# Prove that $L^{\nsubseteq}_{M'}=\{ \langle M \rangle \ | \ L(M) \nsubseteq L(M')\}$ with $M'$ a TM that always halt is undecidable

I've done some other problems by reduction, but I'm quite stuck here. I'm not really sure what to do with $M'$. I know that because $L(M) \nsubseteq L(M')$ there exists $w \in L(M)$ and $w \notin L(M')$ and I'm trying to create the machine that ignores the input and accepts only when M accepts $w$, but I'm kinda stuck on what to do next. Any suggestions?

Don't be confused by this question. Instead of studying $L^{\nsubseteq}_{M'}=\{ \langle M \rangle \ | \ L(M) \nsubseteq L(M')\}$ you might consider as well the language $L_G=\{ \langle M \rangle \ | \ L(M) \nsubseteq G\}$ for $G$ being any language. It is important to notice that there is always a Turing machines $M$ with $\langle M \rangle \in L_G$ but always a machine $M'$ with $\langle M' \rangle \not \in L_G$. So you are looking for a nontrivial set of Turing machines. This allows you to apply Rice's theorem and you are done.