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33 votes
33 votes

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 $P \vee Q$?

  1.    $\neg Q \Box \neg P$
  2.    $P\Box \neg Q$
  3.    $\neg P\Box Q$
  4.    $\neg P\Box \neg Q$
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8 Answers

Best answer
38 votes
38 votes
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 $\bar{ \bar{Q}}+P = P + Q$
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6 votes
6 votes

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’ $

 

$Correct \ Ans \ : Option \ B$

5 votes
5 votes

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

2 votes
2 votes
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.
Answer:

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