Consider $(\mathbb N, +)$ as a model for the language with one binary function $+$ . Are the following statements true?

Question

Consider $(\mathbb N, +)$ as a model for the language with one symbol $+$ for a binary function. Are the following statements true?

1. $(\mathbb N, +) \vDash \forall x \exists y \forall z\ x + y\neq z$
2. $(\mathbb N, +) \vDash \exists x \forall y \exists z\ x + y\neq z$

I have put true for for both based mainly on intuition. I reasoned that the statements could be translated into English as (1) For any $x$ and $z$ you give me (from the natural numbers) there exists a $y$ such that $x + y\neq z$. Which I think is true. For the second one, I interpreted as for any y you give me, there exists an x and z such that $x + y\neq z$ is true, which I think is true as well. Is this the correct way to approach the problem?

Answer 1

Order of quantifiers matters! We claim that the first statement is false. To see this, we will prove its negation, namely: $$ (\mathbb N, +) \models \exists x \, \forall y \, \exists z \, [x + y = z] $$ Let $x = 3 \in \mathbb N$. Given any $y \in \mathbb N$, let $z = 3 +y \in \mathbb N$. Then $x + y = 3 + y = z$, as desired.

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The second statement is true. Let $x = 3 \in \mathbb N$. Given any $y \in \mathbb N$, let $z = 4 + y \in \mathbb N$. Then $x + y = 3 + y \neq 4 + y = z$, as desired.

Answer 2

The first is false. It says that for any natural number $x$, there is a natural number $y$ such that $x+y$ is not a natural number (for all natural numbers $z$, $x+y$ differs from $z$). Note that in your explication of the sentence, you implicitly read $\exists y\forall z$ as $\forall z\exists y$.

Now perhaps you can deal with the second.