# UGC NET COMPUTER SCIENCE SOLVED PAPERS 2014-16 - UGC NET Computer Science Paper 2 December 2015

>>>>>>>>UGC NET Computer Science Paper 2 December 2015

(a) "If Gora gets the job and works hard, then he will be promoted. If Gora gets promotion, then he will be happy. He will not be happy, therefore, either he will not get the job or he will not work hard"

(b) not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty”.

• Option : A
• Explanation :
(a) P: Gora gets the job
Q: Gora works hard
R: Gora gets promotion
S: Gora will be happy
The argument can bet written as
(P˄Q)→R
R→S
¬S
Therefore ¬P˅¬Q
(b) Let P: Puneet is not guilty
Q: Pankaj is telling the truth The argument can bet written as
P˅Q
~Q
Therefore, P
Thus by disjunctive syllogism, the argument ia valid
Disjunctive Syllogism:
The disjunctive syllogism rule may be written as:
P˅Q, ¬P ⊢Q
It may also be written as :
((PvQ)˄¬P)→Q
Where P, and Q are propositions expressed in some formal system.

• Option : A
• Explanation :
∃m ∀n P(m, n) : There exist some m which divides all n.True
∀n P(1, n) Every n divided by 1.True
∀m ∀n P(m, n) Every m divides every n False
So, option (A) is correct.

 List I List II (a) Vacuous proof (i) A proof that the implication p→q is true  based on the fact that p is false (b) Trivial proof (ii) A proof that the implication p→q is true  based on the fact that q is true (c) Direct proof (iii) A proof that the implication p→q is true  that proceeds by showing that q must be true when p is true. (d) Indirect proof (iv) A proof that the implication p→q is true  that proceeds by showing that p must be false when q is false.
• Option : A
• Explanation :
• Vacuous proof is a proof in which the implication p → q is true based on the fact that p is false.
• Trivial proof is a proof in which the implication p → q is true based on the fact that q is true.
• Direct proof is A proof in which the implication p → q is true that proceeds by showing that q must be true when p is true.
• Indirect proof a proof in which the implication p → q is true that proceeds by showing that p must be false when q is false. So, option (A) is correct.

• Option : D
• Explanation :
• p ∨ ~(p ∧ q) = p + (pq)` = p + p` + q` = 1 + q` = 1.This is a tautology.
• (p ∧ ~q) ∨ ~(p ∧ q) = pq` + (pq)` = pq` + p` + q` = p` + q`.
• This is not a tautology. p ∧ (q ∨ r) = pq + pr. This is not a tautology.
So, option (D) is correct