Variables bound by two quantifiers and quantifers binding non-existent variables

Question

Just wondering if these formulae are valid in classical logic:

1. $\forall x \exists x (Px) \leftrightarrow \exists x (Px)$
2. $\exists x \forall x (Px) \leftrightarrow \forall x (Px)$
3. $\forall x (Py) \leftrightarrow Py$
4. $\exists x (Py) \leftrightarrow Py$

I 'm a beginner (using Suppes'Introduction to Logic). Couldn't find any sources on this. Thanks!

Answer 1

See in Patrick Suppes, Introduction to Logic (1957 - Dover reprint), the recursive definition of formulas [page 52] :

(a) Every atomic formula is a formula.

[...]

(d) If $R$ is a formula and $v$ is a variable then $(v)R$ and $(\exists v)R$ are formulas.

Thus, according to the syntax, $Px$ is an atomic formula; so $(∃x)Px$ is a formula, and also $(x)(∃x)Px$ [i.e. $∀x∃x(Px)$] is a formula.

Now for the "meaning": of course quantifying a variable which is not present in a formula does nothing.

In formula $∃x(Px)$ there are no a free variable $x$; thus the "meaning" of $∀x∃x(Px)$ is exactly $∃x(Px)$.

How to prove it ? See BASIC RULES OF INFERENCE, page 99.

(a) Start with $∀x∃x(Px)$ and apply US :

from $∀xS$, derive $S(t)$, provided that no free occurrence of $x$ within scope of quantifier using variable of $t$ [i.e. we have to avoid that the term $t$, if containing a variable $y$, will be "captured" by a quantifier $\forall y$ or $\exists y$ already present into $S$].

In our case, $∃x(Px)$ has no free occurrences of $x$; thus, applying US we simply obtain $∃x(Px)$.

Thus, by Rule C.P [page 29] we may conclude with :

$∀x∃x(Px) \rightarrow ∃x(Px)$.

(b) Now consider $∃x(Px)$ and apply UG :

from $S$ derive $∀xS$, provided that $x$ is not flagged.

In our case, $∃x(Px)$ has no free occurrences of $x$; thus, $x$ is not flagged and we may apply UG to obtain $∀x∃x(Px)$.

Again by Rule C.P :

$∃x(Px) \rightarrow ∀x∃x(Px)$.

Finally, apply the definition of $\leftrightarrow$ to conclude with :

$∀x∃x(Px) \leftrightarrow ∃x(Px)$.

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Note

The same approach, using the "existential" rules, applies to : $∃x∀x(Px)↔∀x(Px)$.