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+15 votes

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. $\sim(\sim A\odot B)$
asked in Set Theory & Algebra by Active (3.7k points)
retagged by | 914 views
plz someone answer it, using truth table method
Oher way

Use kmap to obtain expression, then try to match with options.
We can also used truth table bcz they have only two variables

5 Answers

+22 votes
Best answer

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. $\neg B\wedge  A$
answered by Boss (42.6k points)
edited by
+6 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

answered by Boss (31.4k points)
@pooja, Option b = ~b ∧ ~a.
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
+6 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.

\begin{align} A\odot B &= AB+A\bar B+\bar A \bar B\\ &= A+\bar B \\ &=\overline{\bar{A}B}  \end{align} $$ \overline{A\odot B} = \bar AB \implies AB = \color{red}{\overline{\bar A \odot B}}$$
answered by Active (4.3k points)
edited by
nice thinking ....
0 votes

therefore option D is correct

answered by (29 points)
–1 vote
Option d
answered by Active (4.1k points)

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