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  "body": "\n## Proof\n\nThe requirements are still the same, with the same priority: $R_{0,A}\\prec R_{0,B} \\prec R_{1,A} \\prec R_{1,B} \\prec \\cdots$.\n\nIn the original construction for the Friedberg–Muchnik theorem, if it were true that every time we add a witness via $A_{2n} := A_{2n-1} \\cup \\{w_{i,A}\\}$, we also have $w_{i,A} \\geq \\min(C_{2n+2} \\setminus C_{2n})$, then by the permitting lemma, $A \\leq_T C$. Similarly for $B \\leq_T C$. However, this is not true in general.\n\nIn order to bypass the difficulty, we define 3 types of witnesses, 2 types of injuries, and 1 type of permissive moves, that a requirement can have.\n\n- Type 0 witness: $(w_{i,A, 0}, -, -)$. This tentatively witnesses $w_{i,A, 0} \\not\\in A$ and $TM_i^B(w_{i,A, 0}) \\neq 0$.\n- Type 1 witness: $(w_{i,A, 0}, w_{i,A, 1}, -)$. This tentatively witnesses $w_{i,A, 0} \\not\\in A$ and $TM_i^B(w_{i,A, 0}) \\neq 0$, and $w_{i,A, 1} \\not\\in A$ and $TM_i^B(w_{i,A, 1}) = 0$.\n- Type 2 witness: $(-,-,w_{i,A, 2})$. This tentatively witnesses $w_{i,A, 2} \\in A$ and $TM_i^B(w_{i,A, 2}) = 0$.\n- Type 0 injury: Because a prior requirement has made a permitted move, this requirement's witness has *possibly* been invalidated. The witness updates into a type 0 witness.\n- Type 1 injury: Because of a previous update to $B$, or because the computation has finally halted, a type 0 witness has been invalidated, in that $TM_i^{B_{2n}}(w_{i,A, 0}) =0$. The witness updates into a type 1 witness.\n- Permitted move: Because $w_{i,A,1} \\geq \\min(C_{2n+2} \\setminus C_{2n})$, the type 1 witness updates into a type 2 witness.\n\nAt stage *2n*,\n\n- We check if any type 1 witness is permitted to be moved into a type 2 witness.\n  - If so, then let $i$ be the smallest $i$ such that $w_{i,A,1} \\geq \\min(C_{2n+1} \\setminus C_{2n})$, and we perform the update: $$A_{2n+1} := A_{2n} \\cup\\{w_{i,A,1}\\},\\; w_{i,A} \\leftarrow (-,-, w_{i,A,1})$$ and simultaneously incur type 0 injury on all $w_{i,B}, w_{i+1,B}, \\dots$ by updating each to $$w_{j,B} \\leftarrow (\\min\\{ r_{jk} : k \\in \\mathbb{N}, r_{jk} \u003e \\max(w_{j,B}, \\phi_i^{B_{2n-1}}(w_{i,A}))\\}, -, -)$$\n- Otherwise, we check for all type 1 injuries.\n  - For each $i\\in 0:2n-1$, such that $w_{i,A}$ is a type 0 or type 1 witness,\n    - Simulate $TM_i^{B_{2n-1}}(w_{i,A,0})$ for up to $2n-1$ steps.\n    - If $TM_i^{B_{2n-1}}(w_{i,A,0})[2n-1] = 0$, then a type 1 injury has occurred. Update the witness to $$w_{i,A} \\leftarrow (\\min\\{ r_{ik} : k \\in \\mathbb{N}, r_{ik} \u003e w_{i,A,0}\\}, w_{i,A,0}, -)$$\n\nSimilarly for stage *(2n+1)*.\n\nTo show that the construction works, we need to show that each requirement's witness can only change state a finite number of times, by induction on the requirements.\n\nBy inductive hypothesis, since $R_{0,A}, R_{0,B}, \\dots, R_{n,A}, R_{n,B}$ would only change their witnesses a finite number of times, $R_{n+1,A}$ can only be type-0 injured a finite number of times. Consequently, the only way $R_{n+1,A}$ can change its witness is if it is type-1 injured an infinite number of times. Suppose this is the case, then its witness will eventually settle into an infinite process of being repeatedly type-1 injured, never permitted to move into a type 2 witness.\n\nSuppose this is the case, then let $f(k)$ be $w_{n+1,A,1}$ at stage *2k*. By construction, $f(k)$ is a function that is computable, non-decreasing, and because the witness is injured infinitely many times, $\\lim_k f(k) = \\infty$. By assumption, this witness will eventually be stuck forever as a type 1 witness, never permitted to move, so\n\n$$f(k) \u003c \\min(C_{2k+2} \\setminus C_{2k})$$\n\nfor all large enough *k*. However, if this is the case, then $C$ is decidable, contradicting the original assumption that $C$ is enumerable but undecidable.\n",
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