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As per the title, why does the ability to generate a model from axioms prove they are consistent?

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3 Answers 3

I would just like to mention that one generally does not construct a model of a theory $T$ from the axioms of $T$ itself. Instead, one uses external means to produce a structure that one can show is a model of $T$.

Here is where Ben's answer comes into play, although in my opinion it obfuscates matters: if you construct a model of $T$ using some other theory $T^\prime$, then what you have really done is shown that the consistency of $T^\prime$ implies the consistency of $T$. There are times where this distinction is important. For example, from the axioms ZFC of set theory one can construct a model of Peano Arithmetic, PA. It is unknown whether ZFC is actually consistent, though widely believed, and therefore we cannot immediately conclude that PA is consistent, though this is also widely believed.

But let's assume that we have constructed a model $M$ of $T$ using methods that are above reproach. For example, $M$ might be a finite $\mathcal{L}$-structure of specific size $n$. The claim that $T$ is consistent then comes down to two facts about the underlying logic.

Fact 1: If $M$ is a structure in some language $\mathcal{L}$, then $M \models \varphi$ and $M \models \neg \varphi$ cannot both be true for every $\mathcal{L}$-sentence $\varphi$. (In fact, exactly one of $M \models \varphi$ or $M \models \neg \varphi$ holds.)

Fact 2: If $T$ is a theory in a language $\mathcal{L}$ and $\varphi$ is a $\mathcal{L}$-sentence such that $T \vdash \varphi$, then $M \models \varphi$ for all models $M$ of $T$.

Now, if $\varphi$ were an $\mathcal{L}$-sentence such that $T \vdash \varphi$ and $T \vdash \neg \varphi$, then by Fact 2 we conclude that $M \models \varphi$ and $M \models \neg \varphi$, but this contradicts Fact 1!

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One thing that may be confusing matters here is that we can define consistency either syntactically ($T$ can't be used to prove a contradiction) or semantically ($T$ has a model in some theory $T'$). The former is absolute, while the latter is relative to $T'$. –  Ben Crowell Mar 9 '12 at 16:06

Given a language $\mathsf{L}$, we want to construct of a model from a set of given axioms. Let us call the $\mathsf{L}$-sentences $A =\{\varphi_{1},...,\varphi_{n}\}$ the axioms. An $\mathsf{L}$-theory (call it $\mathbf{T}$) is a set of $\mathsf{L}$-sentences, so we have that $A$ is an $\mathsf{L}$-theory and so is any other set of $\mathsf{L}$-sentences. It is defined that $\mathcal{M}$ is a model for $\mathbf{T}$ if $\mathcal{M} \vDash \psi$, for all $\mathsf{L}$-sentences $\psi \in \mathbf{T}$.

When a theory $\mathbf{T}$ is inconsistent, it means that there is at least one $\psi \in \mathbf{T}$ such that $\mathbf{T} \vdash \psi \wedge \mathbf{T} \vdash \neg \psi$. Yet, our definition of a model $\mathcal{M}$ for an $\mathsf{L}$-theory $\mathbf{T}$ is exactly that $\mathcal{M} \vDash \psi$, for all $\psi \in \mathbf{T}$.

Hence it follows that for our set of axioms $A = \{\varphi_{1},...,\varphi_{n}\}$, that if we can construct a model $\mathcal{M}$ for $A$ it means that $\mathcal{M} \vDash \varphi_{i}$, where $1 \leq i \leq n$. Therefore, it must necessarily be the case that if we can construct a model $\mathcal{M}$ for our set of axioms $A$, that $A$ is consistent, otherwise we would not be able to construct such a model $\mathcal{M}$.

If you have any other interesting questions about model theory feel free to ask, I'm glad to help! Let me know if you need any clarification about the terminology or you aren't acquainted with what some of the symbols are.

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A point to make here is that Tarski's definition of truth precludes models from satisfying both a statement and its negation, and implies that the set of truths in a model is closed under logical deduction, so if ${\mathcal M}$ is a model of a theory $T$, it is a model of all the consequences of $T$ and, in particular, $T$ cannot prove an inconsistency. –  Andres Caicedo Mar 9 '12 at 2:39
    
That's exactly what I was trying to say, maybe it was too obfuscated with symbols. –  Samuel Reid Mar 9 '12 at 2:53
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This may be pedantic, but I find your use of quantifiers in the meta language a bit distracting. One cannot formally write $\exists \psi$, nor $T\vdash\psi \wedge T\vdash \neg \psi$, nor $\forall \psi$, instead opting for the english: There is a formula $\psi$, $T\vdash \psi$ and $T\vdash\neg \psi$, and for all formulas $\psi$. I guess I just feel as though someone learning logic may think these logical connectives have the same meaning in the meta language as in the actual language, and this is simply false, at least syntactically. –  Jason DeVito Mar 9 '12 at 4:22
    
@JasonDeVito: Thank you for pointing that out, I was being a bit too informal. I have edited my answer to include more english phrases as opposed to some sort of meta quantification. –  Samuel Reid Mar 9 '12 at 5:03

Creating a model doesn't show that a theory is consistent. What it shows is that, if we can create a model of theory A within theory B, then we know that if A is inconsistent, B must be inconsistent as well.

For example, say A is elliptic geometry. At one time, people thought that A was inconsistent, and they tried really really hard to prove that it was inconsistent. But let B be Euclidean geometry. We can construct a model of A in B, by taking A's points to be points on a sphere, identifying points that lie at each other's antipodes, and taking lines to be great circles. This shows that if A was inconsistent, then B would as well.

In most cases, there is no hope of proving B's consistency, due to Godel's theorem. However, it often happens that B is a theory that people don't think is likely to be inconsistent, e.g., ZFC set theory or Peano arithmetic. Therefore A has a model in B, everybody realizes that it's a waste of time to worry about inconsistencies in A.

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(Of course, I understand your point, but) I'll have to disagree here. If you have a model, you have consistency, period. This is just a matter of definitions. If you use a theory $B$ to prove that there is a model of a theory $A$ then, sure, this is a conditional consistency statement, but not an equiconsistency, as you claim. ZFC proves that PA is consistent, but PA and ZFC are not equiconsistent in any non-trivial sense. Also, if you go this route, then you have to explain what you mean, and this is subtle, as $A$ is different from its formalization within $B$. –  Andres Caicedo Mar 9 '12 at 4:00
    
I'm down voting this answer because I believe it to be misleading and incorrect; I think you are using the idea of model too loosely. My concerns are similar to Andres'. –  Samuel Reid Mar 9 '12 at 4:10
    
@AndresCaicedo: Thanks for your comment. You're correct that equiconsistency is not implied. I've edited my answer appropriately. –  Ben Crowell Mar 9 '12 at 4:19
    
However, I don't see how you can assert that the existence of a model in B implies consistency of A. If B is inconsistent, then the existence of a model of A in B does not prove the consistency of A. –  Ben Crowell Mar 9 '12 at 4:26
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Ben, I never claimed that. –  Andres Caicedo Mar 9 '12 at 5:13

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