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I wish to undestand how to proof a function is/is not computable. I found this example online (without solution) beacuse I was thinking was easy to understand, but I am stuck in understanding how to set the proof.

Given $M_n$ the nth Turing Machine produced in output by an enumeration algorithm for Turing Machines. Let $f_n$ the computable partial function compute by $M_n$.

Now, we consider the $g:\mathbb{N}^3 \longrightarrow \mathbb{N}$. Proof g is not computable.

$g(x,y,z) = \left\{\begin{matrix} 1 & ,if \ f_z(x)=y\\ 0 & ,otherwise \end{matrix}\right.$

My solution idea is based on the Church-Turing Thesis for Computable Functions: A function $f$ is effectively computable is, and only if, there is a Turing Machine that computes $f$.

Gianni Spear
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    If that function were computable, the Halting problem would be decidable. Can you see how? (Also, what do you mean in the first case "$\text{if }f_z(x) = y$" if $M_z$ does not even halt on input $x$?) – ComFreek Nov 05 '19 at 17:33
  • Do you have a specific question? This is a question-and-answer site, so we need you to articulate a specific question about your situation or your approach. If you haven't already, I suggest reading some background on computability/decidability and reductions. – D.W. Nov 05 '19 at 17:58

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