Sign up ×

In pretty much any Model Theory or Logic textbook you will find the following claim, where $T$ is a theory (a set of $\mathsf{L}$-sentences),

$T$ is consistent if and only if $T$ is satisfiable.

Sometimes it is proved and sometimes it is not, I have read the proof from a few books and I have tried to shorten it as much as possible to what follows,

Assume to the contrary that there exists a theory $T$ such that $T$ is consistent and $T$ is not satisfiable. Since $T$ is not satisfiable, there does not exist any models $\mathcal{M}$ of $T$. So, any $\mathsf{L}$-structure which attempts to model $T$ is a model of $\perp$. Then, $T \vDash \perp$ so by the Completeness Theorem, $T \vdash \perp$; yet this contradicts the assumption that $T$ us consistent. Therefore, $T$ is consistent if and only if $T$ is satisfiable.

My question is as to why we can make the jump from "there does not exist any models of $T$", to "every model of $T$ is a model of $\varphi \wedge \neg \varphi$."

I thought I understood this stuff, but I am reviewing Godel's Completeness Theorem, the Compactness Theorem, and a couple related corollaries and found myself confused on this point. I'm sure it is very simple, but I thought I'd ask. Thanks.

EDIT: In some of the comments it has been indicated that the proof is circular because it uses the completeness theorem but is attempting to prove that direction. I will literally quote exactly from David Marker's, "Model Theory: An Introduction"

Corollary: $T$ is consistent if and only if $T$ is satisfiable.


Suppose that $T$ is not satisfiable. Because there are no models of $T$, every model of $T$ is a model of $(\phi \wedge \neg \phi )$. Thus, $T \vDash (\phi \wedge \neg \phi)$ and by the Completeness Theorem $T \vdash (\phi \wedge \neg \phi)$

What is the difference between Marker's proof than the one I gave?

share|cite|improve this question
Why would you think you can't? And furthermore, the negation of such a statement would be obviously false. (since it starts with "there exists a model of T such that....") –  Hurkyl Mar 24 '12 at 0:23
It might help you to rephrase "For every model X of T, <comment about X>" into "For every X, if X is a model of T, then <comment about X>". (assuming, of course, you are comfortable with vacuous implications) –  Hurkyl Mar 24 '12 at 0:27
One direction is easy: If $T$ is satisfiable then $T$ is consistent. The other direction is the Completeness Theorem, definitely not easy since we have to construct a model from syntactic elements. –  André Nicolas Mar 24 '12 at 0:56
The larger problem in your proof is that you are proving the completeness theorem ("if $T$ is consistent then $T$ is satisfiable") and you use the completeness theorem in contrapositive form in the middle of the proof... –  Carl Mummert Mar 24 '12 at 0:56
@Samuel: I think the reason people are objecting is that many people consider that "consistent => satisfiable" is the Completeness Theorem. It looks like the textbook you're quoting from proves an equivalent statement to be the Completeness Theorem (for all statements $\phi$ and sets of statements $T$, we have $T \vDash \phi$ implies $T \vdash \phi$) and then uses it to get a one-line proof of "consistent => satisfiable". The hard work goes into the proof of the Completeness Theorem. –  Ted Mar 24 '12 at 4:39

1 Answer 1

up vote 1 down vote accepted

In order to grasp the fact that there are two equivalent versions of the same theorem, I think it can be useful to review the details of a typical proof of it. See Joseph Shoenfield, Mathematical Logic (1967), page 43 :

Completeness Theorem (First Form : Gödel). A formula $A$ of a theory $T$ is a theorem of $T$ iff it is valid in $T$ [see page 22 : By a model of a theory $T$, we mean a structure for $L(T)$ in which all the non-logical axioms of $T$ are valid (true). A formula is valid in $T$ if it is valid in every model of $T$; equivalently, if it is a logical consequence of the nonlogical axioms of $T$.]

This theorem has a second form, which concerns consistency.

Completeness Theorem (Second Form). A theory $T$ is consistent iff it has a model.

We first show that the second form of the completeness theorem implies the first.

In view of the closure theorem and its corollary, it suffices to prove the first form for a closed formula $A$.

By the corollary to the reduction theorem for consistency [see page 42], $A$ is a theorem of $T$ iff $T \cup \{ \lnot A \}$ is inconsistent.

By the second form of the completeness theorem, this holds iff $T \cup \{ \lnot A \}$ has no model. Now since $A$ is closed, a model of $T \cup \{ \lnot A \}$ is simply a model of $T$ in which $A$ is not valid. Hence $A$ is a theorem of $T$ iff $A$ is valid in every model of $T$.

share|cite|improve this answer

Your Answer


