GATE CSE 2006 | Question: 28

Question

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 $A\wedge B$ ?

• 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

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Detailed Video Solution

Answer 1

therefore option D is correct

Answer 2

Make the K-Map , then write the expression and check the options.

Answer 3

A	B	~A	~B	(~A op B)	~(A op ~B)	~(~A op ~B)	~(~A op B)	A ⋀ B
T	T	F	F	F	F	F	T	T
T	F	F	T	T	F	T	F	F
F	T	T	F	T	F	F	F	F
F	F	T	T	T	T	F	F	F

op is the binary logical realtion mentioned in the question.

So second last column i.e. option D matches AND operation of A,B.

Answer 4

.