Timeline for Closure property of recursively enumerable language
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2017 at 6:28 | vote | accept | Zephyr | ||
| Oct 24, 2017 at 6:27 | comment | added | fade2black | @Zephyr That is not true. A subset of a R.E. set does not have to be R.E. For example $N$ is a r.e. set (even recursive), but any non r.e. set is a subset of $N$, for example the complement of the Halting problem set. | |
| Oct 24, 2017 at 6:23 | comment | added | Zephyr | I have one more doubt. R.E is not closed under Set difference property. We know that A - B will always be a subset of A. If we go by that, then as A is R.E, A-B is also R.E as it is subset of A. | |
| Oct 23, 2017 at 20:07 | comment | added | fade2black | @Zephyr If r.e. sets are not closed under a specific operation $\star$, that does not mean that r.e. sets are (or are not) closed under another operation $*$ even if $*$ can be represented in terms of $\star$. | |
| Oct 23, 2017 at 19:57 | history | edited | fade2black | CC BY-SA 3.0 |
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| Oct 23, 2017 at 19:50 | comment | added | Zephyr | wikipedia has used a similar argument (union of complements)for proving closure of CFL 's under intersection en.m.wikipedia.org/wiki/Context-free_language | |
| Oct 23, 2017 at 19:39 | comment | added | fade2black | @Zephyr See updates. Also, please decide on what question you want to ask, and do not change your OP too often. It may make people's answers obsolete and inconsistent. | |
| Oct 23, 2017 at 19:37 | history | edited | fade2black | CC BY-SA 3.0 |
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| Oct 23, 2017 at 19:24 | comment | added | Zephyr | See my edit to the question. | |
| Oct 23, 2017 at 19:17 | history | edited | fade2black | CC BY-SA 3.0 |
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| Oct 23, 2017 at 19:07 | history | edited | fade2black | CC BY-SA 3.0 |
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| Oct 23, 2017 at 19:00 | history | answered | fade2black | CC BY-SA 3.0 |