How can I express the "uniqueness quantifier" without "$\exists!$"?

Question

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.

Answer 1

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.

$$ \exists! x \mathop. P(x) \iff (\exists z \mathop . P(z)) \mathop\land (\forall x \mathop. \forall y \mathop. (P(x) \land P(y)) \to x =y)$$

Answer 2

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:

$$\exists x (D(x) \land \exists y \forall z (F(x,z) \leftrightarrow z =y))$$

Answer 3

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.

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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}