GATE2006-28

Question

A logical binary relation $\odot$, is defined as follows: 

$A$	$B$	$A\odot B$
True	True	True
True	False	True
False	True	False
False	False	True

Let $\sim$ be the unary negation (NOT) operator, with higher precedence then $\odot$.

Which one of the following is equivalent to $A\wedge B$ ?

• A. $(\sim A\odot B)$
• B. $\sim(A \odot \sim B)$
• C. $\sim(\sim A\odot\sim B)$
• D. $\sim(\sim A\odot B)$

Comments

plz someone answer it, using truth table method

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Oher way

Use kmap to obtain expression, then try to match with options.

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We can also used truth table bcz they have only two variables

Answer 1

This question is easier to answer With Boolean Algebra.

$A\odot B \equiv B\to A$, i.e. $(\neg B\vee A).$

Now, lets look at Option D

$\neg (\neg A\odot B)$
$\quad \equiv \neg (B\to \neg A)$
$\quad \equiv \neg (\neg B\vee \neg A)$
$\quad \equiv B\wedge A$

So Answer is D.

Other options:

• A. $\neg B\vee \neg A$
• B. $\neg B\wedge \neg A$
• C. $\neg B\wedge  A$

Answer 2

Instead of checking all the answers, we can approach similar problems by finding out the relationship between the given operation and the one which we have to express.

Here,
\begin{align} A\odot B &= AB+A\bar B+\bar A \bar B\\ &= A+\bar B \\ &=\overline{\bar{A}B}  \end{align} $$ \overline{A\odot B} = \bar AB \implies AB = \color{red}{\overline{\bar A \odot B}}$$

Comments

nice thinking ....

Answer 3

This truth table is of B->A ie B' V A

so operator given in question is implication

we want A ∧B

which is B'->A'

so ans is d

Comments

@pooja, Option b = ~b ∧ ~a.

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Similar que was asked In gate2009 it was asked for or there...i have answered it for gate 2009 i think answer was uploaded at wrong place..sry for that yes ans should be d for this que

Answer 4

we can also Answer this question by putting TRUE for both A and B, and the result should be TRUE for AND Operation,

By this method, only D will qualify.

Answer 5

therefore option D is correct

Answer 6

Option d