Propositional Logic | Self doubt

Question

Please give one example of each following logic in english sentence -

1. ($\forall$xP(x) -> $\exists$xQ(x)) == $\exists$x(P(x) -> Q(x))

2.  ($\exists$xP(x) -> $\forall$xQ(x)) -> $\forall$x(P(x) -> Q(x))

3. $\forall$x(P(x) -> R)  -> ($\exists$xP(x) -> R)

Answer 1

1. I'm parsing this as $(\forall x P(x)) \to \exists Q(x)$
LHS: If every one is robbed it means there is some robber.
RHS: If someone is robbed, then he is a robber.

Both are not same rt?

2.
LHS: If someone is robbed (say in a bus), everyone is searched (including himself).
RHS: If anyone in the bus is robbed, he is searched
So, LHS implies RHS but not otherway.

3.
LHS: If any Indian wins India gets a medal.
RHS: There exists an Indian who if wins, India gets a medal.
Here also LHS implies RHS but not necessairly otherway.

Comments

Is this possible in First order logic?
 ∃x (¬P(x) V Q(x) ) = ∃x (P(x) -> Q(x) )

---

not sure sir... about first order logic but seems possible to me  :)

---

sir if seprators are not there as  in 2. then what priority and associtivity we should take .

as in this question what is associvity of implication ?

Answer 2

Say P(x) = where x divisible by 2

Q(x)=where x divisible by 3

(∀xP(x) -> ∃xQ(x)) == ∃x(P(x) -> Q(x))

Here LHS= there exists a x not divisible by 2 or there exists a x divisible by 3

RHS=there exists a x not divisible by 2 or divisible by 3

both are equivalent

Say for set (2,3,6,8,9); LHS and RHS both satisfiable

(∃xP(x) -> ∀xQ(x)) -> ∀x(P(x) -> Q(x))

LHS=For all x P(x) is not divisible by 2 or  for all x Q(x) divisible by 3

RHS=for all x P(x) not divisible by 2 or Q(x) not divisible by 3

Say for set (3,6,9) P->Q satisfied

but imagine a set (3,5,6,7,9) here Q->P not satisfied

∀x(P(x) -> R)  -> (∃xP(x) -> R)

LHS=for all x, if P(x) is divisible by 2 ,then R is also divisible by 2

RHS= there exists x , P(x) is divisible by 2 then R also divisible by 2

Similarly it is also satisfies one direction

Comments

x=10 is not divisible by 2?

it is false, rt?

 u proved beautifully the

 (∀xP(x) -> ∃xQ(x)) == ∃x(P(x) -> Q(x))

But it always not true in propositional logic.

see this link

---

chk it
, where is fault

https://gateoverflow.in/3454/gate2007-it_21

not getting (A) as answer

@Arjun Sir

I have written stmt (A) like u but getting some error

---

@Srestha, 

LHS= there exists a x not divisible by 2 or there exists a x divisible by 3.

Now see, 

x=10 is not divisible by 2?

it is false, rt?

P -> Q  results false if P is false right ??

LHS= True.