Contraposition and contradiction I have some questions about mathematical logic.
1) About reasoning by contraposition :
If we have two predicates $P(x)$ and $Q(x)$ on a set $A$, do we have the following equivalence between the following propositions :
$(\forall x \in A, P(x) \Rightarrow Q(x)) \Leftrightarrow [(\forall x \in A, P(x)) \Rightarrow (\forall x \in A, Q(x))]$ ?
From that, when we say that we do a reasoning by contraposition, what do we do exactly :
$(\forall x \in A, \neg P(x) \Rightarrow \neg Q(x))$
or
$[(\forall x \in A, \neg Q(x)) \Rightarrow (\forall x \in A, \neg P(x))]$ ?
2) About reasoning by contradiction :
Again, if we have for example two predicates $P(x)$ and $Q(x)$ on a set $A$, and if we have the proposition $(\forall x \in A, P(x) \Rightarrow Q(x))$, when we do a proof by contradiction, what do we assume exactly :
$\neg (\forall x \in A, P(x) \Rightarrow Q(x)) \Leftrightarrow (\exists x \in A, P(x) \wedge \neg Q(x))$
or
$[\forall x \in A, \neg (P(x) \Rightarrow Q(x))] \Leftrightarrow (\forall x \in A, P(x) \wedge \neg Q(x))$ ?
3) More generally :
If we have two predicates $P(x)$ and $Q(x)$ on a set $A$, when we state : "$P(x) \Rightarrow Q(x)$", are we agree that we cannot assign the value "true" or the value "false" to this ?
In general, are we agree that (when we read that kind of notation in a book for example), we can understand : "$\forall x \in A, P(x) \Rightarrow Q(x)$" ?
Thank you for your future answers.
 A: (1) No, they are not equivalent. For example take $A$ to be $\mathbb N$ and $P(x)$ to mean "$x$ is odd" and $Q(x)$ to mean "$x$ is even".
Then $(\forall x\in A:P(x))$ and $(\forall x\in A:Q(x))$ are both false, and therefore $(\forall x\in A:P(x)) \Rightarrow (\forall x\in A:Q(x))$ is true.
But $(\forall x\in A: P(x)\Rightarrow Q(x))$ is false, because $P(x)\Rightarrow Q(x)$ is false when $x=5$.

(2) $\neg(\forall x\in A: P(x)\Rightarrow Q(x))$ is equivalent to $\exists x\in A:P(x)\land\neg Q(x)$.
These are not equivalent to $\forall x\in A:\neg(P(x)\Rightarrow Q(x))$. For example, if we assume the above meanings of $A$, $P$, and $Q$, then $\neg(\forall x\in A: P(x)\Rightarrow Q(x))$ is true, but $\forall x\in A: \neg(P(x)\Rightarrow Q(x))$ is false because $P(x)\Rightarrow Q(x)$ does hold for $x=28$.

It is not clear to me what either of these have to do with reasoning by contraposition and/or contradiction.

(3) You are right that $P(x)\Rightarrow Q(x)$ does not in itself have a truth value until it is specified what $x$ is.
But it is not the case that you can therefore reinterpret it willy-nilly to mean $\forall x\in A: P(x)\Rightarrow Q(x)$ in an arbitrary context.
Some texts do unfortunately leave it to the reader to insert quantifiers that way, but that is fraught with opportunities for misunderstanding, and is a poor model for how to express mathematical relationships readably and unambiguously.
