91Ó°ÊÓ

Let's define languages A and B as follows: \(A = \{w \mid w \text{ is the encoding of a Turing machine } M_w \text{ where } M_w \text{ halts on input } w \}\) \(B = \{w \mid w \text{ is the encoding of a Turing machine } M_w \text{ where } M_w \text{ does not halt on input } w \}\) Notice that A and B are disjoint because a Turing machine either halts or does not halt on a given input. Both A and B are Turing-recognizable, as we can construct Turing machines that recognize each: - For A, the recognizer can simulate the input Turing machine on the input and accept if it ever halts. - For B, the recognizer can similarly simulate the input Turing machine on the input and accept if it can determine that it will never halt.

Assume, for the sake of contradiction, that there exists a decidable language C that separates A and B, so: 1. \(A \subseteq C\) 2. \(B \subseteq \bar{C}\) By definition, a decidable language has a Turing machine that decides it, meaning it halts with an answer (accept or reject) for every input.

Since we assume the existence of a Turing machine that decides C, we can use C to show the existence of a Turing machine to decide the Halting problem as follows: Given a Turing machine M and input x: 1. Construct a new Turing machine \(M'\) that, on any input w, simulates M on x and accepts if M halts on x. 2. Encode \(M'\) as a string, \(w'\). 3. Run the decider for C on input \(w'\). 4. If C accepts \(w'\), then M halts on x (since \(w' \in A\)). If C rejects \(w'\), then M does not halt on x (since \(w' \in B\)). Since we have constructed, using the decider for C, a Turing machine that decides the Halting problem, which is known to be undecidable, this contradicts our assumption.

Since our assumption led to a contradiction, it must be false. Therefore, no decidable language C exists that can separate the disjoint Turing-recognizable languages A and B, as stated in the exercise.