Difference between if A, then B, and A and B I am reading the book How to prove it, and doing some of the exercise.
In section 2.2, it asked us to negate the statement: Everyone has a roommate who dislike everyone. And then reexpress the results as equivalent positive statements.
My trial:
Let $R(x,y)$ means $x$ is a roommate of $y$. And L(x,y) means $x$ likes $y$.
$$\forall x \exists y (R(x,y) \to \forall z(\neg L(y,z)))$$
negate this and get
$$\neg \forall x \exists y (R(x,y) \to \forall z(\neg L(y,z))$$
The answer is 
$$\exists x \forall y(R(x,y) \land \exists z (L(y,z)))$$
The answer is very similar to the correct answer, $\exists x \forall y (R(x,y) \to \exists z(L(y,z)))$
After some investigation, I found that if I set the first equation to be
$$\forall x \exists y (R(x,y) \land \forall z(\neg L(y,z)))$$
then I will get the correct answer.
So, my question is what is the difference between the equation that used "if" and the one used "and".
I am confused because in another question: Everyone who is majoring in math has a friend who needs help with his homework. I wrote the logical form as 
$$\forall x (M(x) \to \exists y(F(x,y) \land H(y)))$$.
where $M(x)$ means $x$ is a math major, $F(x,y)$ means $x$ is a friend of $y$ and $H(y)$ means $y$ needs help on homework.
Can I write it as 
$$\forall x (M(x) \land \exists y(F(x,y) \land H(y)))$$.
 A: $\forall x ~\exists y ~(R(x,y) \to \forall z(\neg L(y,z)))$ says "For everyone ($x$) there exists someone ($y$) who, if they ($y$) are the first person's ($x$) roommate, then they ($y$) will dislike everyone." &nbsp Thus it is not what you want.


In section 2.2, it asked us to negate the statement: Everyone has a roommate who dislike everyone. 

Recall, universal statements are restricted by conditional and existential statements are restricted by a conjunction. $$\forall x\in A: P(x)\iff \forall x~(x\in A\to P(x))\\\exists y\in B: Q(y)\iff \exists y~(y\in B\land Q(y))$$
You want to say "For every person there exists a person, who is their roommate and dislikes everyone".   Here the implcit domain is people and the restriction is on the existential (to roommates of the first entity).   It is, basically: $$\forall x{\in}\text{People}~\exists y{\in}\text{RoommatesOf}(x)~\forall z{\in}\text{People}:\lnot L(y,z)$$
Hence:$$\forall x~\exists y~(R(x,y)\land\forall z~(\lnot L(y,z)))$$


I am confused because in another question: Everyone who is majoring in math has a friend who needs help with his homework. I wrote the logical form as 
  $$\forall x (M(x) \to \exists y(F(x,y) \land H(y)))$$.
  where $M(x)$ means $x$ is a math major, $F(x,y)$ means $x$ is a friend of $y$ and $H(y)$ means $y$ needs help on homework.

Now, here you are restricting both the universal entities and the existential entities:
$$\forall x{\in}\text{MathMajors}~\exists y{\in}\text{FriendsOf}(x): H(y)$$ 
