How can I express the "uniqueness quantifier" without "$\exists!$"? I am trying to express the sentence as a logical statement using quantifiers and without using "$\exists!$":
"There is a dog who has exactly one favorite toy."
My first question that has me stuck is should I make my propositional function a function of two variables $L(x,y)= \text{"$x$ has a favorite toy $y$}."$ ($x$ for all dogs and $y$ for all toys), or is there a way to do this by just using one variable $L(x)=\text{"$x$ has one favorite toy"}$. I don't know why I'm so stumped by this but I have no idea where to start.
 A: Let $x$, $y$, and $z$ be variables and $P$ be a predicate.
You can split up there exists a unique $x$ such that $P(x)$ into two claims.


*

*There exists at least one $x$ such that $P(x)$ .

*If $P(x)$ and $P(y)$, then $x=y$ .


More concretely.
$$ \exists! x \mathop. P(x) \iff (\exists z \mathop . P(z)) \mathop\land (\forall x \mathop. \forall y \mathop. (P(x) \land P(y)) \to x =y)$$
A: An efficient way to express $\exists ! x \ P(x)$ is:
$$\exists x \forall y (P(y) \leftrightarrow y =x)$$
So, if we use $D(x)$ for '$x$ is a dog', and $F(x,y)$ for '$x$ has $y$ as a favorite toy', your sentence can be symbolized as:
$$\exists x (D(x) \land \exists y \forall z (F(x,z)  \leftrightarrow z =y))$$
A: It is possible to use just 1 variable "$L(x)$=$x$ has 1 favourite toy", or even two variables "$L(x,y)$=$x$ has 1 favourite $y$" however it is a better learning experience to specify what you mean by exactly 1. 
A sufficient way to express "There is a dog who has exactly one favorite toy," in symbolic logic is the following $(1)$.
\begin{equation}\tag{1}
\exists x\big(\exists y(D(x)\land T(y)\land F(x,y)\land\forall z((F(x,z)\land T(z))\rightarrow z=y))\big),
\end{equation}
where $D(x)$ expresses "$x$ is a dog", $T(y)$ "$y$ is a toy", and $F(x,y)$ "$x$'s favorite toy is $y$."
The way $(1)$ works is that it asserts that there is a dog $x$, a toy $y$ and the dogs favorite toy is $y$. Then it goes on to assert, that any toy which is the dogs favorite, is the same toy as the first. Thus there is only 1 toy. 

For your future reference, I have included the definitions of less than, greater than or equal to, and exactly 1 below this answer.
Less than n:
\begin{equation}\tag{2}
\forall x_1\dots\forall x_{n-1}\forall x_n((x_1\neq x_2\land\dots\land x_{n-1}\neq x_n)\rightarrow \lnot P(x_1)\lor\dots\lor\lnot P(x_{n-1})\lor\lnot P(x_n))
\end{equation}
Greater than or equal to n:
\begin{equation}\tag{3}
\exists x_1\dots\exists x_{n-1}\exists x_n(x_1\neq x_2\land\dots\land x_{n-1}\neq x_n\land P(x_1)\land\dots\land P(x_{n-1})\land P(x_n))
\end{equation}
Exactly 1
\begin{equation}\tag{4}
\exists x(P(x)\land\forall y(P(y)\rightarrow y=x))
\end{equation}
