14 views
$(i) \exists(R\rightarrow I)\equiv \forall R\rightarrow \exists I$

$(ii)\forall (R\rightarrow I)\equiv \exists R\rightarrow\forall I$

Are both of them coreect??

I have doubt in second one...
| 14 views
+1
1st one is correct

2nd  one not true.

By going meaning of the statements, I am getting this

+1 vote
$\exists (R \rightarrow I ) \equiv \exists (R' \vee I) \equiv \exists R' \vee \exists I$

$\forall R \rightarrow \exists I \equiv \exists R' \vee \exists I$

$\forall (R \rightarrow I ) \equiv \forall (R' \vee I) \equiv \sim \exists (R \wedge I')$ It means that there does not exists a case where $R$ and $I'$ holds true togather.  like $R$ is getting heads and $I'$ is getting tails then both head and tails can't occur togather.

$\exists R \rightarrow \forall I \equiv \forall R' \vee \forall I \equiv \sim \exists R \ \wedge \sim \exists I'$  It means that " there does not exists an $R$ for a case and there does not exists an $I'$ for a case." Clearly it is different from above example since heads and tails occur for all cases of coin flip.
by (4.1k points)
0
Sir, Your example of head and tail is not correct here.. R and I are should be independent(if nothing is given about them).
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2nd one RHS is telling , for some R , there exists all I

But Say

$R=\left \{ 1,2 \right \}$

$I=\left \{ 3,4 \right \}$

Mapping here

$1\rightarrow 3$

$2\rightarrow 4$

Now, If we take, some R means only $1$, it doesnot have a mapping for $4.$

So, RHS fails
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Yes i understand..

We can give analogy as--

$\forall (R\rightarrow I)$ Every rich person is intelligent.

$\exists R\rightarrow \forall I$ if there is atleast one rich person then all are intelligent.

If there exist only one person which is rich and intelligent. Then LHS willl be true. If other persons are not intelligent then RHS becomes false.. So (ii) is not coreect.
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Yes, you are correct but don't relate it to real world , i just gave it so that i can show both c and d are different meaning.

If you relate these things to real world then in exam you can get confused, like they will give "every flower is a rose" , and "every rose is a flower." You would get confuse if you start relating it to real world.

You can make ball and bucket example.

another example can be R -> ram goes to school, I' -> monu goes to job

LHS = there does not exists a case when ram goes to school and monu goes to job (at same case)

RHS = There does not exist any case when ram goes to school and there does not exists any case when monu goes to job.(at different cases)
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Yes yes..

But we shouldn't have to use complementary propositions..