# Propositional and First Order Logic.

Question 1 |

A | |

B | |

C | |

D |

**GATE CS 2013**

**Propositional and First Order Logic.**

**Discuss it**

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 | |

B | |

C | |

D |

**GATE CS 2013**

**Propositional and First Order Logic.**

**Discuss it**

**¬ ∃ x ( ∀y(α) ∧ ∀z(β) )**

where ¬ is a negation operator, ∃ is Existential Quantifier with the meaning ofNow we need to use these results as shown below:"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 applyingHence, the correct answer is Option A and Option D. But in GATE 2013, marks were given to all for this question.Result 1&Result 2] => ∀ x ( ∀y(α) -> ¬∀z(β) ) [after applyingResult 3] =>∀ x ( ∀y(α) -> ∃z(¬β) )[after applyingResult 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 applyingResult 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 isNOTlogically equivalent to the statementsAandD.

Question 3 |

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

A | |

B | |

C | |

D |

**GATE CS 2012**

**Propositional and First Order Logic.**

**Discuss it**

(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))

P(x) being true means that x is a number other than 1 | |

P(x) is always true irrespective of the value of x | |

P(x) being true means that x has exactly two factors other than 1 and x | |

P(x) being true means that x is a prime number |

**GATE CS 2011**

**Propositional and First Order Logic.**

**Discuss it**

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.

Question 5 |

Everyone can fool some person at some time | |

No one can fool everyone all the time | |

Everyone cannot fool some person all the time | |

No one can fool some person at some time |

**GATE CS 2010**

**Propositional and First Order Logic.**

**Discuss it**

Question 6 |

∀x(P(x)→(G(x)∧S(x))) | |

∀x((G(x)∧S(x))→P(x)) | |

∃x((G(x)∧S(x))→P(x) | |

∀x((G(x)∨S(x))→P(x)) |

**GATE-CS-2009**

**Propositional and First Order Logic.**

**Discuss it**

=> 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?

¬Q₡¬P | |

P₡¬Q | |

¬P₡¬Q | |

¬P₡¬Q |

**GATE-CS-2009**

**Propositional and First Order Logic.**

**Discuss it**

P₡¬Q = Q' ⇢ P = (Q')' v P = Q v P

Question 8 |

II and III | |

I and IV | |

II and IV | |

I and III |

**GATE-CS-2009**

**Propositional and First Order Logic.**

**Discuss it**

According to negation property of universal qualifier and existential quantifier

Question 9 |

A | |

B | |

C | |

D |

**GATE CS 2008**

**Propositional and First Order Logic.**

**Discuss it**

(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?

Only I, II and III | |

Only I, II and IV | |

All of I, II, III and IV | |

Only I and II |

**GATE CS 2008**

**Propositional and First Order Logic.**

**Discuss it**

I and II are same by Demorgan's law

The IIIrd can be simplified to I.

since is true always