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GATE CSE 2006 | Question: 28

• edited by
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44 44 votes

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

$$\begin{array}{|l|l|l|} \hline \textbf{A} & \textbf{B}& \textbf{A} \odot \textbf{B}\\\hline \text{True} & \text{True}& \text{True}\\\hline \text{True} & \text{False}& \text{True}\\\hline \text{False} & \text{True}& \text{False}\\\hline \text{False} & \text{False}& \text{True}\\\hline   \end{array}$$

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. $\sim(\sim A\odot B)$

8 Answers

Best answer
68 68 votes

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:

  1. $\neg B\vee \neg A$
  2. $\neg B\wedge \neg A$
  3. $\neg B\wedge  A$
• edited by
19 19 votes
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
9 9 votes

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

6 6 votes
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.
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