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I'm interested in Fuzzy logic. I have read that the compactness theorem holds for predicate Lukasiewicz logic, with semantics over $[0,1]$.

However I found the following question on mathoverflow about Godel logic

Compactness and completeness in Gödel logic

While I realize that the comments under the "EDIT" doesn't apply, it seems to me that you could take a structure, expand the language by adding uncountably many nullary predicates, and $\{P_{a}|a\in{\omega_{1}+2}\}$ and show that the theory $T$ axiomatized by $\{(P_{b}\rightarrow{P_a})\rightarrow{P_b}|a<b\}$ is finitely satisfiable. But this should lead to a contradiction as in the above question; paraphrasing the answer given there slightly;

However, any model of $T $ where $P_{\omega_{1}+1}$ has truth value $x$, with $x<1$ in its ordering, must have the valuation of $P_{a}$ is less than the valuation of $P_{b}$ for $a<b$ (as the linear order has the property that the truth value of $p\implies{q}=1$ iff $p\leq{q}$ where the ordering here is the ordering on the truth value of algebras), and hence the valuation $v$ provides an embedding of $\omega_{1}+1$ into the algebra of truth values, which should be a contradiction.

My questions are:

Where am I going wrong? If I were to hazard a guess, I would say that since we cannot add $\neg(P_{\omega + 1}\leftrightarrow{1})$ to $T$, any model of $T$ over $[0,1]$ would necessarily have all of the $P_{a}$'s having truth value $1$. Am I correct on this?

Am I correct in stating "as the linear order has the property that the truth value of $p\implies{q}=1$ iff $p\leq{q}$ where the ordering here is the ordering on the truth value of algebras"? This is true in $[0,1]$, and I think it should be true in general, but I'm not a 100%.

Edit: Here by compactness I mean the following: If $T$ is finitely satisfiable over $[0,1]$ then $T$ has a model over $[0,1]$.

Edit2: I would also like to know if the following is true in general: If $T$ is finitely satisfiable over $K$, where $K$ is some $MV$-algebra, then $T$ has a model over $K$.

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  • $\begingroup$ @Emil Jeřábek: Your answer is what inspired the question. I thought I'd tag you here to see if you had further thoughts on the matter. $\endgroup$
    – user75685
    Oct 7, 2015 at 11:43
  • $\begingroup$ Does a tag like this work? Or can you only tag someone already involved in this question... $\endgroup$ Oct 7, 2015 at 13:00
  • $\begingroup$ (@GeraldEdgar is perfectly correct. The @ notification didn’t work, but I’ve noticed the question regardless.) “Compactness” can mean several different things that are not necessarily equivalent for nonclassical logics, so you first need to state exactly what you mean by compactness theorem, and what is meant by it in the reference you read. For instance, see Theorem 2 in logica.dmi.unisa.it/lucaspada/wp-content/uploads/… : as you can see, three formulations of compactness hold for Łukasiewicz logic, but the fourth one does not; beware that the paper only ... $\endgroup$ Oct 7, 2015 at 14:12
  • $\begingroup$ ... considers finite languages with relations and constants, in particular all theories in the statement are countable. The argument outlined in the question can indeed be used to refute (iv). $\endgroup$ Oct 7, 2015 at 14:15
  • $\begingroup$ @Gerald Edgar: Ah, I didn't know that, thank you for letting me know. $\endgroup$
    – user75685
    Oct 7, 2015 at 15:08

1 Answer 1

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[It is not an answer to your questions, but it is too long for a comment]

This compactness is a well-known fact of Lukasiewicz.

Such compacteness is a rather straightforward consequence of (the non-trivial result) Theorem 5.4.24 published in Hájek's book "Metamathematics of Fuzzy Logic" https://books.google.es/books?id=JUb9sywOIeYC&pg=PA135 (the result goes back to Chang & Belluce in the 60s)

Let me point out that the previous theorem by Chang & Belluce reduces your compactness question to checking that

  • every "finitely satisfiable theory over some MV-chain" is also "satisfiable over some MV-chain";

and this is a particular case of the usual ultraproduct construction in first-order classical logic (notice that Lukasiewicz under the general semantics can be seen as 2-sorted classical first-order theory, one sort deals with the domain of the structure and the other sort deals with the MV chain)

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  • $\begingroup$ Thanks for the answer. But it isn't quite what I'm looking for. $\endgroup$
    – user75685
    Oct 7, 2015 at 21:00
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    $\begingroup$ It is not clear to me what you are asking, but nevertheless let me add thatmy guess is that the statement inside your "Edit 2" is false and that a counterexample can be obtained using the MV-chain known as Chang-algebra [i.e., the one corresponding, using Mundici's functor, to the strong unit $(1,0)$ inside the $\ell$-group $(\mathbb{Z}^2,+)$ with the lexicographic order] $\endgroup$
    – boumol
    Oct 7, 2015 at 21:26
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    $\begingroup$ Because of a homomorphism from the Chang algebra onto the Boolean algebra on {0,1}, the two have the same set of satisfiable formulas. I think you guess wrongly. $\endgroup$
    – anemone
    Oct 7, 2015 at 21:41
  • $\begingroup$ @anemone: You guess right, my intuition was wrong (let me add that in your "proof" you also use the fact that your homomorphism preserves all existing suprema and infina, not just the finite ones; but this is not important in the case of the Chang's algebra) $\endgroup$
    – boumol
    Oct 8, 2015 at 6:55

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