2

I know that given a sentence or formula of a formal system, this formula is a logical truth if it is true under all interpretations.

Is it possible to define this same concept of logical truth without the reference to models and interpretations?

Thanks!

user62358
  • 23

2 Answers2

3

For first-order logic this is essentially the completeness theorem.

The completeness theorem tells us that if $T$ is a first-order theory, then $\varphi$ is provable from $T$ if and only if $\varphi$ is true in every model of $T$.

If a formula is logically true it means that it is true in every interpretation. Every interpretation is a model for the empty theory, and so by the completeness theorem we can say that something is logically true if and only if it is provable from $\varnothing$.

Asaf Karagila
  • 405,794
  • What does the quantifier "every" run over? – alancalvitti Feb 15 '13 at 04:28
  • @alancalvitti: Where? – Asaf Karagila Feb 15 '13 at 04:33
  • "true in every interpretation. Every interpretation is a model..." - is there a set of all interpretations or similar quantifiable object? – alancalvitti Feb 15 '13 at 15:35
  • @alancalvitti: No. But there is no set of ordinals when I quantify over "Every ordinal is a transitive set". The class is definable and therefore we can quantify over its elements. – Asaf Karagila Feb 15 '13 at 15:40
  • When I wrote similar, I meant perhaps class or category if not set. Are interpretations well-ordered and can we do induction over them like ordinals? Do they form a category? – alancalvitti Feb 15 '13 at 15:51
  • @alan: If we assume choice then every interpretation is well-ordered; if we assume global choice then the class of interpretation is well-ordered. In either case we don't really care about doing an induction over these structures. We have other uses for them. As for a category? Sure, everything forms a category. Simply have no morphisms. In a more meaningful way, I suppose one can consider interpretation for the language $\cal L$ as objects (or isomorphism equiv. classes, using Scott's trick) and $\cal L$-homomorphisms as morphisms. I don't really know why this would be useful, though. – Asaf Karagila Feb 15 '13 at 16:03
  • It's not useful to my work, but I'm curious how such abstractions are quantified. Maybe the key question for me is, how is it decidable that a statement is true in every interpretation? – alancalvitti Feb 15 '13 at 16:08
  • You can ask whether or not a particular object is an interpretation for a language, and if so you can ask whether or not a particular sentence (or formula) are true in that interpretation. There's not much to it. This is not constructive mathematics, I don't need to show that the algorithm for verifying stops in $O(n)$ or whatever. It is a far more obvious shtick: we can write down a formula for what is an interpretation, and another for when something is true in an interpretation. – Asaf Karagila Feb 15 '13 at 16:25
  • But regardless of the computational complexity, if the "class" of interpretations is open-ended, how can the truth or falsehood be determined for all interpretations. I have in mind Euclid's parallel lines postulate and the non-Euclidean models developed in the 19th century. Is this an apropos example? – alancalvitti Feb 15 '13 at 18:15
  • @alancalvitti: I'm not sure what "open-ended" means. It's perfectly definable. Much like you can ask whether or not every set can be well-ordered, you can ask whether or not every set which happens to encode (via a specific coding, of course) an interpretation for a particular language satisfies a certain formula. You can write a very long and complicated formula in the language of $\in$ which has parameters for a language and a formula and it asks whether or not there is an interpretation where the formula is false. – Asaf Karagila Feb 15 '13 at 19:27
  • Ok, I think I understand this part: "an interpretation for a particular language satisfies a certain formula". But to then quantify over every interpretation including ones that will be defined in the future (hence the non-Euclidean geometry example)? – alancalvitti Feb 15 '13 at 21:33
  • @alan: From a logical and set theoretical point of view there is no "future". You are given a language, and you have all its interpretations. This is even before talking about models, theories, consistency and so on. This is just ways to interpret the language into a structure. – Asaf Karagila Feb 16 '13 at 00:23
  • Well, the intuitionists might disagree. For instance Goldblatt explicitly mentions the time aspect in Topoi. Aren't there languages for which there are so many (unbounded) interpretations that it is undecidable whether all interpretations of a given statement are true? – alancalvitti Feb 16 '13 at 00:35
  • @alan: Everything I wrote on this page is in classical logic. If we went this far just so you could tell me that you assume that I am talking about non-classical logics then we are wasting time. – Asaf Karagila Feb 16 '13 at 00:36
  • Forget the first 2 sentences in my last comment - I only mentioned intuitionist logic w.r.t. the time factor ("no future"). Even in classical systems, what about my last sentence above? – alancalvitti Feb 16 '13 at 02:02
  • @alan: I don't know the answer to this question. I imagine it may be possible to come up with languages and formulas that it would be undecidable whether or not something is logically true. But surely you must agree that the statement $\forall x(x=x)$ is always true in every interpretation, regardless to whether or not we can decide if something is a structure. Much like we cannot prove everything, there shouldn't be too much shock if this is the case here. Alas, I don't know much more on the topic, and as I said at the beginning of this comment, this is mainly speculations. – Asaf Karagila Feb 16 '13 at 02:13
  • Nobody even seems to know (or apparently even care about) what properties "=" enjoys: http://math.stackexchange.com/questions/285484/distinguishing-equality-and-isomorphism-as-relations – alancalvitti Feb 16 '13 at 02:41
  • @alan: I can't testify for category theorists, but in logic $=$ is a logical relation and if $x=y$ then for every predicate $R(x)\leftrightarrow R(y)$ and for every function symbol $F(x)=F(y)$, and so on. – Asaf Karagila Feb 16 '13 at 04:06
  • Categorically, Barry Mazur has proposed that $x=y$ when they are the same up to unique isomorphism. But even with your definition above, how is it decidable that $x=y$? Blass et al have a paper "When are two algorithms the same". It is difficult to determine. You wrote "for every predicate", "for every function symbol" - how is it decidable, these are generally unbounded. (Similarly, object, arrow, sum, product, monic, epic, iso, equivalence, etc vary greatly. MacLane called it the "protean" nature of math) – alancalvitti Feb 16 '13 at 15:44
  • @alan: Not everything in classical logic is decidable. It doesn't mean that a definition shouldn't be given. – Asaf Karagila Feb 16 '13 at 15:48
  • Can I guess that it is never decidable when the universal quantifier ranges over structures that are unlike ordinals in the sense that we can't do induction on them. You may be able to enumerate, eg finite categories, finite programs, but their properties will be all over the place. In this context, it is their properties that are used, not just the enumeration. – alancalvitti Feb 16 '13 at 15:52
2

The common distinction is between syntactical truth and semantic truth. Given a deduction system (i.e., some rules telling us what which strings are allowed and how to deduce new the syntactical truth of a sentence given the syntactical truth of others) you get a well-defined notion of syntactic truth as those statements that are derivable in the deduction system from a given theory.

In contact, semantic truth relates to a given model and is means those statements that when interpreted in $M$ are true.

Goedel's completeness theorem states that (for first order logic) in a deduction system, a statement is syntactically true given a theory $T$ if, and only if, it is semantically true in all models of $T$. So, this answers your question.

The fact that syntactic truth implies semantic truth is quite easy to prove. The other direction is involved and requires a slightly weaker axiom than the axiom of choice.

Ittay Weiss
  • 81,796