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There is an equivalence between cellular automata and formal systems, you can code one into the other and vice versa. But in the the cellular automata (CA) the rules of inference are fixed and are pretty simple (for instance, the universal two neighbor, two state rule 110). So all the rules of the formal system that is coded into a rule 110 CA (also a formal system) must be coded mostly in the input. And vice versa, if you have a formal system resembling rule 110, you can code your axioms into inference rules in a different formal system.

Is this correct? Do this exchangeability between axioms and rules have any meaning for the foundations of logic and/or mathematics?

noelia
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2 Answers2

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In axiomatic and natural deduction proof systems, axiom schemata and inference rules are interchangeable, for the most part. In a system with modus ponens and a conditional $\to$, for instance, any inference rule of the form $$\Lambda\,/\,\Gamma$$ could be replaced by the axiom schema $$\Lambda \to \Gamma$$ In the first case, if you have $\Lambda$, then by the inference rule, you may infer $\Gamma$. In the second case, if you have $\Lambda$, then instantiate the axiom schema to yield $\Lambda \to \Gamma$ and use modus ponens to infer $\Gamma$. I think that shows how inference rules can be replaced by axiom schemata (as long as you have modus ponens or an equivalent).

Replacing axiom schemata by inference rules is a little bit easier, though it may feel like a bit of hack. If you have the axiom schema $$\Lambda$$ just replace it by an inference rule with no preconditions $$/\,\Lambda$$

This approach should work regardless of the particulars of your axiom schemata and inference rules, since checking whether a step in a derivation is a valid application of an inference rule based on some set of premises should not be any more or less difficult than checking whether a formula is an instance of an axiom schema.

Joshua Taylor
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    There is a slight issue in the first half. The inference rule $A \vdash B$ is only applicable if one can derive $A$, while the axiom $A \to B$ must be true in every model. Thus, if $A$ is not derivable, it may be that a model does not satisfy $A \to B$ even though those true sentences are closed under the inference rule $A \vdash B$. This phenomenon does occur in certain formal systems that arise in mathematical logic, although it is obscure enough that introductory texts usually do not mention it. – Carl Mummert Apr 23 '13 at 03:18
  • @CarlMummert, That's a very interesting point. Off the top of my head, I'm not aware of any such systems, but I'd be very interested to hear some examples. Is my intuition correct that such systems don't have a deduction theorem (since, if they did, one could assume $A$, derive $B$, and then, by the deduction theorem, conclusion $A \to B$)? – Joshua Taylor Apr 23 '13 at 03:23
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    Exactly - these systems do not have a deduction theorem. – Carl Mummert Apr 23 '13 at 03:24
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    My other comment, which I deleted, was worded poorly. Let $A$ be $c = 0$ and $B$ be $d = 1$ in the language of arithmetic with new constant symbols $c,d$. Let $M$ be the standard model of arithmetic, with $c$ interpreted as $0$ and $d$ interpreted as $0$. Let $T$ be the closure of $PA$ under the usual rules of inference. Then $T$ is also closed under the rule $c = 0 \vdash d = 1$, because $PA$ can't prove $c = 0$ under the usual rules of inference. Hence $M$ is a model of $T$. But $M$ is not a model of $PA$ plus the additional axiom $c = 0 \to d = 1$. – Carl Mummert Apr 23 '13 at 03:29
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    (I was able to follow the first example, but the rewrite is fine, too.) So, in general such a system would not have deduction theorems. It seems like this system can't have a semantically oriented soundness proof either, since there are models on which $c = 0 / d = 1$ introduces false sentences. (At least, the soundness of that rule won't be shown by demonstrating that for every set of formulae $\Phi$, $\Phi \models c = 0$ implies $\Phi \models d = 1$.) But, as you point out, since $c = 0$ will not be provable, neither will $d = 1$. Does a different type of soundness apply here? – Joshua Taylor Apr 23 '13 at 03:38
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    In general, proving a soundness theorem with extra inference rules requires restricting the set of "allowable" models to only include models where the true sentences are closed under the new inference rules. This is automatic for the usual set of inference rules - every model works for those, by the usual completeness theorem. But additional inference rules, as you point out, may not be sound for the set of all models. The example I gave is an example of a new inference rule that is not sound, where the usual form of the soundness theorem that quantifies over all models would be false. – Carl Mummert Apr 23 '13 at 03:56
  • Modal logics are an example where the deduction theorem does not hold (answering a question above. For example, see TLA+2 https://research.microsoft.com/en-us/um/people/lamport/tla/tla2-guide.pdf – 0 _ Nov 14 '16 at 07:08
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Cellular automata are considered a subset of the formal systems, so, what you say about axioms and rules being relative is correct. I actually remember reading something about that a long time ago in a book about automatic theorem proving, but I cant remember the details. Regarding the philosophical meaning of this, my view is that it only shows how many options you have when choosing a formal system to code your model of interest.

Wolphram jonny
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