Logical Connectives and Quantifiers

Question

So I have been told that the rule of thumb when symbolizing Existence statements is to use the "and/$\wedge$" logical connective.

$$\exists x (P_x \wedge Q_x)$$

Whereas symbolizing a Universal statement usually uses a "conditional/$\rightarrow$".

$$\forall x(P_x\to Q_x)$$

But when we compound the quantifiers, should we base the main logical connector on the innermost or the outermost quantifier?

$$\forall x \exists y(P_x\wedge Q_y)\text{ or }\forall x \exists y (P_x\to Q_y)$$

If there is no general rule, what is the reasoning behind choosing the correct logical connective for compound statements?

Answer 1

What he is likely referring to is sometimes you will have relations appear informally inside the quantifier, and those are the rules for how it is translated. For instance,

$(\forall x \in I)P(x)$ informally expresses $(\forall x)(x\in I \to P(x))$

and $(\exists x \in I)P(x)$ informally expresses $(\exists x)(x \in I \land P(x))$

EDIT-

In response to your comment, if he/she is referring to the informal notation above the correct way would be something like the following:

$(\forall x \in I)(\exists y >0)(P(x,y))$ would be translated

$(\forall x)\Big((x \in I) \to \exists y\big((y > 0) \land P(x,y)\big)\Big)$

So you use both. Absent this, I don't know which convention he could be referring to because both can be correct depending on what the statement itself is trying to express. There are plenty of times when you want weaker statements. For instance, you generally want the weakest possible assumptions that suffice to prove a given theorem. This allows you to use the theorem more broadly/frequently as the conditions that must be met to employ it are easier to satisfy.

Answer 2

There are no rules of thumb when it comes down to expressing logical statements - depending on what you want to express you may need to use one connective or another.

For example, suppose you want to express that for every number $n$ which is not prime there is another number $m$ different from $n$ and $1$ such that $m$ divides $n$.

Then you may express this as $\forall n \exists m . \neg isPrime(n) \implies Divides(m,n) \wedge m\neq n $