How do I negate the following statement? How do I negate the following statement?

What are the steps which should be taken to negate an argument such as the one given?

$$\exists y \forall x ((y &gt; 0) \land (x &lt; y))$$
Essentially the $\forall$ becomes $\exists$ and vice-versa, and $\land$ becomes $\lor$. Hence, we have:
$$\begin{align}\lnot\left(\exists y\forall x((y&gt;0)\land(x&lt;y))\right)&amp;\iff\forall y\exists x(\lnot(y&gt;0)\lor\lnot(x&lt;y))\\&amp;\iff\forall y\exists x((y\leq 0)\lor(x\geq y))\end{align}$$
Theorem. The negation of $$\exists y \forall x ((y &gt; 0) \land (x &lt; y))$$ is $$\forall y \exists x ((y \le 0) \lor (x \ge y))$$ 

Proof.
By Lemma 1, 2, 3, 4, and 5.


  Lemma 1. The negation of $\exists y~ P(y)$ is $\forall y,~\lnot P(y)$


Proof.
Calling $Y$ the universe
$$\exists y~ P(y)\equiv\exists y\in Y,~ P(y)\equiv Y\cap P\neq\emptyset$$
where $P=\{z|~P(z)\}$.

So its negation is: $$Y\cap P=\emptyset\equiv Y\subset P^c\equiv\forall y\in Y,~\lnot P(y)$$
where $P^c=\{z|~\lnot P(z)\}$


  Lemma 2. The negation of $\forall y~ P(y)$ is $\exists y,~\lnot P(y)$


Proof.
Calling $Y$ the universe
$$\forall y~ P(y)\equiv\forall y\in Y,~ P(y)\equiv Y\subset P$$

So its negation is: $$Y\not\subset P\equiv Y\cap P^c\neq\emptyset\equiv\exists y\in Y,~\lnot P(y)$$


  Lemma 3. The negation of $a\land b$ is $\bar a\lor\bar b$


Proof.
By de Morgan $$a\land b=\overline{\bar a\lor\bar b}$$


  Lemma 4. The negation of $y&gt;0$ is $y\le 0$


Proof.
Here it is understood that $y\in \mathbb{R}$, though not explicitly stated.
$$y&gt;0\equiv y\in\{z|~z&gt;0\}$$
So its negation is
$$y\not\in\{z|~z&gt;0\}\equiv y\in\{z|~z&gt;0\}^c$$
where $\{z|~z&gt;0\}^c=\mathbb{R}\setminus\{z|~z&gt;0\}=\{z|~z\le0\}$


  Lemma 5. The negation of $x&lt;y$ is $x\ge y$


Proof.
Here it is understood that $x,~ y\in \mathbb{R}$, though not explicitly stated.
$$x&lt;y\equiv (x,y)\in\{(w,z)|~w&lt; z\}$$
So its negation is
$$(x,y)\not\in\{(w,z)|~w&lt; z\}\equiv y\in\{(w,z)|~w&lt; z\}^c$$
where $\{(w,z)|~w&lt; z\}^c=\mathbb{R}^2\setminus \{(w,z)|~w&lt; z\}=\{(w,z)|~w\ge z\}$