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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)

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.

by

So is the correct option 'D' ?
Yes

@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.