The best really introductory textbook on FOL (first-order logic) that I have ever seen is "Language, Proof and Logic". It is really designed for beginners; it teaches not only the syntax and semantics of FOL (not as mathematical objects but truly at a beginner's level), but also teaches a complete Fitch-style deductive system for FOL. Beware of using any 'introductory texts' without checking thoroughly, as many of them that I have come across have fundamental conceptual flaws! I did not check 100% of LPL, but I didn't notice any significant issue so far.
After your target audience has learnt everything in LPL, they can easily move on to the other texts mentioned in this thread. Hannes' text, for example, starts by giving some simple structures from ordinary mathematics and explaining how we use FOL to axiomatize them.
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 ^_^
Best Answer
No, the formula is not valid.
And, in general, we cannot test a formula of predicate logic with truth-table.
For a counter-example, consider the interpretation with domain $\mathbb N$ and interpret formula $\alpha$ with the binary relation $\lt$, i.e.
Obviously, it is not true that $n < m \lor m < n$, for every $n,m$.