Partially decidable is same as semi-decidable language which means that there exists a Turing Machine for such languages(given an input, the machine can accept/reject/loop) but we are not sure whether there is any Total Turing Machine(given an input, the machine can accept or reject but does not loop) or not.

Ref: https://www.youtube.com/watch?v=v2VuM0AIZVU&t=13s **<2:47 to 4:30>**

Here the question is whether M loops on a given string 'w' which is same as saying whether M does not halt on w. As Arjun Sir said, this is just the complement of Halting problem.

Halting problem is an undecidable problem. Given an input x to a machine M, there is no way to decide whether M will halt(accept or reject) or not(loop). The language L={M#x | M halts on x} is **R.E but not Recursive. **

So** complement of Halting problem** would be **complement of language which is (R.E. but not Recursive)** which is **Non-R.E.**

Why is it so? Ref: https://www.youtube.com/watch?v=VqDMudEzZCo **<0:00-7:00>**

Non R.E. means unrecognizable meaning there does not exist any Turing machine also for such languages hence they are not even partially decidable.