Interpreting the meaning of $\exists x \ni \forall y \ \exists z \ni x+y=z$

Question

In exercise (11a) of Lay's Analysis with an introduction to proof, one is asked to determine the truth value of the following statement

$\exists x \ni \forall y \ \exists z \ni x+y=z\ .$

I'm having trouble getting started.

I've tried looking at its negation $$\sim[\exists x \ni \forall y \ \exists z \ni x+y=z] \Leftrightarrow \forall x \ \exists y \ni \forall z,\ x+y \neq z$$ , but this doesn't lead to more insight.

How does one approach this particular problem?

$\ni$ is to mean "such that", according to the text.

Answer 1

The (contrived) statement is true.

Let the domain for the variables $x,y$ and $z$ range over, say, the integers $\Bbb Z$. Addition is a binary operation and so

$\tag 1 (\forall x, y \in \Bbb Z)\, ( \exists z \in \Bbb Z ) \;[z = x+y]$

is true.

Since $\Bbb Z \ne \emptyset$, if we fix any $x_0 \in \Bbb Z$ then for any $y \in \Bbb Z$ there exist a $z \in \Bbb Z$ such that $z = x_0+y$.

So if a logical statement starting with $\forall u$ is true and the domain is nonempty, we can change the quantifier to get another true statement starting with $\exists u$.

Answer 2

I suspect the $\ni$ symbol may be used by your text to indicate "such that" after an existential quantifier.   This is not usually used.

Rather [quantifier][term] [predicate] is written for both universal and existential quantifiers.   Optionally, a comma, dot, or colon may be used to separate the term and predicate if a space does not feel sufficient.   Parenthesis may also be placed around the predicate clarify the scope of the quantifier.

So $∃x∋∀y~∃z∋x+y=z $ would mean "there is some $x$ such that forall $y$ there is some $z$ such that the sum of $x$ and $y$ equals $z$", and be more usually written simply as: $\exists x~\forall y~\exists z~x+y=z$ or perhaps: $$\exists x~\forall y~\exists z~(x+y=z)$$

The negation is "for all $x$ there exists some $y$ such that forall $z$ the sum of $x$ and $y$ does not equal $z$"$$\forall x~\exists y~\forall z~(x+y\neq z)$$