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

Hint --> Just obtain Sum of Product expression from there answer becomes trivial.

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Yes it is very easy

Thanks @Chhotu sir

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The SOP expression will be :

SOP: (P □ Q )  + (P’ □ Q ) + (P’ □ Q’ ) 

From this , we can obtain the corresponding POS form 

POS: (P □ Q’ ) 

 

Is this what u were mentioning @Chhotu Sir ?

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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

Answer is B because the truth values for option B is same as that of P "or" Q.

The given truth table is for $Q \implies P$ which is $\bar Q+P$.

Now, with B option we get $\bar{ \bar{Q}}+P = P + Q$

Comments

@Arjun Sir please explain the question ?

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ans is 'c' ,    given operation in the truth table is implication ,  Q---->P .    option c is equivalent to P+Q.

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Hello set2018

Question is , there is some operator $\square$ , for which truth table is given , like when first operand is T and second operand is T , o/p will be T. When first operand is T and second is F , o/p will be T...as so on..now just on the bases of this truth table , you have to find truth tabes for Q'$\square$P' , P$\square$Q' , P'$\square$Q , P'$\square$Q' ,now compare those truth tables with truth table for P∨Q , see which one exactly matches with that one.

Answer 2

Lets draw the truth table for $P$ ∨ $Q$  along with $P \ □ \ Q :$ 

 

If we observe the truth table, $P$ ∨ $Q$  and $P \ □ \ Q $ is differ in only last two cases. 

 

So if we replace $Q’$ in place of $Q$ in $P$ ∨ $Q$ truth table, it will be like $:$

 

P	Q	P V Q’
T	T	T
T	F	T
F	T	F
F	F	T

         

This is exactly same truth table as of  $P \ □ \ Q $ 

         

So, from here we can conclude $:$            

         $P$ ∨ $Q’$ $=$ $P \ □ \ Q $

==>  $P$ ∨ $(Q’)’$   $=$ $P \ □ \ Q’ $  [Replacing Q by Q’ in both side]

==>  $P$ ∨ $Q$   $=$ $P \ □ \ Q’ $

 

$Correct \ Ans \ : Option \ B$

Answer 3

Here (P □ Q ) is equivalent to (Q-->P) so by this (P v Q) is equivalent to (P□¬Q).

So the ans is (B) P□¬Q

Answer 4

The answer will be A.

The given truth table is of the form P ⇐ Q which is Q'+P.

and option A satisfies that as Q' ⇐ P' is Q' + P.

Comments

Answer is C as □ is equal to implication .On applying implication in C we get P or Q.

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We have to find P V Q.