Properties of bound variables

Question

I'm wondering if following properties hold for not free variables (bound variables)

Let $x$ be a variable and $\alpha, \beta $ be formulas.

1. If $x$ is not free in $\alpha$, then $(\forall x) (\alpha) \equiv \alpha$
2. $(\forall x)(\lnot\alpha) \equiv \lnot(\forall x)(\alpha)$ and by (1), if $x$ is not free in $\alpha$ then $(\forall x)(\lnot\alpha) \equiv \lnot \alpha$
3. $(\forall x)(\alpha \lor \beta) \equiv (\forall x)(\alpha)\lor(\forall x)(\beta)$ and by (1),
   $(\forall x)(\alpha \lor \beta) \equiv (\forall x)(\alpha)\lor(\forall x)(\beta)$ if $x$ is free in $\alpha, \beta$
   $ \quad \quad \quad \quad\equiv \alpha\lor(\forall x)(\beta)$ if $x$ is not free in $\alpha$ and free in $\beta$
   $ \quad \quad \quad \quad\equiv (\forall x)(\alpha)\lor\beta$ if $x$ is free in $\alpha$ and not free in $\beta$
   $ \quad \quad \quad \quad\equiv \alpha\lor\beta$ if $x$ is not free in $\alpha$, $\beta$

If $x$ is not free in a formula $\alpha$, then $x$ is either in a quantifier or $\alpha$ does not contain $x$, so I guess any additional quantifier is not meaningful and hence we can remove it (e.g. ($\forall x)(\forall x) (\alpha) \equiv (\forall x) (\alpha)$))

The reason why I want to check if these properties hold is I think they are useful when I prove a part of Deduction Theorem:
Let $\Sigma$ be a set of formula, $\theta$ a sentence, $\alpha, \beta$ formulas and assume the $x$ is not free in $\beta$. Show that $\Sigma \vdash \theta \rightarrow (\alpha \rightarrow \beta)$ implies $\Sigma \vdash \theta \rightarrow ((\exists x)(\alpha) \rightarrow \beta)$

Answer 1

First, you don't really need to distinguish between $x$ not being free in $\alpha$ due to it not occurring or due to it being bound. If it is bound in $\alpha$, then you can rename the bound variable and thus it will no longer occur.

1 is correct, albeit this is because in typical formulations of (classical) first-order logic, axioms are added that imply that the (semantic) domain will be non-empty. In free logic, which omits those axioms, these would not be equivalent statements since $\forall x.\alpha$ would always be true in an empty domain regardless of what $\alpha$ was. In particular, even if $\alpha$ was false.

2 is incorrect. Maybe it was a typo. $\forall x.\neg \alpha \equiv \neg\exists x.\alpha$. Consider an example where $\alpha$ holds for only some but not all inputs. You are correct about the case where $x$ is not free in $\alpha$, but this is just a special case of 1 as you state.

3 is also incorrect. Let $\alpha$ be "$x$ is even" and $\beta$ be "$x$ is odd". Again, when $x$ is not free in $\alpha$ or $\beta$ then this is a special case of 1. It's not necessary to analyze 1 again for various connectives.