Quantified Proposition

Question

I've been trying to translate the following sentences into quantified propositions by making sure I state all propositional functions that I use and any assumptions that I make.

There is exactly one person who hates everyone.

Let $H(x, y)$ be '$x$ hates $y$,' where the domain of $x$ is all people in the world.

Then, $\exists x \forall y\ (\ H(x,y)\ \land \forall z\ (z \neq x) \rightarrow \neg H(z,y)\ ) $.

Can you see if I'm on the right track here?

Just to clarify, does that one person hate themself as well? (I would assume so) — Regret, Feb 19 '15 at 08:38
@yellowhellow Are you familiar with $\exists!$, the unique existential quantifier? If not, you may want to check out this post. — Daniel W. Farlow, Feb 19 '15 at 08:48
I know the use of that quantifier. But, I do not want to use it for this example. — yellowhellow, Feb 19 '15 at 08:55

score 2 · Answer 1 · answered Feb 19 '15 at 08:41

2

Your statement, as it reads now, means:

There exists one such person $x$ that for every other person $y$, $x$ hates $y$ and nobody else hates $y$.

That is not what you want to state. For example, if

$A$ hates $A$, $B$ and $C$ and
$B$ hates $C$,

then the original statement (there exists a person who hates everybody) is true, but your statement is false (because it is not true that nobody else hates $C$)

answered Feb 19 '15 at 08:41

5xum

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$\exists x\ (\ \forall y\ H(x,y)\ \land \forall z \exists y\ (z \neq x) \rightarrow \neg H(z,y)\ ) $ Is this correct? – yellowhellow Feb 19 '15 at 08:54
@yellowhellow Yes. That looks better. – 5xum Feb 19 '15 at 08:56
Is there a way to get better at these things? I mean, they always confound me, even after I check and recheck my answers. – yellowhellow Feb 19 '15 at 08:57
@yellowhellow The standard way of saying "there exists exactly one $x$ for which $p(x)$ is true" is to write $\exists x: (p(x) \wedge \forall x': p(x') \implies x=x')$ – 5xum Feb 19 '15 at 09:02
I was wondering if it shouldn't rather be $\exists x \forall y\ (\ H(x,y)\ \land \forall z\ (z \neq x) \rightarrow \exists y\ \neg H(z,y)\ ) $ ? – yellowhellow Feb 19 '15 at 09:57
The OP's proposed solution doesn't exclude the possibility that $x=y$ as you claim. – Git Gud Feb 19 '15 at 09:57
or say $\exists x \forall y\ (\ H(x,y)\ \land \forall z\ (\forall y\ H(z,y))\ \rightarrow (z=x)\ ) $ – yellowhellow Feb 19 '15 at 09:58

score 0 · Answer 2 · answered Feb 19 '15 at 08:43

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$\exists!x\forall yH(x,y)$, where $\exists!$ is the unique existential quantifier.

answered Feb 19 '15 at 08:43

Daniel W. Farlow

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I know the use of that quantifier. But, I do not want to use it for this example. – yellowhellow Feb 19 '15 at 08:56
@yellowhellow Why not? You seem to be unnecessarily complicating things. – Daniel W. Farlow Feb 19 '15 at 08:56
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I want to make sure I fully understand how to do it the hard way, before I do it the easy way. – yellowhellow Feb 19 '15 at 08:58

Graham Kemp · Answer 3 · 2015-02-19T09:35:42.563

To say "there exists a unique $x$ for which the predicate $P$ holds" : we usually say "there exists some $x$ for which the predicate holds and if the predicate holds for anything it is that $x$".

$$ \exists x \Big(P(x)\wedge \forall z\big(P(z)\to x=z\big)\Big) $$

We can use the contrapositive.

$$ \exists x \Big(P(x)\wedge \forall z\big(x\neq z \to \neg P(z)\big)\Big) $$

Or more breifly: $$ \exists x\;\forall z \Big(x=z \leftrightarrow P(z) \Big)$$

And you want to do this for the predicate "$x$ hates every one". $$P(x) \;\equiv \;\forall y\; H(x,y)\\ \neg P(x) \;\equiv\; \exists y\;\neg H(x,y)$$

So: ...

Quantified Proposition

3 Answers3