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I was reading a book where the author converts the sentence

Everyone who loves all animals is loved by someone

to this logical sentence:

∀x [∀y Animal(y) => Loves(x, y)] => [∃z Loves(z, x)

However I came up with this logic, which's different:

∀x [∀y Animal(y) ∧ Loves(x, y)] => [∃z Loves(z, x)]

Note however that the next sentence, is converted similar to second version one instead of the first version. So going by the previous example, why is the sentence

Anyone who kills an Animal is loved by no one

converted to

∀x [∃y Animal(y) ∧ Kills(x, y)] => [∀z ¬Loves(z, x)]

instead of this:

∀x [∃y Animal(y) => Kills(x, y)] => [∀z ¬Loves(z, x)

I'm confused as to why we pick one over the other. Can anyone please explain this?

user97
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  • Your clause "$\forall y$ $Animal(y)$ $\wedge$ [stuff]" doesn't mean what you want it to; it means "Every $y$ is an animal and [stuff]." Presumably your universe has things in it which are not animals. "Every animal [stuff]" is different from "Everything is an animal and [stuff]." – Noah Schweber Mar 07 '22 at 02:50
  • @NoahSchweber So why would the second sentence not use the same format of using => instead of ∧ – user97 Mar 07 '22 at 03:01
  • Because the quantifier involved is different in each case. Think about the difference between "There is something which is an animal and belongs to me" versus "There is something which, if it is an animal, belongs to me." The former basically says I have a pet; the latter says literally nothing if we accept that there are things which aren't animals. $\forall$ and $\exists$ require different Boolean operations to form their corresponding "bounded" versions (e.g. "Every animal" and "Some animal" respectively); see also this old answer of mine. – Noah Schweber Mar 07 '22 at 03:04
  • This summary confirms that your book is correct on both counts. Here are two more explanations in addition to Noah's above link. – ryang Mar 07 '22 at 04:57

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