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I have seen 2 answers in stackoverflow:

  1. A "trivial" property is one that holds either for all languages or for none.

  2. The property is trivial if it contains every TM, or if it is empty.

My problem is: Saying that a property is trivial if it contains every TM is not the same as saying that a property is trivial if it contain all the languages (including non RE languages).

Same as for: Saying that a property is trivial if it is does not contain any TM languages is not the same as saying that a property is trivial if it is empty.

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  • $\begingroup$ Have you read the comments under the second answer you link? You are right, it's not the same thing, and they don't claim that (but wrote the post in a potentially confusing way.) $\endgroup$
    – Raphael
    Aug 31, 2017 at 18:16
  • $\begingroup$ Then what is the currect definition? $\endgroup$
    – Stav Alfi
    Aug 31, 2017 at 18:25
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    $\begingroup$ I find the comments rather clear about that: the first. But that said, there are probably many equivalent definitions. Definitions are never "correct", there are only more or less useful. $\endgroup$
    – Raphael
    Aug 31, 2017 at 18:25

1 Answer 1

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We call a set of languages, $P\subseteq 2^{\Sigma^*}$, a property. If you think of this subset as the set of languages who satisfy some property, then we can simply say that a language $L$ satisfies the property iff $L\in P$.

Rice theorem tells you that you can't, given a Turing machine $M$, check if $L(M)$ satisfies some non trivial property $P$, i.e. $P\neq \emptyset$, $P\nsupseteq RE$. Note that Shaull makes a distinction between semantic and syntactic properties in the comments. To the best of my knowledge, the standard definition of a property in the context of Rice's theorem is a set of languages.

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  • $\begingroup$ courses.engr.illinois.edu/cs373/sp2013/Lectures/lec25.pdf - Definition 3 - You are saying something complitly different. $\endgroup$
    – Stav Alfi
    Aug 31, 2017 at 18:24
  • $\begingroup$ FWIW, it's also common to express properties as predicates, i.e. boolean-valued functions (here of type $\Sigma^* \to \{0,1\}$). This is clearly equivalent, since subsets and indicator functions identify uniquely. $\endgroup$
    – Raphael
    Aug 31, 2017 at 18:27
  • $\begingroup$ @StavAlfi How so? I don't see it. (I think you need some time learning how to read mathematics. I can recommend the Book of Proof, which is available for free.) $\endgroup$
    – Raphael
    Aug 31, 2017 at 18:28
  • $\begingroup$ Saying that a property is trivial if it contains every TM is not the same as saying that a property is trivial if it contain all the languages (including non RE languages)...... $\endgroup$
    – Stav Alfi
    Aug 31, 2017 at 18:29
  • $\begingroup$ Each TM has a corresponding language. So if it contains every TM, then it contains every language. $\endgroup$
    – rus9384
    Aug 31, 2017 at 18:33

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