What happen if a universal quantifier with a conjunction? Theorically, they clearly explained that the universal is the quantifer combining with a multiple conjuction.So if in statements with that quantifier, they have to with a implication not a conjuction, but somehow I still see those can have, specifically is:
All dog owners are animal lovers, is:
∀x (∀y (Dog(y) ∧ Rear(x,y)) → AnimalLover(x)
but, Dogs are all animals, is:
∀y Dog(y) → Animal(y)

 A: Your formula reads "For everyone, if everything in the world is a dog and they own everything, then they are an animal lover". What you want instead is "Everyone who owns some dog is a animal lover, which translates with am existential quantifier for $y$.
In general, it is not forbidden to use $\forall$ with $\land$ or $\exists$ with $\to$ in the sense that it wouldn't be a syntactically well-formed formula, it just doesn't correspond to a situation we'd typically want to express.
A: 
What happens if a universal quantifier with a conjunction?

If we assume that $\exists a: \neg P(a)$ for any predicate $P$, then $\forall a:[P(a) \land Q(a)]$ is a contradiction regardless of the truth value of $Q(a)$. Though technically valid, it is not something to be recommended in applications!
Proof: Using a form of natural deduction, we can formally prove $\exists a: \neg P(a) \implies \neg \forall a:[P(a) \land Q(a)]$ as follows:

A: Alternative answer - after @Pdubya_24 pointed out a possible misunderstanding:
The formula as you wrote it is missing brackets. The first opening bracket one is never closed (there should be another "$)$" at the end), and there should be another pair of brackets for what I believe is the intended interpretation. The way it is written, the structure of the formula is basically (inserting here a redundant pair of brackets for clarity)
$$\forall x ((\forall y (Dog(x) \land Raise(x,y))) \to AnimalLover(x))$$
i.e. the $\forall y $ ranges only over $(Dog(x) \land Raise(x,y))$. This is a well-formed sentence, but not a very useful one, as explained in my other answer.
What I believe the formula is meant to be is
$$\forall x (\forall y ((Dog(x) \land Raise(x,y)) \to AnimalLover(x)))$$
whith the $\forall y$ scoping all over $((Dog(x) \land Raise(x,y))) \to AnimalLover(x))$. In this case, the main operator of $\forall y$ is the implication, and the conjunction occurs only as a subformula on the left-hand side of the implication, not directly under $\forall$. So this is an instance of the common pattern $\forall y (\phi \to \psi)$, and not of the form $\forall y (\phi \land \psi)$, and nothing unusual is going on.
