Propositional and First Order Logic.

F(x) ==> x is my friend
P(x) ==> x is perfect

D is the correct answer.

A. There exist some friends which are not perfect

B. There are some people who are not my friend and are perfect

C. There exist some people who are not my friend and are not perfect.

D. There doesn't exist any person who is my friend and perfect 

where ¬ is a negation operator, ∃ is Existential Quantifier with the 
meaning of "there Exists", and ∀ is a Universal Quantifier 
with the meaning   " for all " , and α, β can be treated as predicates.

here we can apply some of the standard 
results of Propositional and 1st order logic on the given statement, 
which are as follows :

[ Result 1 : ¬(∀x P(x)) <=> ∃ x¬P(x), i.e. negation 
of "for all" gives "there exists" and negation also gets applied to scope of 
quantifier, which is P(x) here. And also negation of "there exists" gives "for all", 
and negation also gets applied to scope of quantifier  ]

[ Result 2 :  ¬ ( A ∧ B ) = ( ¬A  ∨ ¬B )  ]

[ Result 3 :  ¬P  ∨ Q <=> P -> Q ]

[ Result 4 : If P ->Q, then by Result of Contrapositive,  ¬Q -> ¬P  ]

 

¬ ∃ x ( ∀y(α) ∧ ∀z(β) )                 [ Given ]

=> ∀ x (¬∀y(α) ∨ ¬∀z(β) )          [ after applying Result 1 & Result 2 ]

=> ∀ x ( ∀y(α) -> ¬∀z(β) )     [after applying Result 3 ]

=> ∀ x ( ∀y(α) -> ∃z(¬β) )      [after applying Result 1]

which is same as the statement C. 

Hence the Given Statement is logically Equivalent
to the statement C.

Now, we can also prove that given statement is logically equivalent to the statement
 in option  B.

Let's see how !

The above derived statement is :

∀ x ( ∀y(α) -> ∃z(¬β) )

Now this statement can be written as (or equivalent to) :

=> ∀ x ( ∀z(β) -> ∃y(¬α) )     [after applying Result 4 ]

And this statement is same as statement B. 
Hence the Given statement is also logically equivalent 
to the statement B.

So, we can conclude that the Given statement is NOT logically equivalent to the 
statements A and D.

(A) "There exist some numbers which are either real OR rational"

(B) "All real numbers are rational"

(C) "There exist some numbers which are both real AND rational"

(D) "There exist some numbers for which rational implies real" 
(See Propositional Logic for details)

Clearly answer C is correct among all

 So the predicate is evaluated as
    P(x) = (¬(x=1))∧(∀y(∃z(x=y*z)⇒((y=x)∨(y=1))))
 P(x) being true means x ≠ 1 and
 For all y if there exists a z such that x = y*z then
 y must be x (i.e. z=1) or y must be 1 (i.e. z=x)
 
 It means that x have only two factors first is 1 
 and second is x itself.
 
This predicate defines the prime number.

=> This statement can be expressed as => For all X, x can be either gold or silver then the ornament X is precious => For all X, (G(X) v S(x)) => P(X).

This solution is contributed by Anil Saikrishna Devarasetty .

According to negation property of universal qualifier and existential quantifier