Mixing and Distributing Qualifiers ($\forall x$, $\exists x$)

Context

I'm having trouble understanding the limited situations in which qualifiers can be distributed.

I am given that the rules are:

$$\forall x\left[P(x)\land Q(x)\right]\equiv\forall xP(x) \land\forall xQ(x) \tag{1}$$ $$\exists x\left[P(x)\lor Q(x)\right]\equiv\exists xP(x) \lor\exists xQ(x) \tag{2}$$

$$\forall x\left[P(x)\lor Q(x)\right]\not\equiv\forall xP(x) \lor\forall xQ(x) \tag{3}$$ $$\exists x\left[P(x)\land Q(x)\right]\not\equiv\exists xP(x) \land\exists xQ(x) \tag{4}$$

Questions

What are examples of $P(x)$ and $Q(x)$ that illustrate when these statements are true?
How can I show these statements always hold using proofs?
Why can't "$\forall x$" be distributed in example 3 or "$\exists x$" in example 4?

Solution 1:

Examples and counterexamples.

Consider the following statements:

All of my fish are ugly and blow bubbles.
All of my fish are ugly, and all of my fish blow bubbles.

These two statements are clearly equivalent. This is an example of $$\forall x(P(x) \wedge Q(x)) \equiv \forall x P(x) \wedge \forall x Q(x)$$ where here the quantifiers range over my fish, $P(x) \equiv$ "$x$ is ugly" and $Q(x) \equiv$ "$x$ blows bubbles".

Now consider the following two statements:

All of my fish are either goldfish or koi.
All of my fish are goldfish, or all of my fish are koi.

So long as not all of my fish are goldfish, and not all of my goldfish are koi, the first statement will be true while the second will be false. This is an example of $$\forall x(P(x) \vee Q(x)) \not \equiv \forall x P(x) \vee \forall x Q(x)$$ where the quantifiers range over my fish, $P(x) \equiv$ '$x$ is a goldfish' and $Q(x) \equiv$ '$x$ is a koi'.

There are similar statements that illustrate the analogous (in)equivalences for $\exists$.

Proofs.

A proof that $\forall$ distributes over $\wedge$ is as follows:

$$\begin{align} \forall x(P(x) \wedge Q(x)) & \Leftrightarrow P(x) \wedge Q(x) & (\forall\text{-elimination})\\ & \Leftrightarrow P(x)\ \text{and}\ Q(x) & (\wedge\text{-elimination})\\ & \Leftrightarrow \forall xP(x)\ \text{and}\ \forall xQ(x) & (\forall\text{-introduction})\\ & \Leftrightarrow \forall xP(x) \wedge \forall xQ(x) & (\wedge\text{-introduction}) \end{align}$$

Sometimes $\forall$-elimination is called 'generalisation' or something similar... your notation and terminology may differ, but the steps in the proof should be roughly the same.

There is a similar proof for $\exists$.

Abstract nonsense.

The category theorist in me wants to give another proof: quantifiers are really functors (between preorder categories of formulae with certain free variables, ordered by entailment) which satisfy a string of adjunctions $\exists \dashv {*} \dashv \forall$. Now 'or' is a coproduct and 'and' is a product, and since right-adjoints preserve limits (and left-adjoints preserve colimits), we must have that $\forall$ distributes over $\wedge$ and $\exists$ distributes over $\vee$.