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(** #<nav><p class="series">Ltac.html</p>
<p class="seriesnext">LtacPatternMatching.html</p></nav># *)
(** * Ltac is an Imperative Metaprogramming Language *)
(** #<div id="history">site/posts/LtacMetaprogramming.v</div># *)
(** Coq is often depicted as an _interactive_ proof assistant, thanks to its
proof environment. Inside the proof environment, Coq presents the user a
goal, and said user solves said goal by means of tactics which describes a
logical reasoning. For instance, to reason by induction, one can use the
<<induction>> tactic, while a simple case analysis can rely on the
<<destruct>> or <<case_eq>> tactics, etc. It is not uncommon for new Coq
users to be introduced to Ltac, the Coq default tactic language, using this
proofcentric approach. This is not surprising, since writing proofs remains
the main usecase for Ltac. In practice though, this discourse remains an
abstraction which hides away what actually happens under the hood when Coq
executes a proof scripts.
To really understand what Ltac is about, we need to recall ourselves that
Coq kernel is mostly a typechecker. A theorem statement is expressed as a
“type” (which lives in a dedicated sort called [Prop]), and a proof of this
theorem is a term of this type, just like the term [S (S O)] (#<span
class="im">#2#</span>#) is of type [nat]. Proving a theorem in Coq requires
to construct a term of the type encoding said theorem, and Ltac allows for
incrementally constructing this term, one step at a time.
Ltac generates terms, therefore it is a metaprogramming language. It does it
incrementally, by using primitives to modifying an implicit state, therefore
it is an imperative language. Henceforth, it is an imperative
metaprogramming language.
To summarize, a goal presented by Coq inside the environment proof is a hole
within the term being constructed. It is presented to users as:
 A list of “hypotheses,” which are nothing more than the variables
in the scope of the hole
 A type, which is the type of the term to construct to fill the hole
We illustrate what happens under the hood of Coq executes a simple proof
script. One can use the <<Show Proof>> vernacular command to exhibit
this.
We illustrate how Ltac works with the following example. *)
Theorem add_n_O : forall (n : nat), n + O = n.
Proof.
(** The <<Proof>> vernacular command starts the proof environment. Since no
tactic has been used, the term we are trying to construct consists solely in
a hole, while Coq presents us a goal with no hypothesis, and with the exact
type of our theorem, that is [forall (n : nat), n + O = n].
A typical Coq course will explain students the <<intro>> tactic allows for
turning the premise of an implication into an hypothesis within the context.
#<div class="dmath">#C \vdash P \rightarrow Q#</div>#
becomes
#<div class="dmath">#C,\ P \vdash Q#</div>#
This is a fundamental rule of deductive reasoning, and <<intro>> encodes it.
It achieves this by refining the current hole into an anymous function.
When we use *)
intro n.
(** it refines the term
<<
?Goal1
>>
into
<<
fun (n : nat) => ?Goal2
>>
The next step of this proof is to reason about induction on [n]. For [nat],
it means that to be able to prove
#<div class="dmath">#\forall n, \mathrm{P}\ n#</div>#
we need to prove that
 #<div class="imath">#\mathrm{P}\ 0#</div>#
 #<div class="imath">#\forall n, \mathrm{P}\ n \rightarrow \mathrm{P}\ (S n)#</div>#
The <<induction>> tactic effectively turns our goal into two subgoals. But
why is that? Because, under the hood, Ltac is refining the current goal
using the [nat_ind] function automatically generated by Coq when [nat] was
defined. The type of [nat_ind] is
<<
forall (P : nat > Prop),
P 0
> (forall n : nat, P n > P (S n))
> forall n : nat, P n
>>
Interestingly enough, [nat_ind] is not an axiom. It is a regular Coq function, whose definition is
<<
fun (P : nat > Prop) (f : P 0)
(f0 : forall n : nat, P n > P (S n)) =>
fix F (n : nat) : P n :=
match n as n0 return (P n0) with
 0 => f
 S n0 => f0 n0 (F n0)
end
>>
So, after executing *)
induction n.
(** The hidden term we are constructing becomes
<<
(fun n : nat =>
nat_ind
(fun n0 : nat => n0 + 0 = n0)
?Goal3
(fun (n0 : nat) (IHn : n0 + 0 = n0) => ?Goal4)
n)
>>
And Coq presents us two goals.
