Logic question is it true

Question

Exercise: $$\begin{align} (\forall x>2) &~:~ |x|<3 \tag{P1}\\ (\forall x\in\mathbb{R})(\exists\varepsilon > 0) &~:~-\varepsilon <x<\varepsilon \tag{P2}\\ (\forall x\in\mathbb{R})(\exists n\in\mathbb{Z}) &~:~ x<n\text{ or } x+1\geq n \tag{P3}\\ (\forall A>0)(\exists\varepsilon > 0)(\forall x>1)& ~:~x-\varepsilon<A\text{ and } x+\varepsilon\in\mathbb{N} \tag{P4} \\ (\forall A>0)(\exists\alpha>0)(\forall x\in\mathbb{R}^+) &~:~ x\geq A\implies x^2+1>A \tag{P5} \end{align}$$

Find $\neg P1,\neg P2...$ My answer

$$\begin{align} (\exists x>2) &~:~ |x|\geq 3 \tag{not P1}\\ (\exists x\in\mathbb{R})(\forall\varepsilon > 0) &~:~ -\varepsilon \geq x\text{ or } x\geq\varepsilon \tag{not P2} \\ (\exists x\in\mathbb{R})(\forall n\in\mathbb{Z}) &~:~ x\geq n \text{ or }x+1< n \tag{not P3}\\ (\exists A>0)(\forall\varepsilon > 0)(\exists x>1) &~:~ x-\varepsilon\geq A\text{ and } x+\varepsilon\in\mathbb{N} \tag{not P4}\\ (\exists A>0)(\forall\alpha>0)(\exists x\in\mathbb{R}^+) &~:~ x\geq A\implies x^2+1\leq A \tag{not P5} \end{align}$$

Is it true?

Answer 1

Not quite. Recall the following. For $A=B\land C$ is the same as $\neg A=\neg B \lor \neg C$.

Then recheck $P3$ and $P4$.

Answer 2

For P5, note that $\lnot(A \Rightarrow B)$ is not the same as $A \Rightarrow \lnot B$

However $\lnot(A \Rightarrow B)$ is the same as $A \text{ and } (\lnot B)$

This is because $A \Rightarrow B$ is the same as $(\lnot A) \text{ or } B$