last day I faced a task, in which one has to negate and prove the following predicate: $$ \forall x \in \mathbb {N} \exists y \in \mathbb {N} \exists z \in \mathbb {N} : x^2=y^2+z \wedge y \neq x $$ Thus the negated statement shall be: $$\exists x \in \mathbb {N} \forall y \in \mathbb {N} \forall z \in \mathbb {N}: x^2 \neq y^2 + z \vee y = x $$
Now for the proof part:
As far as I know, if one was able to find a number that contradicts this property, the whole statement would be false. Otherwise one shall prove the statement is true. What I'm confused about is the order and the way I can pick numbers to check the statement.
If there exists an $x$ so that for all $z$ and $y$ $x^2\neq y^2+z \vee y=x$ am I about to find at least one that makes it work for all y and z? For example I pick $x=5$ than $25 \neq y^2 + z \vee y=x $ But it doesn't obviously hold for all possible $y$ and $z$. But if I was able to find such $x$ that satisfies all $y$ and $z$ the property holds then, right?
And on the other hand I could interpret the statement as: for all $y$ and $z$ there is an $x$ such that $x^2 \neq y^2 + z \vee y = x $ and in this case i would pick an y and z first and show that there would be an x such that the property holds. But then it will always be an $x$ for that it holds.
I think the first approach makes more sense to me, but correct me please if I was wrong.
HINT: Take $x$ to be the smallest member of $\Bbb N$; is there a $y\in\Bbb N$ such that $y\ne x$ and $x^2=y^2+z$ for some $z\in\Bbb N$?
Your negation is fine.