Suppose we want to represent "nobody loves everybody" and the universe of discourse is all people. Would a correct representation be:
$$\require{enclose}\enclose{horizontalstrike}{(\forall x)(\sim \!x\Rightarrow(\forall y)(y\Rightarrow (\sim\!x) \; \text{loves} \; y )) \; \; \; \text{(Universe is all People)}}$$
Edit: I made a second attempt
$$\sim\!\!(\forall x)(x\Rightarrow(\forall y)(y\Rightarrow x \; \text{loves} \; y )) \; \; \; \text{(Universe is all People)}$$
According to the answer key, the answer is
$$(\forall x)\!(\sim\!\!(\forall y)(x \; \text{loves} \; y))$$
Am I correct? If not then why?
Best Answer
Firstly, $x$ and $y$ as propositions do not make sense. It would if you we using the proposition $x\in\mathcal U$, where $\mathcal U$ is some explicit domain or other restriction. However, as you have an implicit domain of discourse, you do not need the implication. So we shall drop them:
$$\sim\!\!(\forall x)(\forall y)(x \operatorname{loves} y ) \; \; \; \text{(Universe is all People)}$$
Secondly, this says "not everyone loves everyone", which is still not what you require, that is "not someone loves everyone."
Remember this: No-one is short for "Not Someone"; it is the claim there does not exist an example of this.
So we shall change that to an existential. We can also use the duality of quantifies to 'move' the negation inwards, once or twice.
$${\sim\!\!(\exists x)(\forall y)(x \operatorname{loves} y ) \; \; \; \text{(Universe is all People)}\\(\forall x)({\sim}\forall y)(x\operatorname{loves}y)\\(\forall x)(\exists y)\;{\sim}(x\operatorname{loves}y)}$$
Which says, "Everyone has someone they do not love." Which is equivalent to "No-one loves everyone".