Prove or disprove formally: $(\forall x F) \lor G \vDash \forall x (F \lor G)$

Question

I want to formally prove that this statement is either false or true. I think it is false, because $G$ could contain the variable $x$ unbound and for some interpretation $\mathcal A$, $G$ could be true for some $x$ while $(\forall x G)$ is false and $(\forall x F)$ is false. So $(\forall x F) \lor G$ evaluates to true, while $\forall x (F \lor G)$ evaluates to false, so $(\forall x F) \lor G \nvDash \forall x (F \lor G)$ for the same interpretation $\mathcal A$.

However, I am in doubt if my reasoning is correct and I do not know how to prove this formally.

Answer 1

Assuming that in your system, formulas with free variables can be assigned a truth value relative to an assignment function, you are correct with your suspicion that $G$ could be true of one object but not another.

Suppose $F$ has a free variable, $F(x)$, and $G$ has a free variable, $G(x)$. Consider the counter model $\mathcal{M} = \langle D, \mathcal{I} \rangle$ with $D = \{a, b\}, \mathcal{I}(F) = \emptyset, \mathcal{I}(G) = \{a\}$ and assignment function $v : x \mapsto a$.

Then, since $v(x) = a \in \mathcal{I}(G)$, we have that $\mathcal{M}, v \vDash G(x)$, so $\mathcal{M}, v \vDash (\forall x F(x)) \lor G(x)$.
But since $b \not \in \mathcal{I}(F)$ and $b \not \in \mathcal{I}(G)$, it holds for the assignment variant $v'$ of $v$ with $v': x \mapsto b$ that $\mathcal{M}, v' \not \vDash F(x) \lor G(x)$, hence $\mathcal{M}, v \not \vDash \forall x (F(x) \lor G(x))$.
Since there are formula instances, a structure and an assignment function such that the premise formula is true but the conclusion is not, the inference scheme does not hold.

Answer 2

In the statement $(\forall x F) \vee G$, $G$ is not within the scope of a quantifier on $x$. That means that, if this is a sentence which can be evaluated as "true" or "false", $G$ must not include $x$. Which means that when you put a quantifier on $x$ in front of it, as in $(\forall x)G$, the result is counterintuitive - the quantifier has no effect!

As a simple example, $G$ might be "the sky is blue". $(\forall x)G$ would then be "no matter what $x$ is, the sky is blue". But whether or not the sky is blue doesn't depend on which object is named "$x$", so $(\forall x)G$ is true exactly when $G$ is.

I would recommend, for this one, thinking about what happens for a particular $x$, say $a$. Suppose that $(\forall x F) \vee G$ is true. Can you show that $F(a) \vee G$ is true? If so, since $a$ was chosen arbitrarily, it must be that $(\forall x)(F \vee G)$ is true.