First, we need to prove #<span class="dmath">#\mathrm{P}\ 0#</span>#, i.e.,
#<span class="dmath">#0 + 0 = 0#</span>#. Similarly to Coq presenting a goal
when what we are actually doing is constructing a term, the use of #<span
class="dmath">#+#</span># and #<span class="dmath">#+#</span># (i.e., Coq
notations mechanism) hide much here. We can ask Coq to be more explicit by
using the vernacular command <<Set Printing All>> to learn that when Coq
presents us a goal of the form [0 + 0 = 0], it is actually looking for a
term of type [@eq nat (Nat.add O O) O].
[Nat.add] is a regular Coq (recursive) function.
<<
fix add (n m : nat) {struct n} : nat :=
match n with
 0 => m
 S p => S (add p m)
end
>>
Similarly, [eq] is _not_ an axiom. It is a regular inductive type, defined
as follows.
<<
Inductive eq (A : Type) (x : A) : A > Prop :=
 eq_refl : eq A x x
>>
Coming back to our current goal, proving [@eq nat (Nat.add 0 0) 0] (i.e., [0
+ 0 = 0]) requires to construct a term of a type whose only constructor is
[eq_refl]. [eq_refl] accepts one argument, and encodes the proof that said
argument is equal to itself. In practice, Coq typechecker will accept the
term [@eq_refl _ x] when it expects a term of type [@eq _ x y] _if_ it can
reduce [x] and [y] to the same term.
Is it the case for [0 + 0 = 0]? It is, since by definition of [Nat.add], [0
+ x] is reduced to [x]. We can use the <<reflexivity>> tactic to tell Coq to
fill the current hole with [eq_refl]. *)
+ reflexivity.
(** Suspicious readers may rely on <<Show Proof>> to verify this assertion
assert:
<<
(fun n : nat =>
nat_ind
(fun n0 : nat => n0 + 0 = n0)
eq_refl
(fun (n0 : nat) (IHn : n0 + 0 = n0) => ?Goal4)
n)
>>
<<?Goal3>> has indeed be replaced by [eq_refl].
One goal remains, as we need to prove that if [n + 0 = n], then [S n + 0 = S
n]. Coq can reduce [S n + 0] to [S (n + 0)] by definition of [Nat.add], but
it cannot reduce [S n] more than it already is. We can request it to do so
using tactics such as [cbn]. *)
+ cbn.
(** We cannot just use <<reflexivity>> here (i.e., fill the hole with
[eq_refl]), since [S (n + 0)] and [S n] cannot be reduced to the same term.
However, at this point of the proof, we have the [IHn] hypothesis (i.e., the
[IHn] argument of the anonymous function whose body we are trying to
construct). The <<rewrite>> tactic allows for substituting in a type an
occurence of [x] by [y] as long as we have a proof of [x = y]. *)
rewrite IHn.
(** Similarly to <<induction>> using a dedicated function , <<rewrite>> refines
the current hole with the [eq_ind_r] function (not an axiom!). Replacing [n
+ 0] with [n] transforms the goal into [S n = S n]. Here, we can use
<<reflexivity>> (i.e., [eq_refl]) to conclude. *)
reflexivity.
(** After this last tactic, the work is done. There is no more goal to fill
inside the proof term that we have carefully constructed.
<<
(fun n : nat =>
nat_ind
(fun n0 : nat => n0 + 0 = n0)
eq_refl
(fun (n0 : nat) (IHn : n0 + 0 = n0) =>
eq_ind_r (fun n1 : nat => S n1 = S n0) eq_refl IHn)
n)
>>
We can finally use [Qed] or [Defined] to tell Coq to typecheck this
term. That is, Coq does not trust Ltac, but rather typecheck the term to
verify it is correct. This way, in case Ltac has a bug which makes it
construct illformed type, at the very least Coq can reject it. *)
Qed.
(** In conclusion, tactics are used to incrementally refine hole inside a term
until the latter is complete. They do it in a very specific manner, to
encode certain reasoning rule.
On the other hand, the <<refine>> tactic provides a generic, lowlevel way
to do the same thing. Knowing how a given tactic works allows for mimicking
its behavior using the <<refine>> tactic.
If we take the previous theorem as an example, we can prove it using this
alternative proof script. *)
Theorem add_n_O' : forall (n : nat), n + O = n.
Proof.
refine (fun n => _).
refine (nat_ind (fun n => n + 0 = n) _ _ n).
+ refine eq_refl.
+ refine (fun m IHm => _).
refine (eq_ind_r (fun n => S n = S m) _ IHm).
refine eq_refl.
Qed.
(** ** Conclusion *)
(** This concludes our introduction to Ltac as an imperative metaprogramming
language. In the #<a href="LtacPatternMatching.html">#next part of this series#</a>#, we
present Ltac powerful pattern matching capabilities. *)
