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The actual full question: If there are free variables in the premises of a natural deduction that doesn't lead to contradiction, then is there a non-empty domain such that its constant terms satisfy the conclusion?

Well, let's consider a set of premises such that it has free variables somewhere. Using the rules of inference, I can get to a conclusion that also has the free variables. Since these free variables are any arbitrary terms, there has to be a non-empty domain that satisfy it. Is it correct? Does the domain have to be previously defined or can it be arbitrary?

Also, is the following inference ok? $$Px \vdash \exists x (Px)$$

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Yes; in Natural Deduction the above inference is a correct application of the $(\exists \text I)$ rule.

The standard semantics for classical predicate logic assume that the domain of interpretation is not-empty.

Thus, the above inference is sound because in every domain where $P(x)$ holds, it is true that there is some element that is $P$.

What does it mean that a formula with a free variables is true ?

Two possibilities : either that (i) $P(x)$ is a shorthand for $\forall x P(x)$, or (ii) there is an assignment function $s : \text {Var} \to A$, where $A$ is the domain of the interpretation $\mathfrak A$, such that $\mathfrak A \vDash P(x)[s]$.

In both cases, it is not necessary that the language has constant symbols.

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  • $\begingroup$ Perfect answer. Thanks! I was exactly wondering about the existencial introduction when there is a free variable in question. Good! $\endgroup$ – rgpluuuuuuu Apr 8 at 20:33

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