# Simple proof with quantifiers. Help me understand.

I have to negate

$$\forall x \in \mathbb{Z} \space \exists y \in \mathbb{Z} \space (( x \ge y) \land (x + y = 0))$$

and prove either the original proposition or negation is true. I get the negation

$$\exists x \in \mathbb{Z} \space \forall y \in \mathbb{Z} \space ((x > y) \lor (x+y \ne 0))$$

I'm not sure if that is right. I used quantifier negations and DeMorgan's law. Perhaps you could check my work. I was thinking that if I chose $$x= -2, y= 2$$ I can show that the negation is false, but I am confused. Please help.

## 1 Answer

When negating $x\geq y$, you should obtain $y>x$, but maybe that is just a typo in what you wrote. Apart from that, the formula you obtain is correct.

To prove that the negation is true, you must pick an $x\in \mathbb Z$, and prove that for all choices of $y\in\mathbb Z$, the predicate $y>x\lor (x+y\neq 0)$ is true. You don't get to choose what $y$ is.

You are right in choosing a negative number for $x$ though, can you finish the proof? As a hint, you can note that the predicate after the quantifiers can be written as $(x+y=0\Rightarrow y>x)$ (maybe you already have learned that $P\Rightarrow Q$ is logically equivalent to $Q\lor \neg P$, and I use this with $Q\equiv y>x$ and $P\equiv (x+y=0)$).