UGC NET CSE | December 2015 | Part 2 | Question: 6

Question

Which of the following arguments are not valid?

1. "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." 
2. "Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty."
3. If $n$ is a real number such that $n>1$, then $n^2 > 1$. Suppose that $n^2 > 1$, then $n>1$.
   • A. i and iii
   • B. ii and iii
   • C. i,ii, and iii
   • D. i and ii

Answer 1

Option A let

• $p$: Gora get the job
• $q$: he works hard
• $r$: he will be promoted  
• $s$: he will be happy

So we have $p \land q \to r$ , $r \to s$  which will give us $\neg s \to \neg r$   (by contrapositive law) and $\neg r \to  \neg(p \land q)$. Now $\neg s$ is given so it implies $\neg r$ and so we have $\neg p \lor \neg q$ i.e either he does not get the job or he does not work hard. So it is valid 

Option B valid as EXACTLY one of two statements must be true. Given one statement is not true so other must be true 

Option C not valid. Here we have $a \to b$. But this does not always mean $b \to a$. 

Hence A and B are valid C is no  valid so no option is correct . 

Comments

Below in the square box, B is written but the correct answer is none of the options because statement a and b are valid, only C is invalid...

plz, correct it, admin...

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Sir then why everybody is giving answer as option (A) if  argument (a) is valid.

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How (ii) part is valid? explain please. Coz. In both the cases whether Pankaj is telling truth or not, Puneet is not guilty at all. Then how they are dependent or making any relationship?

Answer 2

Statement a

let $G(x)$ : Gora gets the job,

    $P(x)$ : Gora gets promoted,

    $H(x)$: Gora is happy,

    $W(x)$: Gora works Hard.

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.

$\equiv$ 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.
 

$\equiv$ $ [G(x) \Lambda W(x)] \rightarrow P(x) $

       $P(x)\rightarrow H(x)$

       $\sim H(x)$

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$\therefore$ $\sim G(x)\ V \sim H(x)$

 

$\because$ It is given that $\sim H(x)$ is True so $\sim G(x)\ V \sim H(x)$ will always be True

$\therefore$ Statement $a$ is valid.

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Statement b

let $Pu$ : Puneet is guilty.

     $Pa$ : Pankaj is telling truth.

 

Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty."

$\equiv$ Either Puneet is not guilty or Pankaj is telling the truth.

     Pankaj is not telling the truth

____________________________________________________________________________

$\therefore$ Puneet is not guilty.

 

$\equiv$  $\sim Pu\ V Pa$

    $\sim Pa$

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$\therefore\  \sim Pu$

This is the case of disjunctive syllogism

$\therefore$ Statement $b$ is valid.

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Statement c

Let $p$ : $n>1$

      $q$: $n^{2}>1$

where $n$ is a real number.

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

$\equiv$ If $n$ is a real number such that $n>1$, then $n^{2}>1$

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Suppose that $n^{2}>1$, then  $n>1$

 

$\equiv$ $ p\rightarrow q$

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$\therefore$ $ q\rightarrow p$

 

We know that if $ p\rightarrow q$ is true then $ q\rightarrow p$ need not be true.

$\therefore$ Statement $c$ is not valid.

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Hence $a$ and $b$ are valid statements and $c$ statement is not  valid so no option is correct . 

Comments

for a) isn’t the conclusion would be like 

∼G(x) V∼W(x) because the conclusion’s statement is either he will not get the job or he will not work hard. 

 

 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,

_____________________________________________________________________

∴∴ either he will not get the job or he will not work hard.

[JOB Λ WH] → P

P → H

~H

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~JOB V ~WH

To check whether the given argument is valid or not, let’s assume that the conclusion is false because for a valid argument at least one premise should be false(I can’t make all premises true) If all sets of premises are true then it is an invalid argument else valid argument.

for the conclusion to be false; JOB = T ; WH = T

for 3^rd premise, He will not be happy states that ~H is true here hence truth value of H is false

for 1^st premise [JOB Λ WH] → P needs to be true we need to make P true as well

for 2^nd premise P → H; P = T; H = F; T → F results in false. It’s not possible to make this premise true.

Hence all premises aren’t true here. Therefore, it is a valid argument.

 

 

 

Answer 3

Answer (1)

(a) not happy may cause he is not promoted

(c) p -> q , not always mean q-> p

Answer 4

P(x) = Gora gets the job.

Q(x) = Gora works hard.

R(x) = Gora will be promoted.

S(x) = Gora will be happy

a) P(x) ^ Q(x) ==> R(x) ------------- (1)

     R(x) ==> S(x).          -------------- (2)

     ~S(x).                       -------------- (3)

     Conclusion: ~P(x) v Q(x)

From (1) & (2),

P(x) ^ Q(x) ==> S(x).  ----- (4)  Hypothetical Syllogism

From (3) & (4),

~S(x) ^[P(x) ^ Q(x) ==> S(x)] gives ~[P(x) ^ Q(x)] --- (5) Modus Tollens

So finally, ~P(x) v ~Q(x). So option a is valid.

b) P(x) = Puneet is guilty.

     Q(x) = Pankaj is telling truth.

     ~P(x) v Q(x) ---------- (1)

     ~Q(x) ------- (2)

Conclusion to be shown : ~P(x)

From (1) & (2)  

~Q(x) ^ [~P(x) v Q(x)] gives ~P(x) using Disjunctive Syllogism. 

So option b is valid.

c) P(x) = n is real number

    Q(x) = n is greater than 1

    R(x) = n square is greater than 1

P(x) ^ Q(x) ==> R(x)  ------- (1) 

R(x) ---- (2)

conclusion: Q(x)

No rule of inference exists to prove the conclusion. so option c is not valid.

Here only the statement c is invalid

Comments

option B is valid. It is ~P(x) as dipika told.

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P(x) = Gora gets the job.

Q(x) = Gora works hard.

R(x) = Gora will be promoted.

S(x) = Gora will be happy

(P(x) ^ Q(x)) -> S(x) .................i

S(X) -> R(x) ...........................ii

from i and ii we can say

(P(x) ^ Q(x)) -> R(x) //hypothetical syllogism

But from it can we say R(x) -> (~P(x)  v ~Q(x))

 means R(x) -> ~(P(x) ^ Q(x)) ?

I think for being valid there should be "if and only if" is the clause

Am I wrong?

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~P(x) v ~Q(x) is conclusion in a. 

rt?

i think there is some typo in a.

btw good explanation...