Lean4双曲几何不可判定性证明的语法与对角化问题求助

我正尝试用Lean4，通过将停机问题归约到双曲几何的方法证明双曲几何的不可判定性，目前遇到以下问题：

• 图灵机对角化步骤存在逻辑障碍，证明中仍有未解决的目标
• 双曲几何部分缺失必要的公理定义
• 代码出现unexpected token语法错误，移除多处逗号后仍无法解决

完整代码 #

import Mathlib.Tactic
import Mathlib.Data.Real.Basic
import Mathlib.Data.List.Basic
import Mathlib.Data.Finset.Basic

-- Suppress linter warnings for unused variables
set_option linter.unusedVariables false

open Classical -- Use classical logic for decidability

-- Define a Turing Machine
structure TuringMachine (states symbols : Type) where
  transition : states → symbols → states × symbols × Bool -- Transition function
  startState : states -- Start state
  haltState : states -- Halt state

-- Define finite states and symbols
inductive TMState
| q0 | q1 | halt
deriving DecidableEq, Fintype

inductive TMSymbol
| zero | one | blank
deriving DecidableEq, Fintype

-- Example Turing Machine
def tm_example : TuringMachine TMState TMSymbol := {
  transition := fun state symbol =>
    match state, symbol with
    | TMState.q0, TMSymbol.zero => (TMState.q1, TMSymbol.one, false)
    | TMState.q1, TMSymbol.one => (TMState.halt, TMSymbol.one, true)
    | _, _ => (TMState.halt, TMSymbol.blank, true),
  startState := TMState.q0,
  haltState := TMState.halt
}

-- Execution of a Turing Machine for a number of steps
noncomputable def TuringMachine.execute {states symbols : Type} [Fintype states] [Fintype symbols]
    [DecidableEq states] [DecidableEq symbols]
    (tm : TuringMachine states symbols) (steps : ℕ) (input : List symbols) :
    Option (states × List symbols × Bool) :=
  let rec step (currentState : states) (tape : List symbols) (remainingSteps : ℕ) :
      Option (states × List symbols × Bool) :=
    match remainingSteps with
    | 0 => none
    | n + 1 =>
      match tape with
      | [] => none
      | symbol :: rest =>
        let (newState, newSymbol, halt) := tm.transition currentState symbol
        let newTape := newSymbol :: rest
        if halt then
          some (newState, newTape, true)
        else
          step newState newTape n
  step tm.startState input steps

-- Input for the Turing Machine
def input_example : List TMSymbol := [TMSymbol.zero, TMSymbol.one]

-- Explicit Definitions for States and Symbols
def orderedStates : List TMState := [TMState.q0, TMState.q1, TMState.halt]
def orderedSymbols : List TMSymbol := [TMSymbol.zero, TMSymbol.one, TMSymbol.blank]

-- Poincaré Disk Model
structure PoincareDisk where
  x : ℝ
  y : ℝ
  hyp_condition : x^2 + y^2 < 1 -- Point lies strictly inside the unit disk

namespace PoincareDisk

-- Geodesics
structure Geodesic where
  center : ℝ × ℝ -- Center of the defining circle
  radius : ℝ -- Radius of the circle
  is_diameter : Bool -- True for diameters, False for orthogonal arcs

-- Membership of a point on a geodesic
def onGeodesic (p : PoincareDisk) (g : Geodesic) : Prop :=
  if g.is_diameter then
    p.x = g.center.1 -- Diameter case
  else
    (p.x - g.center.1)^2 + (p.y - g.center.2)^2 = g.radius^2 -- Orthogonal arc case

end PoincareDisk

-- Encode TuringMachine states and symbols as hyperbolic geometry configurations
noncomputable def TuringMachine.encodeAsGeometry (tm : TuringMachine TMState TMSymbol) (input : List TMSymbol) :
    List PoincareDisk × List PoincareDisk.Geodesic :=
  -- Map states and symbols to unique ℝ values
  let stateToReal (state : TMState) : ℝ :=
    orderedStates.indexOf state + 1
  let symbolToReal (symbol : TMSymbol) : ℝ :=
    orderedSymbols.indexOf symbol + 1

