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Express the quantification $∃!xP (x)$,using universal quantifications, existential quantifications, and logical operators.?
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$\exists !xP(x)$  = There is exactly one (one and only one) element in the domain for which property $P$ is True.

$\exists !$ is  uniqueness quantifier or unique existential quantifier.  
Uniqueness quantification ($\exists !$) can be expressed in terms of the existential and universal quantifiers as follows :

1. $\exists x (P(x) \wedge \forall y ((x \neq y) \rightarrow \,\,\sim P(y)))$ 
interpretation : There is some element $x$ in the domain for which Property $P(x)$ is True And For any(every) element $y$, if $y$ is not equal to $x$ then $P(y)$ is Not True.

Since we know that Every implication statement is equivalent to its contrapositive. We can write the above expression in different way, as following :

 $\exists x (P(x) \wedge \forall y(P(y) \rightarrow (x = y)))$

2. Another idea to find expression for "Exactly One" is as following :

Exactly One = At least One - At least Two (In logic, Minus can be expressed by ($\wedge \sim$))

$\exists x P(x) \wedge (\sim \exists x \exists y ((x \neq y) \wedge P(x) \wedge P(y)))$

Above expression can also be written as following :

$\exists x P(x) \wedge ( \forall y \forall z( (P(y) \wedge P(z)) \rightarrow (y = z)))$

3. Another interpretation could be as following : There is some element $x$ in the domain for which $P(x)$ is True and There is No such element $y$ in the domain which is--- "Not equal to $x$ and for which $P(y)$ is True".

$\exists x (P(x) \wedge \sim \exists y (P(y) \wedge (x \neq y)))$

4. The last but the most concise way (yet most confusing way) is as following :

$\exists x \forall y (P(y) \leftrightarrow (x = y))$  .. Try it once and try to see why it is Correct expression. And Let me know if doubtful.... Will add "why it is correct" reason/interpretation  in the answer. 

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