GATE CSE 2013 | Question: 27

Question

What is the logical translation of the following statement?

"None of my friends are perfect."

• A. $∃x(F (x)∧ ¬P(x))$
• B. $∃ x(¬ F (x)∧ P(x))$
• C. $ ∃x(¬F (x)∧¬P(x))$
• D. $ ¬∃ x(F (x)∧ P(x))$

Comments

A) "Some of my friends are not perfect"

B) "Some are not my friends and are perfect"

C) "Some are not my friends and are not perfect"

D) "There exist no one who is my friend and is perfect"

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can we solve it using truth table?

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Yes but it'll take lot of time

Answer 1

I think there should not be any confusion if you know this :

The first-order logic binary form of de Morgan's law is

 

 

reference: https://oeis.org/wiki/De_Morgan%27s_laws

Answer 2

option d is correct

Comments

Ur calculation is wrong ... correct it ..

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Hello set 2018

No , you're doing a common mistake.

It's a common mistake for a person who didn't study 'Logic theory'. Before engineering i also thought like the negation of 'Everyone is intelligent' is 'None is intelligent'.But that's wrong logically.

in logic theory the negation of 'everyone is intelligent' is 'There exists atleast someone who is not intelligent'.

So respectively the negation of None of my friends are perfect  is 'Atleast some of my friends are perfect'

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None of my friend is perfect

$\equiv \forall x( F(X)\Rightarrow \sim P(X) )$

$\equiv \forall X (\sim F(X)\vee \sim P(X))$

$\equiv \sim \sim \forall x(\sim F(X)\vee \sim P(X))$

use one negation to apply demorgan law

$\sim \exists x(F(X)\wedge P(X))$

Answer 3

 

Answer (D)

"None of my friends are perfect."

Its negation is atleast one of my friends is perfect  read the negation again and think

atleast one of my friends is perfect

∃x(F(x)∧P(x))

we use ∧ because I want to test my friends and true value should come from them not from outside people

if we used implication we will get true value for a person who is not even my friend and ∃x to be true only needs one true value.

now negate ∃x(F(x)∧P(x))

=> ¬∃x(F(x)∧P(x))

Answer 4

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