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???
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
$\sim \exists x(F(X)\wedge P(X))$
"None of my friends are perfect."
Gatecse