Premise: Let $F(x,y)$ be the statement $x$ can fool $y$, where the domain for discourse for both $x$ and $y$ is all people.

I converted the following corresponding statements:

1) Nobody can fool me

2) Anybody can Fool Fred

3) Everyone can fool someone


1) $\neg \exists xF(x,me)$

2) $\forall xF(x,Fred)$

3) $\forall x\exists yF(x,y)$

Now I am trying to negate these and I can't just throw in a not in front of the statements and I don't know how to go about this.

I looked at the negation rules and came up with this but I doubt it's correct:

1) $\neg\exists xF(x,me) \iff \forall x\neg F(x,me)$

2) $\forall xF(x,Fred) \iff \forall x\neg\neg F(x,Fred) \iff \neg[\exists x\neg F(x,Fred)]$

3) $\forall x\exists yF(x,y) \iff \neg\forall x\exists yF(x,y)$

  • 2
    $\begingroup$ Well, $\lnot (\lnot Q)$ is $Q$ so the negation of $\lnot\exists xF(x,me)$ would be ... $\exists xF(x,me)$, right? $\endgroup$ – fleablood Aug 15 '17 at 22:17
  • 1
    $\begingroup$ And the negation of $\forall x A$ is $\exists x \lnot A$ so the negation $\forall x F(x, fred)$ is $\exists x \lnot F(x, fred)$. $\endgroup$ – fleablood Aug 15 '17 at 22:19
  • 1
    $\begingroup$ 3 is a little harder. As with 2) is $\lnot(\forall x \exists y F(x,y))$ will be $\exists x (\lnot \exists y F(x,y))$ but we need to find the negation of $\exists y F(x,y)$. The negation of $\exists x Q$ is $\forall x \lnot Q$ so the negation is $\exists x \forall y \lnot F(x,y)$. i.e. There is someone who can not fool anybody. $\endgroup$ – fleablood Aug 15 '17 at 22:25
  • $\begingroup$ I think you're spot on, @fleablood. Please post your comments in an answer! I'll upvote it. $\endgroup$ – Namaste Aug 15 '17 at 22:29
  • $\begingroup$ Wait... you aren't actually negating them. You are stating the equivalence of them. You can't "just" put a not in front, but you must start by putting a not in front. $\endgroup$ – fleablood Aug 15 '17 at 22:30

Except for what I assume to be a typo on the answer to 3, what you've written is correct (or more accurately, not wrong), although it seems you've missed out the actual negation of the statements which is what you're after.

Essentially, all you need is the rule

$$\neg((\forall x)\quad P(x))\iff((\exists x)\quad \neg P(x))$$

as well as the more basic $\neg\neg P\iff P$.


$$S_1\equiv¬((∃x)\quad F(x,\text{me}))$$

$$S_2\equiv(∀x)\quad F(x,\text{Fred})$$

$$S_3\equiv(∀x)\quad((∃y)\quad F(x,y))$$

Then you're after $\neg S_1,\neg S_2,\neg S_3$.

As has been noted in the comments, the first is simple

$$\neg S_1\equiv\neg\neg((∃x)\quad F(x,\text{me}))$$

$$\neg S_1\equiv(∃x)\quad F(x,\text{me})$$

Your working for the second is helpful and we get

$$\neg S_2\equiv\neg((∀x)\quad F(x,\text{Fred}))$$



Finally, as has been pointed out by @fleablood, the last is slightly harder, but is just applying the above rule twice

$$\neg S_3\equiv\neg((∀x)\quad((∃y)\quad F(x,y)))$$

$$\equiv(∃x)\quad\neg((∃y)\quad F(x,y))$$

$$\equiv(∃x)\quad((∀y)\quad \neg F(x,y))$$


Okay. First to translate:

1) "Nobody can fool me". "Nobody" can be thought of as either "Everyone can not...", or as "There doesn't exist anyone who can..."

So either $\forall x\lnot F(x,me)$ or $\lnot \exists x F(x,me)$ will both be correct and are equivalent. So yes you are are correct.

As for 2) and 3)... I have nothing to add. You did those just fine. (You did 1) just fine as well, but I did have something to add.)

So the negations:

The rules are simple:

A) $\lnot$(That's not true) is, of course, "That is true" so $\lnot(\lnot Q) \iff Q$.

B) $\lnot$(Everbody does something) is, there is at least one person who doesn't do it. So $\lnot \forall x Q \iff \exists x \lnot Q$.

C) $\lnot$(Somebody does something) is. Nobody does something. So $\lnot \exists x Q \iff \forall x \lnot Q$.


1) $\lnot(\lnot \exists x F(x,me)) \iff \exists x F(x,me)$


$\lnot \forall x\lnot F(x,me) \iff \exists x\lnot(\lnot F(x,me)) \iff \exists x F(x, me)$.

i.e. It is not the case that noone can fool me $\iff$ someone can fool me.

2)$\lnot(\forall x F(x, fred)) \iff \exists x \lnot F(x,fred)$.

Not everbody can fool Fred = there is someone who can't fool Fred.

3) $\lnot (\forall x \exists y F(x,y)) \iff$

$\exists x \lnot(\exists y F(x,y)) \iff$

$\exists x \forall y \lnot F(x,y)$

i.e. not(Everbody can fool somebody) = somebody can not fool anybody.


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