Register

GATE CSE 1996 | Question: 2.3

• edited by
8,412 views
29 votes
29 votes

Which of the following is NOT True?

(Read $\wedge$ as AND, $\vee$ as OR, $\neg$ as NOT,  $\rightarrow$ as one way implication and $\leftrightarrow$ as two way implication)

  1. $((x \rightarrow y) \wedge x) \rightarrow y$

  2. $((\neg x \rightarrow y) \wedge (\neg x \rightarrow \neg y)) \rightarrow x$

  3. $(x \rightarrow (x \vee y))$

  4. $((x \vee y) \leftrightarrow (\neg x \rightarrow \neg y))$

• edited by
User Avatar

10 Answers

3 votes
3 votes
option D.
((x or y) <-> (x' -> y' ))
=((x+y) <-> (x+y'))
=(x+y)'.(x+y')' + (x+y).(x+y') (as we know <-> evaluated as Xnor gate like A<->B = A'B' +AB)
=x'y'.x'.y + x+yy'
=0+x+0
=x
value of x may be 0 or 1 so it need not be false always
but all other option are true ( getting 1 after evaluating ) so closest possible ans is D
User Avatar
1 votes
1 votes

The question is saying which of the following is false not contradiction ..so by making the truth values u will see that options A,B and C are tutology . Here I'll prove C  (make truth table or follow the properties)  :

X->(XvY)=> ~XvXvY  => ~XvX is T and TvY=>T so option C is  a tutology

now coming to D make truth table(here only 2 variables so it's not that difficult)  n you will see that when x=F and y=F .It is False .

NOTE: option D is not a contradiction but it is a contigency 

User Avatar
0 votes
0 votes
option (d) because bi-implies means LHS should be equivalent to RHS, but here : L.H.S = (X V Y)

 

R.H.S = (~X ---> ~Y) = (~ ~X V ~Y) = (X V ~Y).
• edited by
User Avatar
0 votes
0 votes

Let us assume the given proposition is false first then if we get any contradiction, our assumption is wrong i.e., the given proposition is not false

 

User Avatar
Page:
Voting & Accepting an answer helps improve the community
Answer:
← PreviousNext →
← Previous in categoryNext in category →

Related questions

27 votes
27 votes
2 answers
1
Kathleen asked Oct 9, 2014
5,293 views
GATE CSE 1996 | Question: 3
Let $f$ be a function defined by $f(x) = \begin{cases} x^2 &\text{ for }x \leq 1\\ ax^2+bx+c &\text{ for } 1 < x \leq 2 \\ x+d &\text{ for } x>2 \end{cases}$ Find the values for the constants $a$, $b$, $c$ and $d$ so that $f$ is continuous and differentiable everywhere on the real line.
Let $f$ be a function defined by$$f(x) = \begin{cases} x^2 &\text{ for }x \leq 1\\ ax^2+bx+c &\text{ for } 1 < x \leq 2 \\ x+d &\text{ for } x>2 \end{cases}$$Find the va...
User Avatar
34 votes
34 votes
6 answers
2
Kathleen asked Oct 9, 2014
8,762 views
GATE CSE 1996 | Question: 1.3
Suppose $X$ and $Y$ are sets and $|X| \text{ and } |Y|$ are their respective cardinality. It is given that there are exactly $97$ functions from $X$ to $Y$. From this one can conclude that $|X| =1, |Y| =97$ $|X| =97, |Y| =1$ $|X| =97, |Y| =97$ None of the above
Suppose $X$ and $Y$ are sets and $|X| \text{ and } |Y|$ are their respective cardinality. It is given that there are exactly $97$ functions from $X$ to $Y$. From this one...
User Avatar
33 votes
33 votes
6 answers
3
Kathleen asked Oct 9, 2014
24,878 views
GATE CSE 1996 | Question: 2.25
A micro program control unit is required to generate a total of $25$ control signals. Assume that during any micro instruction, at most two control signals are active. Minimum number of bits required in the control word to generate the required control signals will be: $2$ $2.5$ $10$ $12$
A micro program control unit is required to generate a total of $25$ control signals. Assume that during any micro instruction, at most two control signals are active. Mi...
User Avatar
40 votes
40 votes
5 answers
4
Kathleen asked Oct 9, 2014
12,977 views
GATE CSE 1996 | Question: 2.23
Consider the following state table for a sequential machine. The number of states in the minimized machine will be ... $4$ $3$ $2$ $1$
Consider the following state table for a sequential machine. The number of states in the minimized machine will be$$\begin{array}{|l|l|ll|}\hline \text{} & \text{} & \tex...
User Avatar
Link Copied!