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I constructed this lean proof to the Generalization Rule in the Hilbert System. I have marked the passage where I am not sure if I overdo. It may not be proper language but I want it to be better readable for me in the future.

Theorem: Let $\Sigma$ be an axiom system of the Hilbert calculus in the language $L$, and let $\varphi$ be a formula of $L$ with $\Sigma \vdash \varphi$. Furthermore, let $x$ be a variable that does not occur freely in any formula of $\Sigma$. Then $\Sigma \vdash \forall x \varphi$.

Proof: We know that the Hilbert calculus is sound, i.e., $\Sigma \vdash \varphi \rightarrow \Sigma \vDash \varphi$. Furthermore, we know that $\Sigma \vdash \varphi$, so $\Sigma \vDash \varphi$, i.e. $\varphi$ holds in all models of $\Sigma$ for any interpretation over any domain $D$ (= $\varphi$ is true under $\Sigma$, $D$). Let $x$ be a variable that does not occur freely in $\Sigma$, i.e. $x$ cannot cause certain premises in $\Sigma$ to become true in a way that makes $\varphi$ false, so that $\varphi$ holds for any arbitrary assignment to $x$. Hence, $\Sigma \vDash \forall x \varphi$. By the completeness of the Hilbert calculus, it follows that $\Sigma \vdash \forall x \varphi$.

Mauro ALLEGRANZA
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God
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  • I changed the proof because originally I said $\varphi$ had to be a tautology which is false. It is just a true formula but it does not change much I think. This other site has not really something to do with what I ask. I wanna use soundness and completeness to prove a syntactical theorem. – God Dec 12 '24 at 22:51
  • "But then a universal quantifier ∀x can be added, because that is what truth means for a formula: holding for all objects of a reference set."

    This can only hold if there is no existential quantifier. If the reference set is the natural numbers, then we have true statements such as "there exists an even prime" which can get represented in any logic useful for arithmetic. As another example "there exists an identity number" can get represented for any logic sufficient for some understanding of addition or multiplication.

    I think "there exists a reference set" counterexamples also.

     – Doug Spoonwood Dec 12 '24 at 23:36
  • But if „there exists an even prime“ it is also true, in fact that is how truth is defined, that „for all x $\in \mathbb N$, there exists an even prime“. I do not see the problem. – God Dec 12 '24 at 23:46

1 Answers1

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Long comment

Usually, the proof that if $Σ \vdash \varphi$ and x does not occur freely in any formula of $Σ$, then $Σ ⊢ ∀x \varphi$ uses the proof system: axioms+rules. Thus, it can help if you list the axioms and the rules of inference (I expect only MP) used.

Having said that, IMO your argument is intuitively correct.

We say: $\Sigma \vDash \varphi$ iff for every structure $A$ for the language and every function $s : \text {Var } → \text { Dom}(A)$ such that $A$ satisfies every member of $\Sigma$ with $s$, also $\varphi$ is satisfied, i.e. $A,s \vDash \varphi$.

The part "every s" means that also "$s(x|d)$, for every $d \in \text { Dom}(A)$", where $s(x|d)$ is exactly like $s$ except that at the variable $x$ it assumes the value $d$.

But this means: $A,s \vDash \forall x \varphi$.

Mauro ALLEGRANZA
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