\\\hline \text{T} & \text{T}& \text{T}\\\hline \text{T} & \text{F}& \text{T} \\\hline \text{F} & \text{T}& \text{F}\\\hline \text{F} & \text{F}& \text{T} \\\hline \end{array}$$

Which one of the following is equivalent to $P \vee Q$?

• A.    $\neg Q \Box \neg P$
• B.    $P\Box \neg Q$
• C.    $\neg P\Box Q$
• D.    $\neg P\Box \neg Q$

Comments

When P = 0 and Q = 0, $ P \vee Q = 0$

When P = 0 and Q = 1, $ P $ $ \square $ $Q = 0$

Only for option B, does this hold true.

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By looking at the truth table you can easily say that ΁ is ← .or P← Q.Therefore Option B is ~Q→ P which is equivalent to ~(~Q) V P=Q V P

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u can even draw table according to the options given and in all four table option (B) table matches with the given diagram in question

Answer 1

Clearly,

$P\ \square \ Q \equiv Q\rightarrow P \equiv Q' \vee P$

according to the given truth table.

A. $Q' \ \square \ P' \ \equiv \ P' \rightarrow Q' \equiv (P')' \vee Q' \equiv P \vee Q'$

B. $P \ \square \ Q' \ \equiv \ Q' \rightarrow P \equiv (Q')' \vee P \equiv {\color{Green} {Q \vee P}}$

C. $P' \ \square \ Q \ \equiv \ Q \rightarrow P' \equiv Q' \vee P'$

D. $P' \ \square \ Q' \ \equiv \ Q' \rightarrow P' \equiv (Q')' \vee P' \equiv Q \vee P'$

So, Option (B) is correct.

Answer 2

From the truth table we can get that $P\;\square\;Q\;is \;Q\rightarrow P$. So the options will be-

A.  $\neg P\rightarrow\neg Q\equiv\;\neg(\neg P)\vee\neg Q\;\equiv P\vee\neg Q$    $(\because P\rightarrow Q\equiv \neg P\vee Q)$

B.  $\neg Q\rightarrow P\;\equiv\;\neg(\neg Q)\vee P\;\equiv\;$ $P\vee Q$  (correct option)

 

Answer 3

$\begin{array}{|c|c|c|} \hline \textbf{P} & \textbf{Q} & \textbf{P} \Box \textbf{Q}\\\hline \text{T} & \text{T}& \text{T}\\\hline \text{T} & \text{F}& \text{T} \\\hline \text{F} & \text{T}& \text{F}\\\hline \text{F} & \text{F}& \text{T} \\\hline \end{array}$

From the truth table, $ P\Box Q$ can be written as$P\Box Q = \left ( P\wedge Q \right ) \vee \left ( P\wedge \sim Q \right ) \vee \left ( \sim P \vee \sim Q \right )\\ =P\wedge \left ( Q\vee \sim Q \right ) V (\sim P \wedge \sim Q )\\ =P \wedge True \vee (\sim P \wedge \sim Q )\\ =P \vee (\sim P \wedge \sim Q )\\ =\left ( P \vee \sim P \right ) \wedge \left ( P \vee \sim Q \right )\\ = True\wedge \left ( P \vee \sim Q \right ) \\= \left ( P \vee \sim Q \right )$

A.  $\neg Q \Box \neg P\equiv \sim Q \vee \sim\left ( \sim P \right ) \equiv \sim Q \vee P$

B.$\ P \Box \neg Q\equiv \ P\vee \sim\left ( \sim Q \right ) \equiv P \vee Q$

$\therefore$Answer is Option B

 

Comments

source:neso academy

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In this  Question, the process of deducing equivalent formula for $P\Box Q$ becomes easy considering the $false$ case unlike $true$ in the above answer

$P\Box Q$ is $false$ when $P = false$ and $Q=true$

Answer 4

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