2.9k views

What is the logical translation of the following statement?

"None of my friends are perfect."

1. $∃x(F (x)∧ ¬P(x))$
2. $∃ x(¬ F (x)∧ P(x))$
3. $∃x(¬F (x)∧¬P(x))$
4. $¬∃ x(F (x)∧ P(x))$
edited | 2.9k views
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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"

1. some of my friends are not perfect
2. some of those who are not my friends are perfect
3. some of those who are not my friends are not perfect
4. NOT (some of my friends are perfect) / none of my friends are perfect

Correct Answer: $D$

edited
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second part of (D) is correct. But "NOT all of my friends are perfect" is not correct. It means some of my friends are not perfect but actually none of my friends are perfect. So the correct way to put (D) would be "There does not exist any friend of mine who is perfect" or "none of my friends are perfect" .

For (B) and (C) you should replace "those" with "some of those" because by default it will take "All" as per English grammar.
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edited now........ yah you are right i must have written some of those
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ok :) NOT (all of my actually it should be NOT (some of my

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got it atlast
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@ arjun sir,

Can we retransform the given statement like this...."None of my friends are perfect" to "All my freinds are not perfect"  so that we can directly write like this, for all x( F(X)-->~P(X)) ....

What do you say?

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None of my friends are perfect can be written an "all of my friends are not perfect".

Now all of my friends are not perfect could be written as:

$\forall x (F(x) \implies \lnot P(x))$

$\forall x (\lnot F(x) \cup \lnot P(x))$

By de Morgan's law: -

$\lnot \exists(F(x) \cap P(x))$

Which is equivalent to option D.

Sir, what is the right way to interpret it?

$F\left(x\right): x \text{ is my friend.}$

$P\left(x\right):\text{x is perfect.}$

$\text{None of my friends are perfect"}$ can be written like

$\forall x[F(x)\implies\neg P(x)]$
$\equiv \forall x[\neg F(x)\vee \neg P(x)]$
$\equiv \forall x\neg[F(x)\wedge p(x)]$
$\equiv \neg\exists x[F(x)\wedge P(x)]$

edited by
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∀x~[F(x)∧p(x)]  is not equal to ~∃x[F(x)∧P(x)]

both will be equal if the case would be

~[∀x~[F(x)∧p(x)]]  then it would be ~∃x[F(x)∧P(x)]
Isn't it?
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∀x~[F(x)->P(x)] Is this valid for None of My friends???

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@sourav

no
~[∀x~[F(x)∧p(x)]] it would be ∃x[F(x)∧P(x)]
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@Avinesh

No
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@vnc

Can u elaborate after 2nd last step?
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why you again negate theirexist?????in last line???

plz tell me my approach is right or not

for all people in the world who is my friend then not perfect i.e ∀x[F(x)--->~P(x)]

=∀x[~F(x)∨~P(x)]
=∀x~[F(x)∧p(x)] now according to me last line should be...                                                                      =∃x[F(x)∧P(x)] but why you put negation??????
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Wrong calculation ... correct it ...
"None of my friends are perfect."

It is NOT the complement of "All of my friends are perfect"   So A is not the answer. (A frequently done mistake)

It is the complement of "At least one of my friend is perfect"  So D is the answer.
edited by
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If I take complement of "All of my friends are perfect" then what will it mean? What will be the semantic meaning of it.
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@ Abhishek Gupta 1
"At  least one of my friends is not perfect."

"None of my friends are perfect."  option d is correct

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none of my friend are perfect

is the opposite of this "All of my friend are perfect "?correct ot not
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i think opposite of "All of my friend are perfect" is Not all of my friend are perfect.
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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

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

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