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UGCNET-June-2019-I-27

Given below are two premises with four conclusions drawn from them. Which of the following conclusions could be validly drawn from the premises?

Premises:

  1. No paper is pen
  2. Some paper are handmade

Conclusions:

  1. All paper are handmade
  2. Some handmade are pen
  3. Some handmade are not pen
  4. All handmade are paper
in Numerical Ability by Veteran (416k points)
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option 2

1 Answer

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Let $x$ : An object

      $p(x)$: $x$ is a paper 

      $n(x)$: $x$ is a pen  and

      $h(x)$: $x$ is handmade

 

No paper is pen.$\equiv$ There does not exist an $x$ such that  $x$ is a paper and it is a pen $\equiv\ \sim \exists x ( p(x) \wedge n(x)) \equiv \forall x$$(\sim p(x)\  \vee  \sim n(x) )$

Also from $\forall x (\sim p(x)\  \vee  \sim n(x) )$ we can conclude $\exists x (\sim p(x)\  \vee  \sim n(x) ) \equiv \exists x ( p(x) \rightarrow \sim n(x)) $

 

Some paper are handmade$\equiv$ There exists some $x$ such that $x$ is a paper and it is handmade $\equiv \exists x(p(x) \wedge  h(x))$

 

Using the above two equations

$\exists x ( p(x) \rightarrow \sim n(x)) $

$\underline{\exists x(p(x) \wedge  h(x))}$

$\exists x( \sim n(x) \wedge h(x))$

 

  1. All paper are handmade. $\equiv \forall x$( if $x$ is a paper then it is handmade) $\equiv \forall x ( p(x) \rightarrow h(x)) \equiv \forall x (\sim p(x)\ \vee h(x))$
  2. Some handmade are pen$\equiv$ There exists some $x$ such that $x$ is a handmade and it is a pen $\equiv \exists x(h(x) \wedge  n(x))$
  3. Some handmade are not pen$\equiv$ There exists some $x$ such that $x$ is a handmade and it is not a pen $\equiv \exists x(h(x) \wedge \sim n(x))$
  4. All handmade are paper $\equiv \forall x$( if $x$ is a handmade then it is a paper) $\equiv \forall x ( r(x) \rightarrow f(x)) \equiv \forall x (\sim h(x)\ \vee p(x))$

 

$\therefore$ Option $C.$ is correct

by Boss (17.5k points)
0

@Satbir can you please explain the Inference Rules ( if any ) , to deduce the following 

Using the above two equations

∃x(p(x)→∼n(x))∃x(p(x)→∼n(x))

∃x(p(x)∧h(x))

∃x(∼n(x)∧h(x))

And can you kindly share the resources for the same? 

0
in deductions we assume that the given propositions are true and on basis of that we calculate the conclusion.

$p(x) \wedge h(x) $ it means that both $p(x)$ and $h(x)$ are True. .....(1)

$p(x) \rightarrow \sim n(x)$ it means that (if $p(x)$ is true then $\sim n(x)$ will also be true) is true.....(2)

 from (1) $p(x)$ is true so we can put it in (2) and can conclude that $\sim n(x)$ is also true.

$\therefore$ Both $h(x)$ and $\sim n(x)$ are true i.e. $h(x) \wedge \sim n(x)$

refer keneth rosen book.
0

@Satbir thanks for replying, appreciate it. How this will always be true ?
 

p(x)→∼n(x)p(x)→∼n(x) it means that (if p(x)p(x) is true then ∼n(x)∼n(x) will also be true) is true.....(2)

so it's like since we are assuming the 2nd proposition to be true. And it is an implication for that to be true always we need to have ~n(x) = true, ryt?

0
yes...because $p(x)$ is true

according to modus ponens

$p \rightarrow q$

$\underline {p \ \ \ \ \ \ \ \ \ \ \ \ \ } $ ...this we are getting from equation (1).

$q$                                                

in question $q$ is $\sim n(x)$
0
perfect :), thanks

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