(** * TP 7 : Reflection  *)
(* To build an html version of this exercise sheet, use the command
   "coqdoc -utf8 --no-lib-name --no-index --lib-subtitles file.v" *)


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(** *  Proofs by reflection  **)
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(** **  Computation in Coq  **)

(** Computing in Coq means doing one of the following four reductions:
    - beta : apply a function to its arguments
    - iota : pattern matching related reduction
    - delta : unfold a definition 
      (something defined by Definition, Fixpoint etc.)
    - zeta : unfold a local definition
      (something defined by [let ... := ... in])

    The Coq tactics [cbv reduction_name] and [lazy reduction_name]
    provoke normalisation of the goal with respect to the reduction in
    parameter using a call-by-value and lazy strategy respectively.
    The tactic [simpl] performs βι-reductions plus some δ-reductions
    (only when further βι-reductions are possible). **)

(** 1) Observe the behaviour of the four reductions on the following
       examples. **)
Check (1 + 2).
Eval cbv beta in (1 + 2).
Eval cbv delta in (1 + 2).
Eval cbv beta delta in (1 + 2).
Eval cbv delta iota in (1 + 2).
Eval cbv beta delta iota in (1 + 2).

Check ((fun x y => x + y) 1 ).
Eval cbv beta in ((fun x y => x + y) 1).

Eval cbv iota in (match 1 with 0 => false | S _ => true end).
Eval cbv iota beta in (match 1 with 0 => false | S _ => true end).

Eval cbv beta in (let f := (fun x y => x + y) in f 1).
Eval cbv delta in (let f := (fun x y => x + y) in f 1).
Eval cbv zeta in (let f := (fun x y => x + y) in f 1).
Eval cbv beta delta iota zeta in  (let f := (fun x y => x + y) in f 1).
Eval compute in  (let f := (fun x y => x + y) in f 1).

(** Some remarks: 
    - the tactic [compute] is equivalent to [cbv beta delta iota zeta].
    - for the δ-reduction you can specify the definitions that are to be
      unfolded or the ones that are not to be unfolded.
    - [cbv], [lazy] and [compute] can be called on any goal in order to 
      normalise it. **)

(** Now, we will see the interest of using computation in proofs.

    2) Prove the following lemma without automatic tactics. **)
Lemma le10_20 : 10 <= 20.
Proof.
...
Qed.
Print le10_20.

(** We can see that the size of the proof is proportional to the numbers.
    Yet, informally, we would do this proof by computing if [le] was a
    function. Now comes in handy the [leb] function! **)
Require Import Arith.Compare_dec.
Print leb.

SearchAbout leb.
Check leb_correct.
Check leb_complete.

(** 3) Prove the same lemma by computation, using [leb]. **)
Lemma leb10_20 : 10 <= 20.
Proof.
...
Qed.

(** We can see now that this last proof is of constant size, no matter how
    large its inputs are. **)
Print leb10_20.
Print le10_20.


(** **  Associativity by reflection  **)

(** To show that two expressions are equal modulo associativity of natural
    number addition, we can do a proof by rewriting. **)
Require Import Arith.
Example reflection_test1 :
  forall x y z t u, x + (y + z + (t + u)) = x + y + (z + (t + u)).
Proof. intros. repeat rewrite plus_assoc. reflexivity. Qed.

Print reflection_test1.
(** The size of the proof term grows with the number of rewrites needed
    to "normalize" the terms w.r.t. associativity.
    
    Our goal is to replace rewriting by reductions, which are more
    efficient and do not appear in the proof terms.
    The idea is to find a datatype that will represent the terms you want
    to deal with (here, formulas with addition) such that you can replace
    formal manipulation of the terms (like rewriting) with computation on
    the new datatype.

    1) What is a good datatype to represent expressions involving
       additions? Define this datatype as an inductive type [repr]. **)

(** 2) Write the interpretation function [interp : repr -> nat] that 
       translates a term of the type [repr] to an expression of type
       [nat] containing additions. **)

(** 3) Write the function [normal_assoc : repr -> repr] that performs the
       equivalent of "normalisation w.r.t. associativity" on the new type
       [repr]. (this is the computation that will replace rewriting) **)

(** 4) Show the lemmas
       [[
       repr_valid : forall t : repr, interp t = interp (normal_assoc t)
       ]]
       and
       [[
       repr_valid_2 : forall t₁ t₂ : repr,
         interp (normal_assoc t₁) = interp (normal_assoc t₁) ->
         interp t₁ = interp t₂
       ]]
       . **)

(** 5) Do an example proof of associativity of natural number addition by
       moving to [repr] and reducing the corresponding terms. **)
Example reflection_test2 :
  forall x y z t u, x + (y + z + (t + u)) = x + y + (z + (t + u)).
Proof.
...
Qed.

(** It is annoying to have to manually insert [interp] in the goal and
    guess which term will give the correct interpretation. We would like
    to have a function doing it for us but that is impossible: you cannot
    pattern match an expression of type [nat] to test wether it contains
    [+] because [+] is not a constructor.
    The solution is to use the tactic language and perform pattern
    matching there.

    6) Write a tactic that does the work automatically (i.e. introduce
       [interp], then apply [repr_valid_2]) and test it. **)
Ltac nat_to_repr n := ...

Ltac rassoc := ...

Example reflection_test3 :
  forall x y z t u, x + (y + z + (t + u)) = x + y + (z + (t + u)).
Proof. intros. rassoc. Qed.

(** 7) Generalize your work for an arbitrary type [A] with an arbitrary
       operation [op] that is associative. **)
Parameters (A : Set) (op : A -> A -> A)
  (assoc : forall x y z, op x (op y z) = op (op x y) z).

(** 8) Adapt the previous work to the case where [op] has a neutral
       element to the right and to the left. **)

Parameters (zero: A)
  (zero_l : forall (x : A),  op zero x = x)
  (zero_r : forall (x : A),  op x zero = x).

(** 9) How would you deal with equalities modulo associativity and
       commutativity?
       Write the corresponding datastructures, lemmas and tactics. **)


(** 10) Write a predicate [nf : repr -> Prop] to state whether an
        representative is in normal form. Prove your normalizing
        functions only returns normal forms. Then prove that it
        leaves normal forms unchanged. *)

Inductive nf : repr -> Prop := .

Lemma normal_is_nf : forall t, nf (normal t).
Admitted.

Lemma normal_nf_unchanged : forall t, nf t -> normal t = t.
Admitted.

End A_assoc_neutral.


(** Remark: this technology is used in the implementation of the automatic
            tactics [ring] and [field]. **)