tp7

TP 7 : Reflection

Proofs by reflection

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.

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