209 lines
5.4 KiB
Plaintext
209 lines
5.4 KiB
Plaintext
/-! # Exercises.Avigad.Chapter8
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Induction and Recursion
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-/
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namespace Exercises.Avigad.Chapter8
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/-! #### Exercise 1
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Open a namespace `Hidden` to avoid naming conflicts, and use the equation
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compiler to define addition, multiplication, and exponentiation on the natural
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numbers. Then use the equation compiler to derive some of their basic
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properties.
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-/
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namespace ex1
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def add : Nat → Nat → Nat
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| m, Nat.zero => m
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| m, Nat.succ n => Nat.succ (add m n)
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def mul : Nat → Nat → Nat
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| _, Nat.zero => 0
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| m, Nat.succ n => add m (mul m n)
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def exp : Nat → Nat → Nat
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| _, Nat.zero => 1
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| m, Nat.succ n => mul m (exp m n)
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end ex1
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/-! #### Exercise 2
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Similarly, use the equation compiler to define some basic operations on lists
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(like the reverse function) and prove theorems about lists by induction (such as
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the fact that `reverse (reverse xs) = xs` for any list `xs`).
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-/
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namespace ex2
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variable {α : Type _}
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def reverse : List α → List α
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| [] => []
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| (head :: tail) => reverse tail ++ [head]
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-- Proof of `reverse (reverse xs) = xs` shown in previous exercise.
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end ex2
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/-! #### Exercise 3
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Define your own function to carry out course-of-value recursion on the natural
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numbers. Similarly, see if you can figure out how to define `WellFounded.fix` on
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your own.
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-/
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namespace ex3
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def below {motive : Nat → Sort u} (t : Nat) : Sort (max 1 u) :=
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Nat.recOn t
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(zero := PUnit)
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(succ := fun n ih => PProd (PProd (motive n) ih) (PUnit : Sort (max 1 u)))
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/--
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A copied implementation of `Nat.brecOn` with the `motive` properly supplied.
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Notice the `noncomputable` tag; the code generator does not support the `recOn`
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recursor yet, for reasons I haven't fully understood yet. This thread touches on
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the topic:
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https://leanprover-community.github.io/archive/stream/270676-lean4/topic/code.20generator.20does.20not.20support.20recursor.html
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-/
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noncomputable def brecOn {motive : Nat → Sort u}
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(t : Nat) (F : (t : Nat) → @below motive t → motive t) : motive t :=
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(Nat.recOn t
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(motive := fun n => PProd
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(motive n)
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(Nat.recOn n PUnit fun n ih => PProd (PProd (motive n) ih) PUnit))
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(zero := { fst := F Nat.zero PUnit.unit, snd := PUnit.unit })
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(succ := fun n ih => {
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fst := F (Nat.succ n) { fst := ih, snd := PUnit.unit },
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snd := { fst := ih, snd := PUnit.unit }
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})).1
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#check WellFounded.fix
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end ex3
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/-! #### Exercise 4
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Following the examples in Section Dependent Pattern Matching, define a function
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that will append two vectors. This is tricky; you will have to define an
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auxiliary function.
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-/
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namespace ex4
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inductive Vector (α : Type u) : Nat → Type u
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| nil : Vector α 0
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| cons : α → {n : Nat} → Vector α n → Vector α (n + 1)
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namespace Vector
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/--
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As long we flip the indices in our type signature in the resulting summation,
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there is no need for an auxiliary function.
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-/
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def append : Vector α m → Vector α n → Vector α (n + m)
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| nil, bs => bs
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| cons a as, bs => cons a (append as bs)
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end Vector
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end ex4
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/-! #### Exercise 5
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Consider the following type of arithmetic expressions.
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-/
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namespace ex5
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inductive Expr where
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| const : Nat → Expr
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| var : Nat → Expr
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| plus : Expr → Expr → Expr
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| times : Expr → Expr → Expr
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deriving Repr
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open Expr
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def sampleExpr : Expr :=
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plus (times (var 0) (const 7)) (times (const 2) (var 1))
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-- Here `sampleExpr` represents `(v₀ * 7) + (2 * v₁)`. Write a function that
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-- evaluates such an expression, evaluating each `var n` to `v n`.
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def eval (v : Nat → Nat) : Expr → Nat
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| const n => n
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| var n => v n
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| plus e₁ e₂ => eval v e₁ + eval v e₂
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| times e₁ e₂ => eval v e₁ * eval v e₂
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def sampleVal : Nat → Nat
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| 0 => 5
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| 1 => 6
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| _ => 0
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-- Try it out. You should get 47 here.
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#eval eval sampleVal sampleExpr
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/-! ##### Constant Fusion
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Implement "constant fusion," a procedure that simplifies subterms like `5 + 7
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to `12`. Using the auxiliary function `simpConst`, define a function "fuse": to
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simplify a plus or a times, first simplify the arguments recursively, and then
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apply `simpConst` to try to simplify the result.
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-/
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def simpConst : Expr → Expr
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| plus (const n₁) (const n₂) => const (n₁ + n₂)
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| times (const n₁) (const n₂) => const (n₁ * n₂)
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| e => e
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def fuse : Expr → Expr
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| plus e₁ e₂ => simpConst $ plus (fuse e₁) (fuse e₂)
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| times e₁ e₂ => simpConst $ times (fuse e₁) (fuse e₂)
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| e => e
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theorem simpConst_eq (v : Nat → Nat)
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: ∀ e : Expr, eval v (simpConst e) = eval v e := by
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intro e
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unfold simpConst
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repeat { unfold eval }
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match h : e with
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| const n
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| var n
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| plus (const _) (const _)
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| plus (var _) _
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| plus (plus _ _) _
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| plus (times _ _) _
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| times (const _) (const _)
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| times (var _) _
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| times (plus _ _) _
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| times (times _ _) _ => rfl
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| plus _ (var _)
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| plus _ (plus _ _)
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| plus _ (times _ _)
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| times _ (var _)
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| times _ (plus _ _)
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| times _ (times _ _) => simp only
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theorem fuse_eq (v : Nat → Nat)
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: ∀ e : Expr, eval v (fuse e) = eval v e := by
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intro e
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induction e with
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| const n | var n => unfold fuse; rfl
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| plus e₁ e₂ he₁ he₂ | times e₁ e₂ he₁ he₂ =>
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unfold fuse
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rw [simpConst_eq]
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unfold eval
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rw [he₁, he₂]
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-- The last two theorems show that the definitions preserve the value.
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end ex5
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end Exercises.Avigad.Chapter8 |