(** printing \2/ $\cup$ #&cup;# *)
(** printing <2= $\subseteq$ #&sube;# *)
(** printing forall $\forall$ #&forall;# *)
(** printing -> $\rightarrow$ #&rarr;# *)
(** printing /\ $\land$ #&and;# *)
(** * Illustrating the [paco] library  This tutorial shows how to use our [paco] library to reason about coinductive predicates in Coq.  By coinductive predicates we mean objects that one can define using [CoInductive] and whose type ends in [Prop].  The library implements _parameterized coinduction_, as described in our draft paper entitled #<a href=" http://plv.mpi-sws.org/paco/"># _The Power of Parameterization in Coinductive Proof_ #</a>#.  How does parameterized coinduction compare to Coq's existing support for coinductive proofs?  For proofs by _induction_, Coq provides a tactic called [fix], but proofs done directly with [fix] are required to satisfy a _syntactic guardedness_ criterion.  Fortunately, Coq also generates induction _lemmas_ for every inductive definition, which eliminate the need to ever use the low-level [fix] tactic and worry about guardedness.  For proofs by coinduction, Coq provides a tactic [cofix] analogous to [fix], i.e., also based on a syntactic guardedness criterion.  However, Coq does not generate any lemmas for coinductive definitions.  Consequently, to reason about such definitions, the user must use the low-level [cofix] tactic.  Due to its syntactic nature, [cofix] is problematic in several ways (please see the paper linked above for detailed discussion of these points):  - It is non-compositional. - It is inefficient. - It is not user-friendly. - It interacts poorly with builtin automation tactics.   The [paco] library provides a tactic, [pcofix], which can be seen as a replacement for Coq's builtin [cofix] tactic (in the case of predicates) that does not involve any syntactic guardedness criterion and thus avoids all of these problems.  In addition, like Coq's [cofix], [pcofix] provides built-in support for _incremental_ proofs, i.e., proofs in which the coinduction hypothesis is extended gradually as the proof progresses.  We believe this combination of compositionality, incrementality, robustness, and ease of use makes [pcofix] a superior alternative to [cofix] for proofs about coinductive predicates.  The rest of this tutorial illustrates the use of the [paco] library with the help of three examples. *)(* *)
(** ** Example: stream equality  The first example involves streams of natural numbers and an equality thereon.  We prove a particular equality via [cofix], and then explain how that proof can be turned into one via [pcofix]. *)
(** We start by defining a type of streams of natural numbers. *)
CoInductive stream :=
| cons : nat -> stream -> stream.
(** The following, seemingly useless, lemma is a common trick for working with corecursive terms such as our streams.  (It will be used in the example proofs.)  For details, see for instance #<a href="http://adam.chlipala.net/cpdt/html/Coinductive.html">Adam Chlipala's book on Certified Programming with Dependent Types</a>#. *)
Definition sunf s :=
match s with cons n s' => cons n s' end.
(** In order to state our example equality, we define an enumeration stream and a map operation for streams. *)
CoFixpoint enumerate n : stream :=
cons n (enumerate (S n)).
CoFixpoint map f s : stream :=
match s with cons n s' => cons (f n) (map f s') end.
(** *** A proof using [cofix]  We will now prove, using [cofix], that for any [n] the stream [enumerate n] is equal to the stream [cons n (map S (enumerate n))]. *)
(** First, we define an equality [seq] on streams.  Usually, one would do this as follows.  [CoInductive seq : stream -> stream -> Prop :=] #<br># [  | seq_fold : forall n s1 s2 (R : seq s1 s2), seq (cons n s1) (cons n s2).]  Instead, we define the generating function, [seq_gen], beforehand, and then define [seq] as its greatest fixed point using [CoInductive].  The reason is simply that we need [seq_gen] later when using [paco].  If applying parameterized coinduction to an existing development where the generating function is not explicit, it can always easily be made explicit.  Note that, albeit the use of [Inductive], the definition of [seq_gen] is not recursive. *)
Inductive seq_gen seq : stream -> stream -> Prop :=
| _seq_gen : forall n s1 s2 (R : seq s1 s2 : Prop), seq_gen seq (cons n s1) (cons n s2).
