Propositional and First Order Logic.

Question 1

What is the logical translation of the following statement?

  "None of my friends are perfect." 

• A. A
• B. B
• C. C
• D. D

Answer: (D)

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 

Question 2

• A. A
• B. B
• C. C
• D. D

Answer: (A), (D)

  Given statement is : ¬ ∃ x ( ∀y(α) ∧ ∀z(β) )

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  ]

Now we need to use these results as shown below:

 

¬ ∃ 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.

Hence, the correct answer is Option A and Option D. But in GATE 2013, marks were given to all for this question.

Question 3

What is the correct translation of the following statement into mathematical logic? “Some real numbers are rational” 

• A. A
• B. B
• C. C
• D. D

Answer: (C)

(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

Question 4

Which one of the following options is CORRECT given three positive integers x, y and z, and a predicate? 

        
      P(x) = ¬(x=1)∧∀y(∃z(x=y*z)⇒(y=x)∨(y=1))

• A. P(x) being true means that x is a number other than 1
  
• B. P(x) is always true irrespective of the value of x
  
• C. P(x) being true means that x has exactly two factors other than 1 and x 
  
• D. P(x) being true means that x is a prime number
  

Answer: (D)

 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.

Question 5

Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. which one of the statements below expresses best the meaning of the formula ∀x∃y∃t(¬F(x, y, t))?

• A. Everyone can fool some person at some time
• B. No one can fool everyone all the time
• C. Everyone cannot fool some person all the time
• D. No one can fool some person at some time

Answer: (B)

∀ x ∃ y ∃ t(¬ F(x,y,t)) => ∀ x ¬(∀ y ∀ t F(x,y,t))

Question 6

Which one of the following is the most appropriate logical formula to represent the statement? "Gold and silver ornaments are precious". The following notations are used: G(x): x is a gold ornament S(x): x is a silver ornament P(x): x is precious

• A. ∀x(P(x)→(G(x)∧S(x)))
• B. ∀x((G(x)∧S(x))→P(x))
• C. ∃x((G(x)∧S(x))→P(x)
• D. ∀x((G(x)∨S(x))→P(x))

Answer: (D)

=> 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 .

Question 7

The binary operation ₡ is defined as follows 

P	Q	P₡B
T	T	T
T	F	T
F	T	F
F	F	T

Which one of the following is equivalent to P∨Q? 

• A. ¬Q₡¬P
• B. P₡¬Q
• C. ¬P₡¬Q
• D. ¬P₡¬Q

Answer: (B)

P₡¬Q = Q' ⇢ P = (Q')' v P = Q v P

Question 8

Which of the above two are equivalent?

• A. II and III
  
• B. I and IV
  
• C. II and IV
  
• D. I and III
  

Answer: (B)

According to negation property of universal qualifier and existential quantifier

Question 9

Let fsa and pda be two predicates such that fsa(x) means x is a finite state automaton, and pda(y) means that y is a pushdown automaton. Let equivalent be another predicate such that equivalent (a, b) means a and b are equivalent. Which of the following first order logic statements represents the following: Each finite state automaton has an equivalent pushdown automaton.

• A. A
• B. B
• C. C
• D. D

Answer: (A)

Considering each option :
(A) If everything is a FSA, then there exists an equivalent PDA for everything.
(B) It is not the case that for all y if there exist a FSA then it has an equivalent PDA.
(C) Everything is a FSA and has an equivalent PDA.
(D) Everything is a PDA and has exist an equivalent FSA.
Thus, option (A) is correct.

Please comment below if you find anything wrong in the above post.

Question 10

P and Q are two propositions. Which of the following logical expressions are equivalent? 

• A. Only I, II and III
  
• B. Only I, II and IV
  
• C. All of I, II, III and IV
  
• D. Only I and II 
  

Answer: (A)

I and II are same by Demorgan's law

The IIIrd can be simplified to I. 

since is true always