# Is the language below not decidable , if yes , it is then R.E ?

I am given a Turing machine M as input and i have to find out if this language below decidable and if not is it then in this case recursively enumerable .

$$L=\\\{<M> \mid \text{ is there w_1,w_2 for M so that M for both halts but with different results }\}$$

Somehow i ended up finding that the language is not decidable nor R.E .

if you have an other point of view , can you then please give me some hints.

• How is "...$M$ for both halts but with different results" interpreted? Is it "If $M(w_1)$ and $M(w_2)$ halt then $M(w_1) \neq M(w_2)$" or "$M(w_1)$ halts AND $M(w_2)$ halts AND $M(w_1) \neq M(w_2)$"? – fade2black Dec 15 '17 at 18:29
• like for example after halting ,$w_1$ ist accepted and $w_2$ is not – Mohbenay Dec 15 '17 at 19:50