Discrete Math Nested Quantifier

Question

Suppose that L(x,y) means that x loves y. I am to write symbolically, "For every person that loves someone, there exists a unique person that loves them" I feel like there are a hundred different ways to answer this question! The closest I can get is AxEyL(x,y) E!cL(c,x) I feel as if there should be an amplification arrow in between the two though..

Answer 1

First of all, I would use $z$ instead of $c$, since $c$ is typically used for a specific individual, and you want to use a variable. Also, this $z$ will need to love $y$ as well as $x$.

But that is not the biggest issue with your attempt.

The sentence you are trying to symbolize is what is called a 'donkey sentence', so named after the following sentence:

'every farmer who owns a donkey beats it'

If you try to symbolize this, your first attempt might well be something like:

$\forall x ((Farmer(x) \land \exists y (Donkey(y) \land Owns(x,y))) \to Beats(x,y))$

The problem, however, is that in the $Beats(x,y)$ part, the $y$ is free, so this isn't right.

Indeed, in order to make reference to the donkey owned by the farmer, it turns out we need to paraphrase this sentence to: 'for any farmer and any donkey owned by that farmer, the farmer beats the donkey.

The same is true for your sentence. Here, we are tempted to do something like:

$\forall x (\exists y Loves(x,y) \to \exists ! z (Loves(z,x) \land Loves(z,y)))$

And we have the exact same problem: $y$ is free in $Loves(z,y)$

So, we need to rephrase this sentence as:

'For any person and any person loved by that person, there is a unique person that loves those two people'

So that becomes:

$\forall x \forall y (Loves(x,y) \to \exists ! z (Loves(z,x) \land Loves(z,y)))$