Hint Constructors seq_gen.
CoInductive seq : stream -> stream -> Prop :=
| seq_fold : forall s1 s2, seq_gen seq s1 s2 -> seq s1 s2.
(** Second, we do the actual proof. *)
(** _Note_: One might want to eliminate the use of [pattern] in the proof script by replacing the corresponding line with the following:  [rewrite sunf_eq at 1; simpl.]  However, doing so will result in an invalid proof (rejected at [Qed]-time).  The reason is that the proof term created by [rewrite] ... [at 1] involves some lemmas from the [Setoid] library.  Like most lemmas, these are opaque, so guardedness checking cannot inspect their proofs and thus fails.  This is a great example of two of the previously mentioned problems with the [cofix] approach, namely lack of compositionality and poor interaction with standard tactics. *)(* *)
(** *** A proof using our [pcofix]  We will now do the same proof, but using the [paco] library.  We first simply show the new (but equivalent) definition of stream equality and the new proof of the example, and then explain what is going on behind the scenes.  In both, we use [paco] constructs with suffix "2" because we are dealing with predicates of arity 2 here.  _Note_: [Paco] supports predicates of arity up to 8.  Also, it supports up to three mutually coinductive predicates (see the last example of this tutorial).  In either case, extending this is just a matter of copy and paste.  We also have to prove monotonicity of the generating function [seq_gen], which can be discharged by the tactic [pmonauto], and register it in the Hint databse [paco].  _Remark_: Unlike [CoInductive], [paco] does not care whether the generating function is given in a _strictly positive_ syntactic form; all that matters is that the function is monotone. More specifically, [paco2 f] is well defined for an arbitrary generating function [f] regardless of whether it is monotone or not. However, in order to ensure that [paco2 f bot2] is the greatest fixed point of [f], we need monotonicity of [f]. *)
Definition seq' s1 s2 := paco2 seq_gen bot2 s1 s2.
Hint Unfold seq'.
Hint Resolve seq_gen_mon : paco.
(** Observe that the proof script is essentially the same as before (this is no accident).  The main change is the use of [pcofix] instead of [cofix], which allows us to avoid any syntactic guardedness checking at [Qed]-time.  As a minor benefit of that, we are able to use [rewrite] ... [at 1], which we could not do before.  *)
(** **** How it works:  To understand [seq'] and the proof of [example'], we have to explain some background.  Given a monotone function [f], we call [paco2 f] the _parameterized_ greatest fixed point of [f].  For a relation [r], [paco2 f r] is defined as the greatest fixed point of the function [fun x => f (x \2/ r)].  Clearly, [paco2 f bot2] (where [bot2] is the empty relation) is just the ordinary greatest fixed point of [f].  Let us look at our example domain to understand better.  A proof of [paco2 seq_gen r s1 s2] can be seen as a proof of [r <2= seq -> seq s1 s2], where the use of the premise is guarded _semantically_, rather than syntactically.  More precisely, [paco2 seq_gen r] relates two streams iff their heads are equal and their tails are either related by [r] or again related by [paco2 seq_gen r].  Compare this to [seq], the ordinary greatest fixed point of [seq_gen], which relates two streams iff their heads are equal and their tails are again related by [seq].  This should also make it clear why our definition of [seq'] is equivalent to [seq].  To fold and unfold parameterized fixed points, we provide two tactics, where [upaco2 f r := paco2 f r \2/ r]:  - [pfold] : when the conclusion is [paco2 f r] for some [f] and [r], [pfold] converts it to [f (upaco2 f r)] - [punfold H] : when the hypothesis [H] is [paco2 f r] for some [f] and [r], [punfold H] converts it to [f (upaco2 f r)]  Other useful lemmas are:  - [paco2_mon f : monotone2 (paco2 f)] - [paco2_mult f : forall r, paco2 f (paco2 f r) <2= paco2 f r] - [paco2_mult_strong f : forall r, paco2 f (upaco2 f r) <2= paco2 f r]   We will see an example involving [paco2_mult] in a moment.  