What is the logical translation of the following statement? "None of my friends are perfect."
ok :) NOT (all of my actually it should be NOT (some of my
∀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)] Is this valid for None of My friends???
@ Abhishek Gupta 1 "At least one of my friends is not perfect."
"None of my friends are perfect."
option d is correct
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'
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
Gatecse
There is one more problem. Ppl who have...