6. Which of the following arguments are not valid?

(c) If n is a real number such that n>1, then n^{2}>1. Suppose that n^{2}>1, then n>1

- 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.

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- 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.

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- 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.

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- Option : D
- Explanation :

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