-2

The model existence theorem - or whatever it's called in English - is important because it is easy to prove the completeness and compactness theorem for FO1 from it. Of course, I would like to make sure that it is really true and therefore I want to prove it, but what I find is very technical, too technical for me. So I try my best and give you my idea how to prove it in hope you can tell me where the shortcomings are, so that I get a better idea about the proof. Maybe someone can even give the correct proof or some link where it's proved and one can follow easier.

The model existence theorem reads (let K = calculus): K has a model <-> K without contradiction.

->: We accept the negation, i.e. K has a model and K is contradictory. Then because of EFQ in K anything can be inferred, including the theorem "p & ~ p", but with that K no longer has a model, in contradiction to the assumption and thus the negation of the negation, i.e. the implication, holds.

<-: We again accept the negation, i.e. K is consistent and K has no model. We know that only contradictions are unsatisfiable, everything else is satisfiable, i.e. has at least one model because every form except A & ~A has a possible truth assignment that could be made with appropriate interpretation. Well, K is consistent, i.e. there are no contradictions derivable and thus no unfulfillable formulas, so that K has a model in contradiction to the assumption and thus the negation of the negation, ie the implication, applies.

  • "We know that only contradictions are unsatisfiable, everything else is satisfiable" That's part of what we're trying to prove, actually. You have to distinguish between contradictions in the semantic sense (= unsatisfiable sentences) for which the claim is trivial and irrelevant, and contradictions in the syntactic sense (= sentences $\varphi$ such that $\neg\varphi$ is a theorem of the system in question) for which the claim is very relevant but not at all trivial. – Noah Schweber Dec 02 '19 at 04:31
  • A (first order/propositional) theory having a model iff it does not prove a contradiction is usually called the Completeness Theorem. – Reveillark Dec 02 '19 at 04:32
  • 2
    This is a similar issue to your earlier question, incidentally. As to technicality, it is a genuinely hard theorem and there's no way around that - if you want to understand it you really need to put work into following the details. – Noah Schweber Dec 02 '19 at 04:34
  • @Noah: So my proof is correct, it's just is too superficial, assuming things that are not as trivial as I pretend they are? What is the theorem called in English anyway? –  Dec 02 '19 at 04:36
  • 1
    @Pippen No, it's not correct at all, it's circular - you're basically saying "X is true because we know X is true." (Actually, it's even worse than that: you're really only treating a special case of the theorem, when $K$ is assumed to be finite. But that's a side issue.) As to the name, "model existence theorem" is sometimes used but "completeness theorem" is by far the more common one. – Noah Schweber Dec 02 '19 at 04:36
  • Waht's the name of this theorem in English, so that I can look for links? –  Dec 02 '19 at 04:38
  • @Pippen As I said above, "model existence theorem" is sometimes used but "completeness theorem" is by far the more common name. – Noah Schweber Dec 02 '19 at 04:39
  • I thought the completeness theorem says: K |= x <-> K |- x. Isn't that different from what I ask about? –  Dec 02 '19 at 04:42
  • @Pippen The equivalence between the two statements is immediate enough that the name "completeness theorem" is in my experience freely used for each. – Noah Schweber Dec 02 '19 at 04:44
  • K is consistent, so you cannot infer any a, ~a. So with other words, K consists of (finitely or infinitely) many formulas that neither are contradictions nor can one infer contradictions. But then we must be able to always find some interpretation where K's set of formulas get a model. Proof: Let's say there's a formula f of K that is not true there. Then we can always make f true by just extending the interpretation, so that f gets true also and if there are infinitely many such f's in K, we just do it infinitely often. For me it seems trivial, but only technically hard to prove. –  Dec 02 '19 at 09:38
  • See Henkin's theorem and Model existence theorem. – Mauro ALLEGRANZA Dec 02 '19 at 10:33
  • 3
    @Pippen "For me it seems trivial, but only technically hard to prove." Absolutely not. In fact, the argument you've given in your previous comment is completely wrong. In particular, note that it uses no properties of first-order logic in particular - so it would imply that every logic is compact, which is not true (e.g. second-order logic). And again it ignores the point that the connection between contradictions and unsatisfiable sentences is part of what you're trying to prove. This really is a hard theorem - the technicalities aren't just annoying details, they're core ideas. – Noah Schweber Dec 02 '19 at 16:23

1 Answers1

2

Short answer:

No, that doesn't work (nor does the similar proposal advanced in the comment thread).

Indeed, the technicalities you're trying to avoid are actually core ideas (the same issue holds of your earlier question). It is true that sometimes the heart of a hard technical argument can be boiled down to something quite simple, but this isn't one of those times. (That said, I've written a summary of the ideas of the argument which I think is close-to-optimally concise.)

Longer answer:

There are multiple issues (including serious vagueness), but a good starting point is the following:

Suppose $p$ is a sentence such that $p$ does not prove $a\wedge\neg a$ for any $a$ (that is, $p$ does not prove a contradiction). How do we know that $p$ is satisfiable?

You take this for granted ("We know that only contradictions are unsatisfiable, everything else is satisfiable"), but this is in fact part of what you're trying to prove. Remember that there is a priori difference between "prove" ($\vdash$ -which is syntactic) and "entail" ($\models$ - which is semantic). Trivially $p$ is satisfiable iff for no $a$ do we have that $p$ entails $a\wedge\neg a$, but when we replace "entails" with "proves" we get something highly nontrivial.

Put another way, you're equivocating between two possible meanings of "contradiction" - a syntactic one, namely "statement (which implies a statement) of the form $a\wedge\neg a$ for some $a$," and a semantic one, namely "statement which is unsatisfiable." And this isn't allowed (yet).

