Does ∃x (Px⇔Rx) imply (∃xPx ⇔ ∃xRx)?

Question

Does $∃x (Px⇔Rx)$ imply $(∃xPx ⇔ ∃xRx)$ ?

I am trying to make an interpretation that verifies $∃x (Px⇔Rx)$ and falsifies $(∃xPx ⇔ ∃xRx).$ I have been trying for a while now, and I don't think it is possible.

On the other hand, I do know that $(∃xPx ⇔ ∃xRx)$ does not imply $∃x (Px⇔Rx).$

Answer 1

$$∃x\:(Px⇔Rx)\tag1$$ means that some object satisfies either both or neither of $P$ and $R.$

$$∃xPx⇔∃xRx\tag2$$ means that $P$ and $R$ are either each satisfied by some object or each satisfied by no object (i.e., $P$ and $R$ are both either empty or nonempty).

The countermodel

shows that $$(1)\kern.6em\not\kern-.6em\implies(2),$$ while the countermodel

shows that $$(2)\kern.6em\not\kern-.6em\implies(1).$$

Answer 2

∃x (Px⇔Rx) is equivalent to ∃x((-P V R) ^ (-R V P))

∃x((-P V R) ^ (-R V P)) -> ∃xPx ⇔ ∃xRx is false by counter example

Domain: {0, 1} P: {1} R: {}

Answer 3

Nope.

Counterexample:

Domain: Natural numbers

$P(x): x$ is even

$R(x): x$ is odd