Write ‘There is exactly 1 person…’ without the uniqueness quantifier

Solution 1:

Is it just that $\forall z$ means that this statement could be true for any element,

Yes, exactly.

...and if so what's the difference between $\forall z$ and $\exists z$?

If we only asserted the existence of some particular $z$ such that if $y$ loves $z$, then that particular person $z$ would thus be $x$. But then we are not ruling out that there might be another person, different from $x$, that is also loved by everyone. And we have already asserted the existence of someone (namely, x) who is loved by all. So in that sense, $\exists z$ such that... is reduntant.

We need the universal quantifier for $z$ (to assert a statement $\forall z$) in the second clause to indicate that if there is any $z$ (which means that the claim that follows - as it relates to $z$, is true for every z) $z$ such that $L(y, z)$, then any/every such $z$ must be $x$, since there is exactly one person, namely $x$, who is loved by all $y$. I.e., for every $z,$ if every y loves z, then z must be x: i.e., that $z$ is not anyone other than $x$. This gives us that $x$ (whose existence we asserted at the start) is therefore the one unique person loved by all.

Now, just one oversight to "clean up" your expression, which you state as:

$$∃x(\forall y L(y,x))\land ∀z(∀y(L(y,z))⟹x=z)$$

But here we have two independent clauses that creates a problem, because in your second clause, you have $x$ appear outside the scope of its quantifier. I.e., it is a free, unquantified varible.

So we want the scope of $\exists x$ to persist over the entire statement, hence the square brackets below.

That is, $$\exists x \big[\forall y(L(y, x))\land \forall z(\forall y(L(y, z)) \rightarrow z = x)\big]$$

Solution 2:

The idea behind the formula is that, to say exactly one person is loved by everybody, you first say that there is such a person $x$ (this is the part you said you understood), and then you say that there isn't a second such person. That second part is expressed here as saying, if any other $z$ shows up who is also (like $x$) loved by everybody, then this wasn't really another $z$ but just the same $x$.

Solution 3:

The simplest definition for $\langle \exists! x :: P(x) \rangle$ I know, which I've probably learned from Dijkstra et al.'s works, is $$\langle \exists y :: \langle \forall x :: P(x) \equiv x = y \rangle \rangle$$ which does only use $P(x)$ once.

Note how we have the following nice symmetry: $$ \begin{array} \\ \langle \exists! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \equiv & x = y \rangle \rangle \\ \langle \exists\phantom! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \Leftarrow & x = y \rangle \rangle \\ \langle \phantom\exists! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \Rightarrow & x = y \rangle \rangle \\ \end{array} $$ where $\langle ! x :: P(x) \rangle$ means "there is at most one $x$ such that $P(x)$".