edited by
5,772 views
37 votes
37 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. $\sim(\sim A\odot B)$
edited by

8 Answers

Best answer
54 votes
54 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
16 votes
16 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 votes
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 votes
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.
Answer:

Related questions

25 votes
25 votes
2 answers
1
makhdoom ghaya asked Feb 13, 2015
6,307 views
The binary operator $\neq$ is defined by the following truth table.$$\begin{array}{|l|l|l|} \hline \textbf{p} & \textbf{q}& \textbf{p} \neq \textbf{q}\\\hline \text{0} & ...
43 votes
43 votes
2 answers
3
makhdoom ghaya asked Nov 27, 2016
6,401 views
The number of possible commutative binary operations that can be defined on a set of $n$ elements (for a given $n$) is ___________.