f95d8108f4
This PR records the `fib_impl n = fib_spec n` example, and a proof using current technologies, as a test. I'd like to think about eliminating `MProd` from the terms produced by `do` notation; it seems (at least) a simproc would be required.
46 lines
1.7 KiB
Text
46 lines
1.7 KiB
Text
/-!
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This is an example for monadic reasoning.
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The eventual goal is to provide a nice user experience for proving `fib_impl n = fib_spec n`
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and related goals.
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Currently, this file just contains a proof that uses `simp` lemmas to convert the `do` notation
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and for loop into a `List.foldl`, and then gives a "functional" proof.
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(This is *not* the nice user experience we are aiming for!)
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Even in this setup, there is an awkward problem that `do` blocks handle multiple mutable variables
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via the universe monomorphic `MProd` type, to avoid universe unification issues arising when using
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`Prod`. We have to jump through some additional hoops to handle that.
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We could provide simp lemmas, simprocs, and possibly custom tactics to eliminator `MProd` from the
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terms produced by `do` notation.
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-/
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def fib_spec : Nat → Nat
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| 0 => 0
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| 1 => 1
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| n+2 => fib_spec n + fib_spec (n+1)
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def fib_impl (n : Nat) := Id.run do
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if n = 0 then return 0
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let mut a := 0
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let mut b := 0
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b := b + 1
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for _ in [1:n] do
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let a' := a
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a := b
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b := a' + b
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return b
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theorem fib_correct {n} : fib_impl n = fib_spec n := by
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-- The default simp set eliminates the binds generated by `do` notation,
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-- and converts the `for` loop into a `List.foldl` over `List.range'`.
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simp [fib_impl, Id.run]
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match n with
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| 0 => simp [fib_spec]
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| n+1 =>
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-- Note here that we have to use `⟨x, y⟩ : MProd _ _`, because these are not `Prod` products.
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suffices ((List.range' 1 n).foldl (fun b a ↦ ⟨b.snd, b.fst + b.snd⟩) (⟨0, 1⟩ : MProd _ _)) =
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⟨fib_spec n, fib_spec (n + 1)⟩ by simp_all
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induction n with
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| zero => rfl
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| succ n ih => simp [fib_spec, List.range'_1_concat, ih]
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