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$L_c=\{\langle M_1 \rangle, \langle M_2 \rangle):L(M_1)\cap L(M_2)=\emptyset\}$ 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!

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    $\begingroup$ Please don't delete questions after they have been answered. That's very impolite. $\endgroup$ – Raphael Nov 8 '17 at 18:57
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Assume $L_c$ is decidable. We will reduce $A_{TM} =\{x\mid M_x \text{ halts on } x\}$ (the Halting problem language) to $L_c$. On input $x $ create a TM $M'$ which halts on only $x$. Then $L(M') = \{x\}$ and $M_x$ halts on $x$ iff $L(M_x) \cap L(M') = \{x\} \neq \emptyset$ iff $\left(\langle M_x \rangle, \langle M' \rangle \right) \notin L_c$. Thus, to check if $M_x$ halts on $x$ we check if $\left(\langle M_x \rangle, \langle M' \rangle \right) \notin L_c$.

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