e of a propositional function such that the statement ∃!x ∃!y p(x,y) is true but the statement ∃!y∃!x p(x,y) is false

Question

Give an example of a propositional function such that the statement ∃!x ∃!y p(x,y) is true but the statement ∃!y∃!x p(x,y) is false (be sure to specify the domain for each variable)

Answer 1

Let $p(x,y)$ be the assertion $x=y^2$, where $x$ and $y$ range over the integers. Then there is a unique $x$, namely $0$ for which there is a unique $y$ such that $x=y^2$.

But it is not true that there is a unique $y$ for which there is a unique $x$ such that $x=y^2$. Indeed for every $y$ there is a unique $x$ such that $x=y^2$.

Answer 2

Here’s a finite example, more or less specifically constructed to satisfy the desired conditions.

The domain of $x$ and $y$ is the set of five women, Ann, Beatrice, Cathleen, Deborah, and Ellen, and $p(x,y)$ means that $x$ is the mother of $y$. Ann is the mother of Beatrice, and Cathleen is the mother of Deborah and Ellen. There are no other mother-daughter relationships amongst these five women.

In this context $\exists!x\,\exists!y\,p(x,y)$ says that there is exactly one woman who has exactly one daughter, which is true: that woman is Ann, each of the others in the set having either zero or two daughters. But $\exists!y\,\exists!x\,p(x,y)$ says that there is exactly one woman in the set who has exactly one mother in the set, which is false: Beatrice, Deborah, and Ellen all have exactly one mother in the set.