Null Quantification with empty domain

Question

Is Null Quantification false?

Null Quantification state that $\exists x [P \lor Q(x)]\equiv P\lor\exists x[Q(x)]$

; $P$ represents any wff in which $x$ does not occur free.

If we set domain of discourse be empty and let T be tautology

we will get $\exists x [\text{T} \lor Q(x)]$ is false. But $\text{T}\lor\exists x[Q(x)]$ is true.

Answer 1

Long comment

Not all valid sentences are inclusive valid, i.e. true in all interpretations, including the interpretation with an empty domain.

Example: $(\forall x) \bot$ is true in the empty domain (because all universaly quantified formulas are) and thus $(\forall x) \bot \to \bot$ is false.

But it is valid, because all instances of $(\forall x) \varphi \to \varphi$ are.

If so, it seems that you have proved that :

$∃x[ \top \lor Q(x)] \leftrightarrow (\top \lor ∃x[Q(x)])$

is valid but not inclusive valid.

See e.g. E.Mendelson, Introduction to Mathematical Logic (6th ed, 2015) Ch.2.16 Quantification Theory Allowing Empty Domains, page 146-on.