Proving if A for all x then B using contrapositive I'm trying to prove a statement of the form $\forall x \, A \implies B$ aka if A is true for all $x$, then B must be true, using proof by contraposition. As far as I know, this involves proving the following statement:
$$\neg B \implies \neg \left(\forall x \, A \right),$$ which should be equivalent to
$$\neg B \implies \exists x \, \neg A.$$ However, this seems somewhat off to me. To me it looks like I just have to provide an example of $\neg A$, given that $\neg B$, but that doesn't seem right. 
Another possibility was that maybe the universal quantifier shouldn't be "distributed", which would result in
$$\forall x \ \neg B \implies \neg A.$$
This can't be correct either, see below $^1$.
What's the correct way to prove a statement like this using contraposition?

As an example, the actual statement I'm trying to show is that if $\mathbf{x} \cdot \mathbf{y}=0$ for all $\mathbf{x} \in \mathbb{R}^n$, then $\mathbf{y}=\mathbf{0}$. Also $^1$, it can't be the last form either, since clearly it isn't true for all $\mathbf{x}$.
 A: 
However, this seems somewhat off to me. To me it looks like I just have to provide an example of $\neg A$, given that $\neg B$, but that doesn't seem right. 

Why doesn't it seem right?
Take, for example, the statement

If all walls of this room are red, then this is a red room.

In this statement, $A$ is "wall $x$ is red", and $B$ is "this room is red".
then $\neg B\implies\exists x: \neg A$ is equivalent to the statement 

If this is not a red room, then it must have at least one wall which is not red.

which is sounds right, doesn't it?

Or, in your assertion, you are trying to prove that if $x\cdot y=0$ for all $x\in\mathbb R^n$, then $y=0$. Well, in this case, $A$ is "$\forall x: x\cdot y=0$, and $B$ is $y=0$.
The statement $\neg B\implies \exists x:\neg A$ is then

If $y$ is not the zero vector, then there exists at least one vector $x$ such that $x\cdot y$ is not $0$.

This statement is indeed true, and easily shown by taking $x=y$.
A: Indeed, what you have to prove is that $\neg B \implies \exists x \, \neg A$. So, consider the assertion, concerning $\mathbb R^n$,$$\forall x\,x.y=0\implies y=0.$$You wish to prove that if $y\neq0$, then there is some $y$ such that $x.y\neq0$. Indeed this is true: just take $y=x$, and you're done.
