in Analytical Aptitude
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1 vote
1 vote

Choose the set in which the combinations are logically equivalent.

  1. All flowers are roses.
  2. No rose is a flower.
  3. No flower is a rose.
  4. Some flowers are roses.
  5. No rose is not a flower.
  6. All roses are flowers.
  1. (i), (v)
  2. (ii), (iv)
  3. (iii), (iv)
  4. (v),  (vi)
in Analytical Aptitude
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1 Answer

5 votes
5 votes
Best answer

Let $x$ : An object

      $r(x)$: $x$ is a rose  and

      $f(x)$: $x$ is a flower

  1. All flowers are roses. $\equiv$ For all $x$ , if $x$ is a flower then it has to be a rose  $\equiv$ $\forall x$($ f(x) \rightarrow r(x))$$\equiv$  $\forall x$$(\sim f(x)\ V r(x))$
  2. No rose is a flower.$\equiv$ There does not exist an $x$ such that  $x$ is a rose and it is a flower $\equiv$ $\sim \exists x ( r(x) \Lambda f(x))$ $\equiv$ $\forall x$$(\sim r(x)\  V  \sim f(x) )$
  3. No flower is a rose.$\equiv$ There does not exist $x$ such that $x$ is a flower and it is a rose $\equiv$ $\sim \exists x ( f(x) \Lambda r(x))$ $\equiv$ $\forall x$$(\sim f(x)\  V \sim r(x) )$
  4. Some flowers are roses$\equiv$ There exists some $x$ such that $x$ is a flower and it is a rose $\equiv$ $\exists x(f(x) \Lambda  r(x))$
  5. No rose is not a flower.$\equiv$ There does not exist $x$ such that $x$ is a rose and it is not a flower $\equiv$$\sim \exists x( r(x) \Lambda \sim f(x))$ $\equiv$ $\forall x$$(\sim r(x)\ V f(x))$
  6. All roses are flowers. $\equiv$$\forall x$( if $x$ is a rose then it is a flower)$\equiv$ $\forall x$$( r(x) \rightarrow f(x))$ $\equiv$ $\forall x$$(\sim r(x)\ V f(x))$$\equiv$

$\therefore$ Option $D.$ is the correct answer.

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4 Comments

So is the correct option 'D' ?
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Yes
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@Satbir

5 is saying all roses are flowers,i.e rose is a subset of flower.but 6 is saying No rose is not a flower.

there might be some part(flower) which are not belongs to rose but belongs to flowers.then it should be No rose may or may not be a flower.

I think option a should be correct.

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Answer:

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