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For example from the Gödel sentence "G iff ¬P([g])", where g is G's Gödel-number, is it possible to make semantic inference (not syntactic, only at the level of truth between the undecidable sentences "outside" the deduction capability of the system) that due to "G iff ¬P([g])" semantically implying "G and ¬P([g])" we have "G and ¬P([g])".

  • Sorry i made a fallacy. actually (G and no-B([g])) implies semantically (G iff no-B([g])). So, Can we make the following deduction: there is no proof to (G iff no-B([g])), so then, there isn't proof for (G and no-B([g])), then (G and no-B([g]) is also undecidable, und then also for semantic implication G , no-B(g) are undecidable)? – Gabriel Reis May 18 '17 at 1:01
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    Sure, why not, but I wouldn't call it "in a system". Semantic inference simply means that one statement follows from another in a chosen interpretation of the system. We can not conclude much about undecidability from semantic implication though, since it is interpretation specific. Something that implies an undecidable sentence in some interpretation does not have to be undecidable. It may be provable, while its semantic consequence isn't true in other interpretations, and remains unprovable. – Conifold May 18 '17 at 1:10
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    Great, thank you very much! And thank you also for editing some language mistakes, and use and mention mistakes, and no misinterpreting what i meant though. As you could note English is not my mother-tongue. So I should also be more precise and talk about the standard interpretation. I also confused madly my logic teacher using semantic implication like in the first post, instead of inference. Thank you for that, it clarifies a lot. Greetings! – Gabriel Reis May 18 '17 at 1:12
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    Not clear what are you asking... See G's Incompleteness Th : the so-called G-sentence is G and we have F ⊢ G ↔ ¬Bew([g]) ("syntactically"). The theorem proves that (under suitable assumptions about the system F): F ⊬ G and F ⊬ ¬G. Thus, G is the undecidable sentence. – Mauro ALLEGRANZA May 18 '17 at 7:45
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    Obviously, F ⊬ G ∧ ¬Bew([g]), otherwise - by simple tautological inference - we would have F ⊢ G, contradicting the previous result. – Mauro ALLEGRANZA May 18 '17 at 7:47

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