m-completeness

$\text{[math]}$ is complete with respect to $\text{[math]}$ ( $\text{[math]}$ is 1-complete) if

$\text{[math]}$ is recursively enumerable, and
$\text{[math]}$ .

$\text{[math]}$ is complete with respect to $\text{[math]}$ ( $\text{[math]}$ is m-complete) if

$\text{[math]}$ is recursively enumerable, and
$\text{[math]}$ .

[cite:;taken from @computability_rogers_1987 chapter 7.2 complete sets]

$\text{[math]}$ is 1-complete.
[cite:;taken from @computability_rogers_1987 chapter 7.2 complete sets]

if $\text{[math]}$ is $\text{[math]}$ -complete, then $\text{[math]}$ isnt recursive.

the intuition behind such a proof would be that all $\text{[math]}$ -complete sets are reducible to each other, and if one of them was recursive all others would be recursive aswell by turing_reducibility.html.