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'''1.''' If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why? | '''1.''' If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why? | ||
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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 is given (either from <math>L</math> or not from <math>L</math>), give it to both those $TM$s. 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.
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 is given (either from <math>L</math> or not from <math>L</math>), give it to both those $TM$s. 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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