  -- Encode state as a PoincareDisk point
  let encodeState (state : TMState) : PoincareDisk :=
    { x := 0.9 * Real.sin (stateToReal state * 0.1),
      y := 0.9 * Real.cos (stateToReal state * 0.1),
      hyp_condition := by
        have h : Real.sin (stateToReal state * 0.1)^2 + Real.cos (stateToReal state * 0.1)^2 = 1 :=
          Real.sin_sq_add_cos_sq (stateToReal state * 0.1)
        linarith [h] }

  -- Encode symbol as a geodesic
  let encodeSymbol (symbol : TMSymbol) : PoincareDisk.Geodesic :=
    { center := (0, 0),
      radius := 1 / (symbolToReal symbol + 2),
      is_diameter := false }

  let states := orderedStates.map encodeState
  let geodesics := input.map encodeSymbol
  (states, geodesics)

-- Undecidability Proof for Turing Machine Execution
theorem TuringMachineExecutionUndecidable :
  ¬(∀ (tm : TuringMachine TMState TMSymbol) (input : List TMSymbol),
    ∃ steps, tm.execute steps input = some (tm.haltState, [], true)) := by
{
  -- Assume decidability of halting for contradiction
  intro h_decidable,

  -- Define a diagonal Turing Machine that alternates indefinitely
  let diagonal_tm : TuringMachine TMState TMSymbol := {
    transition := fun state symbol =>
      match state, symbol with
      | TMState.q0, TMSymbol.zero => (TMState.q1, TMSymbol.one, false) -- Move to q1
      | TMState.q1, TMSymbol.one => (TMState.q0, TMSymbol.zero, false) -- Back to q0
      | _, _ => (TMState.halt, TMSymbol.blank, true), -- Default to halt
    startState := TMState.q0,
    haltState := TMState.halt
  }

  -- Input for the diagonal machine
  let input := [TMSymbol.zero, TMSymbol.one]

  -- Assume the diagonal machine halts (contradiction setup)
  have h_halts : ∃ steps, diagonal_tm.execute steps input = some (diagonal_tm.haltState, [], true),
  { exact h_decidable diagonal_tm input }, -- From assumption of decidability

  -- Extract steps and result from halting assumption
  cases h_halts with steps h_steps

  -- Prove that the diagonal machine cannot halt
  have h_contradiction : diagonal_tm.execute steps input ≠ some (diagonal_tm.haltState, [], true),
  {
    induction steps with
    | zero =>
      -- Base case: 0 steps mean  no execution, so the machine cannot halt
      simp [TuringMachine.execute] at h_steps,
      contradiction
    | succ n ih =>
      -- Recursive case: analyze the machine's step-by-step execution
      simp [TuringMachine.execute] at h_steps,
      cases input with
      | [] =>
        -- If the input is empty, the machine cannot proceed, so it cannot halt
        simp [TuringMachine.execute] at h_steps,
        contradiction
      | symbol :: rest =>
        -- Simulate the first step of the machine
        cases diagonal_tm.transition diagonal_tm.startState symbol with
        | (new_state, new_symbol, halt) =>
          by_cases halt
          -- Case 1: Halt = true leads to contradiction, as diagonal_tm loops indefinitely
          { simp [*] at h_steps, contradiction }
          -- Case 2: Halt = false, continue recursion using IH
          { apply ih, assumption }
  },

  -- Conclude contradiction
  exact h_contradiction h_steps,
}

-- Test the Geometry Encoding
noncomputable def encoded_geometry := TuringMachine.encodeAsGeometry tm_example input_example

报错信息 #

mathlib-stable.lean:120:0
unsolved goals
h_decidable : ∀ (tm : TuringMachine TMState TMSymbol) (input : List TMSymbol),
  ∃ steps, tm.execute steps input = some (tm.haltState, [], true)
⊢ False

mathlib-stable.lean:122:19
unexpected token ','; expected '}'

内容的提问来源于stack exchange，提问作者João Teixeira