$\forall x\forall y(\dots )$ or $\forall x(\forall y\dots )$.

Question

The sentence that needs translation is this:

Everything hates something, but only scientists hates everything.

With respect to the first part of the sentence I am fairly certain the correct translation is:

$$\forall x\exists yHxy$$

but with respect to the latter part I have come down to two solutions: $$ \forall x\forall y(Hxy \to Sx)$$ or $$ \forall x(\forall yHxy \to Sx)$$

My knowledge tells me these aren't equivalent, but I just can't grasp which one fits the sentence best (I should perhaps add that I do have some trouble grapsing two quantifiers together, in general).

The latter seems to me to be saying something along the lines of, "For everything x, if it hates everthing $y$, then it, $x$, is a scientists" But what about the former? Does it say "for every pair of things, $x$ and $y$, if $x$ hates $y$, then it, $x$, is a scientists", or what?

Are any of them correct, for that matter?

Thank you!

Answer 1

The second version is what you are looking for, and you describe both pretty well (there might be some natural language nuances, but I would disregard those interpretations as irregular or uncommon).

Some other way to spot a difference is to negate both expressions. In such case

$$\neg\forall x\forall y (Hxy \to Sx) \quad \text{ is equivalent to } \quad \exists x \exists y (Hxy \land \neg Sx), $$

that means "there exists a non-scientist that hates something", while

$$\neg\forall x ((\forall y Hxy) \to Sx) \quad \text{ is equivalent to } \quad \exists x ((\forall y Hxy) \land \neg Sx), $$ that is, "there exists a non-scientists that hates everything".

Please be aware that it is the quantifier being on the left side of the implication operator $\to$ that changes everything, e.g.,

• $\forall x \forall y (Px \to Qxy)$ is the same as $\forall x (Px \to \forall y Qxy)$,
• $\forall x \forall y (Px \land Qxy)$ is the same as $\forall x (Px \land \forall y Qxy)$,
• $\forall x \forall y (Qxy \land Px)$ is the same as $\forall x (\forall y Qxy \land Px)$,
• of course there are other operators with similar behavior, e.g. $\gets$ (the backward implication).

This phenomenon is related to covariance and contravariance (i.e. the implication operator is contravariant in its left side). Some similar issue is also discussed in the comments here.

Edit: On your second question whether

$$∀x.\ \Big(∀y.\ H(x,y)\Big)→S(x) \quad \text{ is equivalent to } \quad ∀x.\ ∃y.\ H(x,y)→S(x).$$

In classical logic yes. We have that $$p \to q \iff \neg p \lor q$$ so \begin{align} \forall x\ \big(\forall y\ &Hxy\big) \to Sx, \\ \big(\forall y\ &H\mathtt{x}y\big) \to S\mathtt{x}, \\ \neg\big(\forall y\ &H\mathtt{x}y\big) \lor S\mathtt{x}, \\ (\exists y\ \neg{\!\!}&H\mathtt{x}y) \lor S\mathtt{x}, \\ \exists y\ (\neg{\!\!}&H\mathtt{x}y \lor S\mathtt{x}), \\ \exists y\ &H\mathtt{x}y \to S\mathtt{x}, \\ \forall x\ \exists y\ &Hxy \to Sx. \end{align}

In non-classical logic I don't know. In some yes (e.g. this is how you implement existential types in Haskell), but things get subtle and I'm not expert enough to write about it.

I hope that helps $\ddot\smile$