Can predicate and propositional logic be 'mixed'? Motivation
I wanted to write out a simple statement from linear algebra in a very formal, predicate logic way. The statement is that given two vector spaces $V \subset W$ over a field $\mathbb{F}$, $V$ is a subspace iff for any two vectors, any linear combination of them is contained in $V$.
I initially wrote $$\forall a \in V \forall b \in V \forall \alpha \in \mathbb{F} \forall \beta \in \mathbb{F} \big(V \text{ is a subsapce iff } a\alpha + b\beta \in V\big)$$
I realised this was wrong, and it should be (as confirmed by this set of notes I found)$$V \text{ is a subsapce iff } \forall a \in V \forall b \in V \forall \alpha \in \mathbb{F} \forall \beta \in \mathbb{F} \big( a\alpha + b\beta \in V\big)$$
It felt wrong in the first instance that the universal quantifiers extend over the statement that $V$ is a subspace.
Question
However, my issue is with proving these two are different. The way I see it, it's akin to trying to showing that $$\forall x P(x) \iff Q \text{ and } \forall x (P(x) \iff Q)$$ are different statements. Here I write $P(x)$ to be analogous to arbitrary vectors/scalars, over which we use a universal quantifier, and I write $Q$ as a statement that is independent of the variable $x$, to represent the statement that a vector is a subspace. To my mind, $Q$ is like a propositional statement, that is either simply either true or false.
But I am failing to to prove the two are not equivalent statements (I also tried showing that they ARE the same statements). In particular, I tried writing out truth tables, but got stuck and came across this post which explains that quantifiers do not work with truth tables.
Are these statements equivalent? And how do you go about proving it? (I realise my discussion of the logic here is likely shaky at best, so apologies if I'm under obvious misconceptions.)
 A: You're right that these statements are different.
The quickest way to see this is by looking at the claim
$$
\forall a,b \in V . \forall \alpha, \beta \in \mathbb{F} . (V \text{ is a subspace} \iff \alpha a + \beta b \in V)
$$
If this is true, then (basically by definition) we should have
$$
(V \text{ is a subspace} \iff \alpha a + \beta b \in V)
$$
for every choice of $a,b \in V$ and $\alpha, \beta \in \mathbb{F}$. (Do you see why?)
In particular, this must be true when $a,b = 0$ and $\alpha, \beta = 0$. But in this case the claim reads
$$
(V \text{ is a subspace} \iff 0 \in V)
$$
which we know is false!

Said in the more abstract terms you adopts later, you're exactly right that
$$
\forall x \in X. (P(x) \iff Q) \quad \quad (1)
$$
is different from
$$(\forall x \in X . P(x)) \iff Q \quad \quad (2)$$
and the same argument works (as long as the domain $X$ has more than one element).
In formula $(1)$ we're saying that $P(x) \iff Q$ is true for every $x \in X$. So in particular, this tells us that $P(x_1) \iff Q \iff P(x_2)$ so that $P$ is constant on its domain $X$.
In formula $(2)$ we're saying that $Q$ is true exactly when $P(x)$ is always true. This is what we usually want to talk about.

But how can we formally prove that $(1)$ and $(2)$ are different? We need to exhibit a countermodel. We actually already have a countermodel from our first example of vector subspaces, but here's a small one you can really get your hands on:
Let $X = \{a,b\}$, with $P(a) = \top$, $P(b) = \bot$, and $Q = \bot$.

*

*Can you show that this model thinks $(1)$ is false?

*Can you show that this model thinks $(2)$ is true?

*From here, conclude that $(1)$ and $(2)$ must not be provably equivalent. If they were, they would have to be equivalent in all models, but here we've built a model where they differ.


I hope this helps ^_^
