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I am unable to understand this proof

L is recursively enumerable if and only if L is Turing recognizable

If anyone can prove this, that would be great help

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1 Answer 1

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If $L$ is recursively enumerable then, given $x \in \Sigma^*$, you can iterate over all $y \in L$ and check whether $x=y$. If that's the case accept, otherwise continue with the next $y$. After examining the last $y \in L$ (if any) reject. If $x \in L$, this procedure eventually accepts. If $x \not\in L$ this procedure either never halts or it rejects.

If $L$ is Turing recognizable by some TM $T$ then you can execute $T(x)$ for all $x \in \Sigma^*$ in a dovetail fashion. If $x \in L$, $T(x)$ will eventually halt and accept, and you can output $x$. If $x \not\in L$, $T(x)$ will either never halt or it will reject (and you ignore $x$).

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