Proving if A for all x then B using contrapositive

Question

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?

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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 $\mathbf{x}$.

Answer 1

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?

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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 $x=y$.

Answer 2

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.