Unique Existential Quantifier

Question

The Unique Existential Quantifier states that there exists a unique $x$ which holds for a $P(x)$.

I came up with

$$\exists x\;p(x)\land\neg\exists y\;p(y)\land x\ne y\;.$$

How is this different than the one on Wikipedia which has it as

$$\exists x\Big(p(x)\land\neg\exists y\;p(y)\Big)\land x\ne y\;.$$

My question is how the parenthesis impacts the statement.

Answer 1

Look up the distinction between a formula and a sentence.

Well, first of all, your statement is ambiguous: do you mean $$(\exists x [p(x)])\wedge (\neg\exists y[p(y)\wedge x\not=y])$$ or $$\exists x[p(x)\wedge \neg\exists y[p(y)\wedge x\not=y]]?$$ The latter is correct; the former doesn't really make sense, because $x$ appears unbound in the second clause (we need every instance of $x$ to occur within the scope of some quantifier on $x$).

However, your "wikipedia formulation" also suffers from this problem!

Luckily, this isn't actually how it's formalized on wikipedia: Uniqueness Quantification gives the same version I wrote above (the first expression in section 2), as well as the alternate formulation $$\exists x[p(x)]\wedge \forall y\forall z[(p(y)\wedge p(z))\implies y=z]$$ (the third expression in that subsection).

Answer 2

You’ve misquoted Wikipedia: it actually has

$$\exists x\left(P(x)\land\neg\exists y\Big(P(y)\land y\ne x\Big)\right)\;.\tag{1}$$

It’s crucial that the $y\ne x$ and $P(y)$ both be within the scope of $\exists y$. Moreover, since the $x$ in $y\ne x$ is the same one whose existence is asserted by the initial quantifier, it’s crucial that $y\ne x$ be within the scope of $\exists x$ as well. Thus, all of the parentheses in $(1)$ are required. In words:

There is an $x$ such that

• $x$ has property $P$, but
• there is no $y$ such that $y$ has property $P$ and is not equal to $x$.

A bit closer to natural language:

There is an $x$ such that $x$ has property $P$, but no $y$ distinct from $x$ has property $P$.

In each verbal version the first assertion says $\exists x\;P(x)$ (there is an $x$ with property $P$), and the second, referring to this $x$, says that there is no other object with property $P$.

Your version doesn’t capture the idea, because the $x$ and $y$ in the element $x\ne y$ aren’t limited by the quantifiers: they’re outside the scope of the quantifiers. Your formula would have the same meaning if it were rewritten as

$$\exists x\;p(x)\land\neg\exists y\;p(y)\land w\ne z\;.$$

Moreover, it’s contradictory: it says that there is an object with property $p$ and that there is no object with property $p$.

Answer 3

What you wrote, with implicit bracketing added, was: $\;\color{silver}{\big(}\exists x\; P(x)\color{silver}{\big)} \wedge \color{silver}{\big(}\neg \exists y\;P(y)\color{silver}{\big)} \wedge \color{silver}{\big(}x\neq y\color{silver}{\big)}$

Quantifier binding takes precedence over logical junctions, in the same way multiplication takes precedence over addition.

So, it's like writing $\; 3 \times 4 + 6\;$ when you meant $\;4 \times (3 + 6)\;$.  The first equals $18$ while the second equals $30$.

The placement of brackets are crucial in identifying the scope bound to each quantifiers.   You can't skimp.

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$$\begin{align} \exists ! x \; P(x) \quad & \equiv\quad \exists x \Big(P(x) \wedge \neg \exists y \big(P(y)\wedge y\neq x\big)\Big) \tag{1} \\ &\equiv \quad \exists x\Big(P(x) \wedge \forall y\big(P(y)\to x=y\big)\Big) \tag{2} \\ &\equiv \quad \color{silver}{\big(}\exists x \; P(x)\color{silver}{\big)}\wedge \color{silver}{\Big(}\forall z\neg \exists y \big(P(y)\wedge P(z) \wedge y\neq z\big)\color{silver}{\Big)}\tag{3} \end{align}$$

Note in (3) we could write $\color{silver}{\big(}\exists x \; P(x)\color{silver}{\big)}\wedge \color{silver}{\Big(}\forall x\neg \exists y \big(P(y)\wedge P(x) \wedge y\neq x\big)\color{silver}{\Big)}$.   Although it may be clearer to use distinct variables in distinct scopes whenever possible, it is not strictly required.