If a set of formula is consistent, there exist a model in which every formula is true. This is only if the set is satisfiable. But satisfiability is the fact that it can be true so what is the difference between the 2 notions ?
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Consistency is a syntactic property. It means that there is no proof of contradiction from your axioms.
Satisfiability is a semantic property. It means that there is a model of the axioms.
In first-order logic (as well as propositional logic) the two notions are equivalent because the logic is sound and complete. Meaning a satisfiable theory is consistent, and a consistent theory is satisfiable.
Other logics, however, are not so lucky to have both of these properties and the two notions separate. In fact, if we do not assume the axiom of choice, then it is consistent that there is a theory which is consistent but not satisfiable.
Asaf Karagila
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Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think? – user21820 May 17 '19 at 16:44
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Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $\sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability. – Asaf Karagila May 17 '19 at 16:45
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I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable... – user21820 May 17 '19 at 16:53
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Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular. – Asaf Karagila May 17 '19 at 16:58
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Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said. – user21820 May 17 '19 at 17:02
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I am probably misremembering this. I guess that anything that I might be remembering would appear in Shelah's work on Morley's theorem without choice. – Asaf Karagila May 17 '19 at 17:14
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Consistency is defined syntactically :
a set $\Gamma$ of formulas is inconsistent iff we can derive (in the proof system) a contradiction from it : $\Gamma \vdash \bot$.
In this way, we prove that :
a set $\Gamma$ is consistent iff it is satisfiable.
See also the post : Relationship between consistency, strong completeness and soundness.
Mauro ALLEGRANZA
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"satisfiable": A 1.-order-logic-statement is satisfiable if it can not be simplified to False without knowing anything about the predicates, that it contains. Also a 1.-order-logic-statement is satisfiable if the 2.-order-logic-formula, that is obtained by replacing all predicates in it with fictive variables of existential quantifiers, that are put around the 1.-order-logic-statement, is not false. Note, that since this 2.-order-logic-formula does not contain any predicates it can be only True or False.
"consistent": A 1.-order-logic-statement is consistent if it can not be simplified to False even with using "outside information" about the predicates, that it contains.
Here https://www.physicsforums.com/threads/about-the-undecidability-of-first-order-logic.1015918/post-6641082 is a more exhaustive explanation about it with examples.
user2952903
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