Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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| − | If L and L' are both recursively enumerable, then L is recursive. Why? | + | '''1.''' If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why? |
===Solution=== | ===Solution=== | ||
| − | Given L is RE. So there is a TM, which accepts and halts for all words in L. | + | Given <math>L</math> is RE. So there is a TM, which accepts and halts for all words in <math>L</math>. |
| − | Now, if L' is RE, then there is a TM, which accepts and halts for all words not in L. | + | Now, if <math>L'</math> is RE, then there is a TM, which accepts and halts for all words not in <math>L</math>. |
| − | So, if a word, from L or not from L,is given, give it to both those TMs. If its from L, the first TM will halt and we say it belongs to L. If its not from L, the second one will halt and we say it doesn't belong to L. Thus L becomes recursive. | + | So, if a word, from <math>L</math> or not from <math>L</math>,is given, give it to both those TMs. If its from L, the first TM will halt and we say it belongs to L. If its not from L, the second one will halt and we say it doesn't belong to <math>L</math>. Thus, <math>L</math> becomes recursive. |
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1. If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why?
Given <math>L</math> is RE. So there is a TM, which accepts and halts for all words in <math>L</math>. Now, if <math>L'</math> is RE, then there is a TM, which accepts and halts for all words not in <math>L</math>. So, if a word, from <math>L</math> or not from <math>L</math>,is given, give it to both those TMs. If its from L, the first TM will halt and we say it belongs to L. If its not from L, the second one will halt and we say it doesn't belong to <math>L</math>. Thus, <math>L</math> becomes recursive.
{{subst:wikEd}}
1. If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why?
Given <math>L</math> is RE. So there is a TM, which accepts and halts for all words in <math>L</math>. Now, if <math>L'</math> is RE, then there is a TM, which accepts and halts for all words not in <math>L</math>. So, if a word, from <math>L</math> or not from <math>L</math>,is given, give it to both those TMs. If its from L, the first TM will halt and we say it belongs to L. If its not from L, the second one will halt and we say it doesn't belong to <math>L</math>. Thus, <math>L</math> becomes recursive.
{{subst:wikEd}}
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