1.6k views

The binary operation □ is defined as follows

 P Q P □ Q T T T T F T F T F F F T

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

1.    $\neg Q □ \neg P$
2.    $P□\neg Q$
3.    $\neg P□Q$
4.    $\neg P□ \neg Q$
edited | 1.6k views
+5
Hint --> Just obtain Sum of Product expression from there answer becomes trivial.
0

Yes it is very easy

Thanks @Chhotu sir

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 A option we get $\bar{ \bar{Q}}+P = P + Q$
selected by
0
(a) option is Q' + P which is false

(b) is right one
0

Yes @NItin answer is b) . It can be solved by doing mapping and then using implication as P□ Q = Q---->p

0
@Arjun Sir please explain the question ?
+1
ans is 'c' ,    given operation in the truth table is implication ,  Q---->P .    option c is equivalent to P+Q.
+1
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.

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

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.
–1
Answer is C as □ is equal to implication .On applying implication in C we get P or Q.
0
We have to find P V Q.

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)

answered ago by Active (3.8k points)

1
2