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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 $A\wedge B$ ?

1. $(\sim A\odot B)$
2. $\sim(A \odot \sim B)$
3. $\sim(\sim A\odot\sim B)$
4. $\sim(\sim A\odot B)$
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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

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

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

Other options:

1. $\neg B\vee \neg A$
2. $\neg B\wedge \neg A$
3. $\neg B\wedge A$
edited by
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} $$\overline{A\odot B} = \bar AB \implies AB = \color{red}{\overline{\bar A \odot B}}$$
edited by
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nice thinking ....

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

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@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
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.

therefore option D is correct

–1 vote
Option d

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