bookshelf/exercises/theorem-proving-in-lean/exercises-8.lean

/- Exercises 8
 -
 - Avigad, Jeremy. ‘Theorem Proving in Lean’, n.d.
-/

-- Exercise 1
--
-- Open a namespace `Hidden` to avoid naming conflicts, and use the equation
-- compiler to define addition, multiplication, and exponentiation on the
-- natural numbers. Then use the equation compiler to derive some of their basic
-- properties.
namespace ex1

def add : Nat → Nat → Nat
  | m, Nat.zero   => m
  | m, Nat.succ n => Nat.succ (add m n)

def mul : Nat → Nat → Nat
  | _, Nat.zero => 0
  | m, Nat.succ n => add m (mul m n)

def exp : Nat → Nat → Nat
  | _, Nat.zero => 1
  | m, Nat.succ n => mul m (exp m n)

end ex1

-- Exercise 2
--
-- Similarly, use the equation compiler to define some basic operations on lists
-- (like the reverse function) and prove theorems about lists by induction (such
-- as the fact that `reverse (reverse xs) = xs` for any list `xs`).
namespace ex2

variable {α : Type _}

def reverse : List α → List α
  | [] => []
  | (head :: tail) => reverse tail ++ [head]

-- Proof of `reverse (reverse xs) = xs` shown in previous exercise.

end ex2

-- Exercise 3
--
-- Define your own function to carry out course-of-value recursion on the
-- natural numbers. Similarly, see if you can figure out how to define
-- `WellFounded.fix` on your own.
namespace ex3

def below {motive : Nat → Type} : Nat → Type
  | Nat.zero => PUnit
  | Nat.succ n => PProd (PProd (motive n) (@below motive n)) (PUnit : Type)

-- TODO: Sort out how to write `brecOn` and `WellFounded.fix`.

end ex3

-- Exercise 4
--
-- Following the examples in Section Dependent Pattern Matching, define a
-- function that will append two vectors. This is tricky; you will have to
-- define an auxiliary function.
namespace ex4

inductive Vector (α : Type u) : Nat → Type u
  | nil  : Vector α 0
  | cons : α → {n : Nat} → Vector α n → Vector α (n + 1)

namespace Vector

-- TODO: Sort out how to write `append`.

end Vector

end ex4

-- Exercise 5
--
-- Consider the following type of arithmetic expressions. The idea is that
-- `var n` is a variable, `vₙ`, and `const n` is the constant whose value is
-- `n`.
namespace ex5

inductive Expr where
  | const : Nat → Expr
  | var : Nat → Expr
  | plus : Expr → Expr → Expr
  | times : Expr → Expr → Expr
  deriving Repr

open Expr

def sampleExpr : Expr :=
  plus (times (var 0) (const 7)) (times (const 2) (var 1))

-- Here `sampleExpr` represents `(v₀ * 7) + (2 * v₁)`. Write a function that
-- evaluates such an expression, evaluating each `var n` to `v n`.

def eval (v : Nat → Nat) : Expr → Nat
  | const n     => sorry
  | var n       => v n
  | plus e₁ e₂  => sorry
  | times e₁ e₂ => sorry

def sampleVal : Nat → Nat
  | 0 => 5
  | 1 => 6
  | _ => 0

-- Try it out. You should get 47 here.
-- #eval eval sampleVal sampleExpr

-- Implement "constant fusion," a procedure that simplifies subterms like
-- `5 + 7` to `12`. Using the auxiliary function `simpConst`, define a function
-- "fuse": to simplify a plus or a times, first simplify the arguments
-- recursively, and then apply `simpConst` to try to simplify the result.

def simpConst : Expr → Expr
  | plus (const n₁) (const n₂)  => const (n₁ + n₂)
  | times (const n₁) (const n₂) => const (n₁ * n₂)
  | e                           => e

def fuse : Expr → Expr := sorry

theorem simpConst_eq (v : Nat → Nat)
        : ∀ e : Expr, eval v (simpConst e) = eval v e :=
  sorry

theorem fuse_eq (v : Nat → Nat)
        : ∀ e : Expr, eval v (fuse e) = eval v e :=
  sorry

-- The last two theorems show that the definitions preserve the value.

end ex5