Prove that disjointlang := {⟨T , T ′⟩ : L(T ) ∩ L(T ′) = ∅}; is not decidable

Question

Prove that disjointlang := {⟨T , T ′⟩ : L(T ) ∩ L(T ′) = ∅} is not decidable. I was thinking about reducing the problem to some other problem, but I'm not sure which and how.

Answer 1

Trying a reduction is a good instinct! First off just a hint for the reduction:

Suppose $\mathcal{T}_\text{all}$ is a TM accepting every word. Exactly when is $\langle \mathcal{T}, \mathcal{T}_\text{all} \rangle$ in your language?

Does this remind you of an undecidable problem you already know?

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A Solution:

Suppose $\mathcal{T}_\text{all}$ is a machine accepting every word and $f$ is given by $$f(\langle \mathcal{T} \rangle) = \langle \mathcal{T}, \mathcal{T}_\text{all} \rangle$$ for all machines $\mathcal{T}$.Then $$\mathcal{T} \in L_\varnothing \iff L(\mathcal{T}) = L(\mathcal{T}) \cap L(\mathcal{T}_\text{all}) = \varnothing \iff f(\mathcal{T}) = \langle \mathcal{T}, \mathcal{T}_\text{all} \rangle \in \texttt{disjointlang}$$ where $L_\varnothing = \{\langle \mathcal{T} \rangle : \mathcal{T} \text{ is a TM }, L(\mathcal{T}) = \varnothing\}$. Thus $L_\varnothing$ (which is known to be undecidable) reduces to $\texttt{disjointlang}$, so it must be undecidable as well.