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$A=\{\langle M \rangle \mid M \text{ is a turing machine and }|L(M)|\geq3\}$

Since Recursive enumerable languages are turing enumerable, so listing of all strings of the language in finite time is possible. Then deciding whether cardinality is greater than 3 should be decidable. But the material that i have been studying says it is partially decidable. Which one is right?

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    $\begingroup$ You can't necessarily enumerate all the strings in finite time - in particular if there's an infinite number of them, this is impossible. In this case there is an infinite number of Turing Machines in $A$ themselves with infinite languages. $\endgroup$ – Luke Mathieson Feb 5 '15 at 6:46
  • $\begingroup$ What does "partially decidable" mean? $\endgroup$ – Andrej Bauer Feb 5 '15 at 7:59
  • $\begingroup$ @AndrejBauer It's yet another name for RE. $\endgroup$ – Luke Mathieson Feb 5 '15 at 8:07
  • $\begingroup$ Why do we need yet another name for c.e.? Oh well. $\endgroup$ – Andrej Bauer Feb 5 '15 at 8:12
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The language is clearly not decidable; use Rice's theorem. See e.g. our reference question.

We can certainly simulate $M$ on every input by dovetailing and count accepted inputs; if $|L(M)| \geq 3$ our counter hits three after finite time and we accept, otherwise we loop. Therefore, $A$ is semi-decidable.

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You can make a reduction from the HALTING problem({,w| is a turing machine that halts on w}), using pseudocode:

def R(<M>,w):
    def F(x):
        M(w) //run M on w, if it loops we are rejecting everything
        return x in ('1','2','3')
    return F

For every x in HALTING problem, there is some y in A, such that f(x) = y.

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