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An axiom system $\Sigma$ in $L$ is called contradiction-free or consistent if there is a formula $\varphi$ with $\Sigma \not\vdash \varphi$. (For even a single contradiction of the form $\varphi \land \neg \varphi$ would allow everything to be derived from $\Sigma$.)

Theorem: If $\Sigma := \Sigma \cup \{\varphi\}$ ($\Sigma$ is extended with the axiom $\varphi$) is consistent, then $\Sigma \not\vdash \neg \varphi$ and vice versa.

If $\Sigma$ contains the formula $\varphi$ and is consistent, then obviously $\neg \varphi$ cannot also follow from $\Sigma$, because from such a contradiction, it would be possible to derive everything in Hilbert's calculus via axiom xi ($(\varphi \wedge \neg \varphi) \to \psi$), making $\Sigma$ inconsistent, in contradiction to the assumption. And if $\Sigma \not\vdash \neg \varphi$, then $\Sigma$ would be consistent with $\varphi$, alone because not every formula is derivable. This would admittedly be more of a metamathematical idea for a proof; the proof in Hilbert's calculus would look more technical.

Is it true to call my proof metamathematical and distinguish it from a proof in the Hilbert-calculus? Feel also free to point out any mistake in the proof (it is more a sketch) itself.

God
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  • Just to emphasize a point in Mauro's answer: it would be impossible to give a non-metamathematical proof of this theorem, because the very definitions of "consistent" and $\vdash$ are metamathematical. – Alex Kruckman Dec 18 '24 at 17:04
  • "distinguish it from a proof in the Hilbert-calculus" Regardless of which Hilbert system (there are infinitely many) you have in mind, every proof in a Hilbert system is purely formal, and usually you cannot use any rule but modus ponens. You seem to have no good idea of what actual proofs in Hilbert systems look like. Here is a tool to handle them, and this blog post has a formula-based example. – xamid Dec 23 '24 at 12:14

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Is it true to call my proof metamathematical and distinguish it from a proof in the Hilbert-calculus?

Yes, the derivability relation: $\vdash$ is a property of the (sequences) of formulas of the calculus and thus it is meta-.

Compare the symbol $\vdash$ with that for the connective: $\to$.

With the second one we build formulas of the language $L$ like $p \to p$, while the first one is used to express properties of the formula: $\vdash p \to p$ expresses the fact that the formula $p \to p$ can be derived in the calculus using the propositional axioms $\Sigma$.

The derivation will be a sequence of formulas starting with a couple of instances of propositional axioms: $[p \to ((p \to p) \to p)] \to [(p \to (p \to p)) \to (p \to p)]$ and $p \to ((p \to p) \to p)$ and applying Modus Ponens.

Note: also the proof in your previous post is meta-: the Generalization Theorem is not a formula of the calculus but a meta-theorem.

Mauro ALLEGRANZA
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  • Can one say that statements are metamathematical to a mathematical system iff they are not a wff of that system. So every proof of such statement, even if it uses wffs from the formal system to show some property or existence of formulas, is also metamathematical. So for instance the Deduction Theorem cannot be proven in the Hilbert-Calculus at all, right? It can only be proven for the Hilbert-Calculus by applying rules of some other calculus (the one we can do strong induction with) that can talk about wffs of Hilbert‘s calculus and how they behave. – God Dec 19 '24 at 04:20
  • @God - correct. – Mauro ALLEGRANZA Dec 19 '24 at 06:47
  • Thx. Last question to this: I like my proof more than the proofs usually given in textbooks because it is pretty simple and straightforward. Why do they not use it in textbooks instead using more technical and formal proofs that use induction which makes it overcomplicated in my eyes? – God Dec 19 '24 at 18:51
  • @God your proof is fine, but there are cases where induction is needed e.g. Deduction Theorem. – Mauro ALLEGRANZA Dec 19 '24 at 21:41