$Q)$ Write out $\exists!x\;P (x)$, where the domain consists of the integers $1, 2,$ and $3,$ in terms of negations, conjunctions, and disjunctions where $∃!x \; P (x)$ denotes “There exists a unique $x$ such that $P(x)$ is true.”
My Attempt :- Since, either $P(1)$ is true (or) $P(2)$ is true (or) $P(3)$ is true but all are mutually exclusive. So, I can write it as :- $(P(1) \;\wedge \sim P(2)\; \wedge \sim P(3)) \vee (P(2) \;\wedge \sim P(1)\; \wedge \sim P(3)) \vee (P(3) \;\wedge \sim P(1)\; \wedge \sim P(2))$
Please tell me whether it is correct or not. I also have seen one interpretation online which is $(P(1) \vee P(2) \vee P(3)) \wedge (\sim\; (P(1) \wedge P(2))) \wedge (\sim\; (P(2) \wedge P(3))) \wedge (\sim\; (P(3) \wedge P(1))) $. It also seems correct to me. So, my doubt is both are correct or only one is correct here.
Please help. Thank you.