Machine-Checked Mathematics

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Exercise 1:

Improve the last symplification function of lesson 5 by adding the simplification rule $$x - 0 = x$$

Module Ex1.

Inductive expr :=
  | Zero
  | Minus (x : expr) (y : expr)
  | Var (n : nat).

Fixpoint interp (e : expr) (c : list nat) :=
  match e with
  | Zero => 0
  | Minus x y => (interp x c) - (interp y c)
  | Var x => nth 0 c x
  end.

Lemma simplify_correct (e : expr) (c : list nat) : interp e c = interp (simplify e) c.
Proof.
elim: e => //= x Hx y Hy.
case: (simplify x) Hx => [|x1 x2|n] -> ; case: (simplify y) Hy => [|y1 y2|m] -> //.
(* fill in *)
by case: eqP => [->|//]; rewrite subnn.
Qed.

Lemma test (x : nat) : x - (x - x) = x.
Proof.
pose AST : expr := Minus (Var 0) (Minus (Var 0) (Var 0)).
pose CTX : list nat := [:: x].
rewrite -[LHS]/(interp AST CTX).
rewrite simplify_correct.
rewrite /=.
by [].
Qed.

End Ex1.

Exercise 2:

Improve on exercise 1 by adding to the expr inductive type a constructor for multiplication and the simplification rules $$x * 0 = 0$$ and $$0 * x = 0$$

Module Ex2.

Inductive expr :=
  | Zero
  | (* fill in *)
  | Minus (x : expr) (y : expr)
  | Var (n : nat).

Fixpoint interp (e : expr) (c : list nat) :=
  match e with
  | Zero => 0
  | Minus x y => (interp x c) - (interp y c)
  | (* fill in *)
  | Var x => nth 0 c x
  end.

Lemma simplify_correct (e : expr) (c : list nat) : interp e c = interp (simplify e) c.
Proof.
elim: e => //= x Hx y Hy.
- case: (simplify x) Hx => [|x1 x2|x1 x2|n] -> ; case: (simplify y) Hy => [|y1 y2|y1 y2|m] -> //.
  (* fill in *)
- case: (simplify x) Hx => [|x1 x2|x1 x2|n] -> ; case: (simplify y) Hy => [|y1 y2|y1 y2|m] -> //.
  (* fill in, like in exercise 1 *)
  by case: eqP => [->|//]; rewrite subnn.
Qed.

Lemma test (x : nat) (y : nat) : x - x - (y * 0) = 0.
Proof.
pose AST : expr := (* fill me *)
pose CTX : list nat := (* fill me *)
rewrite -[LHS]/(interp AST CTX).
rewrite simplify_correct.
rewrite /=.
by [].
Qed.

End Ex2.

Exercise 3:

Improve on exercise 2 by adding to the expr data type a constructor for One and the simplification rules $$x * 1 = x$$ and $$1 * x = x$$

Module Ex3.

Lemma simplify_correct (e : expr) (c : list nat) : interp e c = interp (simplify e) c.
Proof.
elim: e => //= x Hx y Hy.
- case: (simplify x) Hx => [||x1 x2|x1 x2|n] -> ; case: (simplify y) Hy => [||y1 y2|y1 y2|m] -> //.
  (* fill in *)
- case: (simplify x) Hx => [||x1 x2|x1 x2|n] -> ; case: (simplify y) Hy => [||y1 y2|y1 y2|m] -> //.
  (* fill in *)
Qed.

Lemma test (x : nat) (y : nat) : x - (x * 1) - (y * 0) = 0.
Proof.
pose AST : expr := (* fill me *)
pose CTX : list nat := (* fill me *)
rewrite -[LHS]/(interp AST CTX).
rewrite simplify_correct.
rewrite /=.
by [].
Qed.

End Ex3.

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