While it is not the case that the extension of every decidable theory is decidable, is it true that:

the extension of every undecidable theory undecidable?

In other words, given an undecidable theory A, is it enough to show

$$A \subseteq B$$

to prove that B is undecidable?

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    $\begingroup$ Cross-posted to computer science and CS theory. Please do not do this. It is against site policy because it fragments answers and wastes people's time when they work on something that has already been answered elsewhere. $\endgroup$ Dec 3, 2014 at 9:40
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    $\begingroup$ Properties like undecidability tend to never be closed against sub- or superset. $\emptyset$ and $\Sigma^*$ are trivial counter-examples to any such claim. $\endgroup$
    – Raphael
    Dec 3, 2014 at 11:44

1 Answer 1


No. A theory is a set of theorems. The set of all formulas is a decidable theory and it is an extension of all theories, including undecidable ones. It is also very inconsistent and thus useless in practice, but it is a counter-example to your claim.


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