  • There is a secondary issue as well: namely, the one-sentence version of the model existence theorem does not in general imply the full model existence theorem. The idea you sketch in your comment suggests that if each sentence in a set $\Gamma$ is satisfiable then we can find a single model satisfying them all via some kind of "limit construction," but that's something that takes proof - and is in fact much more complicated than it first appears (to the point that that doesn't at all get to the core ideas).

On a much more technical note, it's worth observing that there are various ways in which my essentiality claim at the top of this answer can be made precise. Here are a few:

  • There is a sentence $p$ such that $p$ is satisfiable but has no computable model. (So building models of satisfiable sentences is genuinely hard.)

    • Namely, let $q$ be (the conjunction of the axioms of) any finitely axiomatizable subtheory of PA to which Tennenbaum's theorem still applies.

  • There is a computable sequence of sentences $p_1,p_2,p_3,...$ such that $(i)$ there is a computable sequence of computable structures $M_1,M_2,M_3,...$ such that $M_n\models p_m$ for all $m<n$ but $(ii)$ the whole set $\{p_1,p_2,p_3,...\}$ has no computable models. (So the "limit construction" of a model of an infinite theory from models of all its finite subtheories is genuinely hard.)

    • Namely, work in the language of arithmetic together with a new constant symbol $c$, let $p_1$ be the sentence $q$ from the previous bulletpoint, and let $p_{i+1}$ be the sentence "$c>1+1+...+1$ (with $i$ many $1$s)." For any $n$ the set $\{p_1,...,p_n\}$ is satisfied in the model consisting of the standard natural numbers with $c$ interpreted as $i+1$, but by Tennenbaum the whole set of sentences has no computable model.

  • Switching over to set theory, the full version of the compactness theorem (for possibly-uncountable languages) is not even provable in ZF - it requires (a weak form of) the axiom of choice! (So the full version really has to use "non-constructive" methods - in fact, ones as nasty as those used to prove the existence of a non-measurable set.)

    • This proof is genuinely hard; sketching it here isn't possible.

  • Finally, blending computability-theoretic complexity and axiomatic strength, a more technical result can be gotten via reverse mathematics: namely, the model existence and compactness theorems for even finite languages are each not provable in RCA$_0$, which is a theory generally understood to capture computable mathematics. In fact, each is equivalent over RCA$_0$ to the stronger theory WKL$_0$. (So even if we ignore the details of any construction and restrict the languages considerably, proving the theorems - no matter how non-constructive we want to be - already takes significant axiomatic power.)

    • While dramatically easier than the proof of the claim in the previous bulletpoint, the proof of this claim is still too hard to include here.
Community
  • 1
Noah Schweber
  • 245,398
  • Can somebody sketch the proof of the model existence theorem (K has a model <-> K without contradiction)? I mean the left-to-right direction should be trivial, so it's all about the right-to-left direction. I'd at least like to understand the rough idea of how it's proven. Or maybe a link where it's explained in a matter so that amateurs can also follow? –  Dec 03 '19 at 12:01
  • @Pippen The second section of this answer of mine (beginning "Having acknowledged") gives an outline of the Henkin-style proof (there are others). The most fundamental idea is that of term models: for any set of sentences $\Gamma$, we can look at the set of terms in the language of $\Gamma$ modulo $\Gamma$-provable equivalence. This can be turned into a structure $M_{term}(\Gamma)$ in the language of $\Gamma$ in a natural way. Unfortunately we don't in general have $M_{term}(\Gamma)\models\Gamma$ even if $\Gamma$ is consistent (continued) – Noah Schweber Dec 03 '19 at 16:33
  • so still more work is needed: we have to find an "improved version" of our set of sentences $\Gamma'\supseteq\Gamma$ such that $M_{term}(\Gamma')$ is guaranteed to satisfy $\Gamma'$ (and hence $\Gamma$ as well). This involves two tricks: passing to a complete theory (easy) and "adding witnesses" (harder). But the full proof really isn't that long, and it's well worth your time to actually dive into - it can be slow going but the new ideas are important and valuable. The above summary is about the simplest it can be made, though: it really is a nontrivial argument, there's no way around it. – Noah Schweber Dec 03 '19 at 16:34
  • First of all I need the very rough idea. Is it like the following? We can't simply prove that a consistent K has a model. So we construct a K where it happens and then we make sure we can infer that if it's possible in this K it's possible in all sound K. –  Dec 04 '19 at 03:27
  • 2
    @Pippen You're not going to get a better sketch of the proof than the one Noah just gave you. If it doesn't make sense to you, then you need to put in the work to understand the proof by actually reading a textbook. Your recent questions suggest that you're trying to think about logic while only having a very sketchy understanding of the definitions of the relevant terminology. This is not how mathematics works. Language is used in very precise ways, and many arguments are technically demanding. – Alex Kruckman Dec 06 '19 at 15:09
  • 1
    @Pippen Not everything can be boiled down to a couple sentences. I'm happy to be proved wrong, but I don't think there's a snappier explanation than the one I wrote in my comments which isn't wildly misleading. – Noah Schweber Dec 06 '19 at 15:12
  • I agree guys. I downloaded "A friendly introduction to math. logic" and may try that one. I am not a mathematician, so I get confused by all the symbols and abstraction without examples and intuitive notions. An example: They often write things like (IR, +, -, 0, 1) and I wonder: why 0 and 1 since they are already in IR? You can probably understand that with these kinds of questions it becomes hard, so I try to throw different things and see what sticks on the wall since even these discussions make me learn new things. –  Dec 07 '19 at 16:15