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Let $L(x,y)$ be the statement "x loves y", where the domain for both x and y consists of all people in the world.

Express the below statement using quantifiers and predicates.

"There is exactly one person whom everybody loves".

My work:

This can be thought of as, There exist a person X, such that all people love him and for all people Z, if everyone love Z, then this Z has to be X.

$\exists x \Biggl(\forall y \biggl(L(y,x) \land \forall z(L(y,z) \rightarrow (z=x)) \biggr) \Biggr)$

Am I in correct direction?

user3767495
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  • This looks pretty good, except the part $\forall z\left( L(y,z)\rightarrow(z=x)\right)$ says that "for each person $z$, if $y$ loves $z$ then $z=x$", but it is not the case that $x$ is the only person who receives any love. What you could do instead is just try to also say that "and for each $z$ also satisfying the condition of $x$, we have $z=x$." – Dave Feb 19 '19 at 04:54
  • Didn't get you Dave :( – user3767495 Feb 19 '19 at 05:20
  • For instance, maybe $$\exists x\left((\forall y L(y,x))\land \forall z\left((\forall y L(y,z))\rightarrow (z=x)\right)\right)$$ – Dave Feb 19 '19 at 15:03

1 Answers1

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No, although close.

$\exists x \Biggl(\forall y \biggl(L(y,x) \land \forall z(L(y,z) \rightarrow (z=x)) \biggr) \Biggr)$

"There is some-x whom every-y loves and any-z that is loved by that y is that x."

"There is exactly one person everybody loves and noone loves anyone else."

Uniqueness is $\exists x~\Big(P(x)\land\forall z~\big(P(z)\to z=x\big)\Big)$ -- "there is someone who satisfies the predicate, and anyone who satisfies the predicate is that someone."

Here, your predicate is that the person is loved by everyone. $~P(x):=\forall y~L(y,x)$

You want to say "There is someone loved by everyone and anyone, who is loved by everyone, is that (first)person." $$\exists x~\biggl(\forall y~\Bigl(L(y,x)\Bigr)\land\forall z~\Bigl(\forall y~\bigl(L(y,z)\bigr)\to z=x\Bigr)\biggr)$$

Graham Kemp
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  • ohh.So, the problem is caused due to $\forall y (P(x) \land \forall z (P(x) \rightarrow z=x))$ right? – user3767495 Feb 20 '19 at 03:42
  • No. The problem is you must to substitute $\forall y~(L(x,y))$ for $P(x)$ in $\exists x~(P(x)\land\forall z~(P(z)\to z=x))$ So notice the bracket placement, and the second $\forall y$ . – Graham Kemp Feb 20 '19 at 05:07