Let $P(x):(x-1=0)$ defined over $\mathbb{Z}^{+}$
then $x=1$ is the only unique solution for which $P(x)$ is true.
$\exists !xP(x)$ means " there exists a unique $x$ such that $P(x)$ is true” and here the unique value of $x=1$
$\Rightarrow$ We can say that there exist a value of $x$ for which proposition holds true.
$\Rightarrow$ $\exists xP(x)$
So if there exists a unique $x$ such that $P(x)$ is true then there will exist atleast $1$ value of $x$ for which proposition holds true.
$\therefore$ $\exists !xP(x)\rightarrow\exists xP(x)$ is true.
Let $P(x):(5/x$ is defined $)$ defined over $\mathbb{R}^{+}$
then $x=0$ is the only unique solution for which $P(x)$ is false since $5/0$ is undefined.
$\exists !x\sim P(x)$ means there exists a unique $x$ such that $P(x)$ is not true i.e. $P(x)$ is false and here the unique value is $x=0$
$\sim \forall xP(x)$ can also be written as $\exists x\sim P(x)$ which means there exists at least one $x$ such that $P(x)$ is not true i.e. $P(x)$ is false and here we are having x=0 which is satisfying.
So if there exists a unique $x$ such that $P(x)$ is not true i.e. $P(x)$ is false then there will definitely exists at least one $x$ such that $P(x)$ is not true i.e. $P(x)$ is false
$\Rightarrow$ $\exists !x\sim P(x)\rightarrow\sim \forall xP(x)$ is true.
