$L_c=\{\langle M_1 \rangle, \langle M_2 \rangle):L(M_1)\cap L(M_2)=\emptyset\}$ prove to be undecidable. My approach: We will prove this by reduction $L_{\emptyset} \leq_T L_c$ We begin by assuming that $L_c$ is decidable and that we have access to a black box $Q$ which decides it. We want to construct another Turing machine $H$ which decides $L_{\emptyset}$.
H = “On input (< M >, w).
Construct the following machine M’
M’ = “On input w.
Ignore input w and reject.
Run decider Q on (< M >,< M’>)
If Q accepts, reject.
If Q reject, accept.
I'm mainly concerned about my choice of $L_{\emptyset}$ (maybe $L_{halt}$ is better suited here?) and the part where I run the decider $Q$ on ($\langle M \rangle, \langle M'\rangle)$. Any help would be much appreciated, Thank you!