How do I negate the following statement?

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

$$\exists y \forall x ((y > 0) \land (x < y))$$


closed as off-topic by user21820, user91500, JonMark Perry, Antonios-Alexandros Robotis, José Carlos Santos Aug 10 '17 at 9:56

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Remember, when you negate a predicate, you negate "each part of it". It's like the negation distributes.

Some rules are: $$\neg\exists=\forall, \neg\forall=\exists,\neg(\lor)=\land, \neg(\land)=\lor,\neg P=\neg P$$ Therefore, distribute the $\neg$ $$\neg(\exists y \forall x ((y > 0) \land (x < y)))\iff \forall y\exists x(\neg((y > 0) \land (x < y))$$

Now, $\neg(y\gt 0)=(y\leq 0)$, "if its not greater, then it must be less than or equal to"

Similarly, $\neg(x<y)=(x\geq y)$

Finishing out negation off, we get:

$$\iff\forall y \exists x ((y\leq 0)\lor(x\geq y))\text{ notice $\land$ changes to $\lor$}$$

  • $\begingroup$ Question, how come the negative of $\forall$ isn't "none"? $\endgroup$ – Andrew Tawfeek Aug 9 '17 at 3:34
  • 1
    $\begingroup$ Consider $P(x):\text{ x is a boy}$. This means, "all $x$ are boys". Negating it, we see, "not all $x$ are boys". If not all of them are boys, there are definitely some that are boys, hence the $\exists$ $\endgroup$ – K Split X Aug 10 '17 at 14:25
  • $\begingroup$ Ahhh that makes things clear, thank you! $\endgroup$ – Andrew Tawfeek Aug 10 '17 at 14:26
  • $\begingroup$ Your welcome @AndrewTawfeek $\endgroup$ – K Split X Aug 10 '17 at 15:44

Essentially the $\forall$ becomes $\exists$ and vice-versa, and $\land$ becomes $\lor$. Hence, we have: $$\begin{align}\lnot\left(\exists y\forall x((y>0)\land(x<y))\right)&\iff\forall y\exists x(\lnot(y>0)\lor\lnot(x<y))\\&\iff\forall y\exists x((y\leq 0)\lor(x\geq y))\end{align}$$

  • $\begingroup$ Hi Dave, thanks for the answer. Just a question, why does > become <= instead of <? $\endgroup$ – Oliver K Aug 9 '17 at 4:46
  • $\begingroup$ Because $>$ does not allow for $=$, so the opposite of $>$ is either $<$ or $=$, and hence, $\leq$. $\endgroup$ – Dave Aug 9 '17 at 10:56

Theorem. The negation of $$\exists y \forall x ((y > 0) \land (x < y))$$ 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 $y>0$ is $y\le 0$

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

Lemma 5. The negation of $x<y$ is $x\ge y$

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


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