But first let us have another look at the proof scripts of [example] and [example'].  In the former, after calling [cofix], the proof state is as follows:  [CIH : forall n : nat, seq (enumerate n) (cons n (map S (enumerate n)))] #<br># [============================] #<br># [forall n : nat, seq (enumerate n) (cons n (map S (enumerate n)))] #<br>#  In this state, the hypothesis precisely matches the conclusion, and there is nothing preventing one from simply concluding the goal directly.  Of course, if one were to do that, one's "proof" would be subsequently rejected by [Qed] for failing to obey syntactic guardedness.  Calling [pcofix] results in a similar state, except that the added hypothesis [CIH] now governs a fresh relation variable [r], which represents the current coinduction hypothesis relating [enumerate n] and [cons n (map S (enumerate n))] for all [n].  The new goal then says that, in proving the two streams equal, we can use [r] coinductively, but only in a semantically guarded position, i.e. after unfolding [paco2 seq_gen r].  In particular, one _cannot_ apply [CIH] immediately to "solve" the goal:  [r : stream -> stream -> Prop] #<br># [CIH : forall n : nat, r (enumerate n) (cons n (map S (enumerate n)))] #<br># [============================] #<br># [forall n : nat, paco2 seq_gen r (enumerate n) (cons n (map S (enumerate n)))] #<br>#  The remaining differences between the proof scripts of [example] and [example'] are as follows.  First, let us examine [example]. After applying [seq_fold] and unfolding [enumerate n] in the proof of [example], we have the following goal:  [CIH : forall n : nat, seq (enumerate n) (cons n (map S (enumerate n)))] #<br># [n : nat] #<br># [============================] #<br># [seq_gen seq (cons n (enumerate (S n))) (cons n (map S (enumerate n)))] #<br>#  By the definition of [seq_gen] and unfolding [map S (enumerate n)], this reduces to showing  [CIH : forall n : nat, seq (enumerate n) (cons n (map S (enumerate n)))] #<br># [n : nat] #<br># [============================] #<br># [seq (enumerate (S n)) (cons (S n) (map S (enumerate (S n))))] #<br>#  which follows directly by applying the coinduction hypothesis [CIH].  In the case of [example'], the proof is slightly different. First, we use the tactic [pfold] rather than [apply seq_fold], simply because we now reason about [seq'] rather than [seq]. After applying [pfold] and unfolding [enumerate n], we have:  [r : stream -> stream -> Prop] #<br># [CIH : forall n : nat, r (enumerate n) (cons n (map S (enumerate n)))] #<br># [n : nat] #<br># [============================] #<br># [seq_gen (paco2 seq_gen r \2/ r) (cons n (enumerate (S n))) (cons n (map S (enumerate n)))] #<br>#  As you see, the relation [r] appears in the argument of [seq_gen]. Thus by definition of [seq_gen] and unfolding [map S (enumerate n)], we need to show [enumerate (S n)] and [cons (S n) (map S (enumerate (S n)))] are related by either [paco2 seq_gen r] or [r]:  [r : stream -> stream -> Prop] #<br># [CIH : forall n : nat, r (enumerate n) (cons n (map S (enumerate n)))] #<br># [n : nat] #<br># [============================] #<br># [paco2 seq_gen r (enumerate (S n)) (cons (S n) (map S (enumerate (S n))))] #<br># [ \/ r (enumerate (S n)) (cons (S n) (map S (enumerate (S n))))] #<br>#  As the coinduction hypothesis gives us that they are related by [r], we first have to select the [right] disjunct, before we apply [CIH].  Summing up, the two proofs are very similar, but in the one using [paco], the guardedness of the coinduction is guaranteed at every step by the way the proof is constructed, rather than by a syntactic check of the whole proof term after the fact. *)(* *)
(** *** Another example  Before moving on to the second part, we briefly demonstrate the use of the tactics [punfold] and [pclearbot]. *)(* *)
(** Here is a proof for [seq]. *)
(** And here is the corresponding proof for [seq'].  Note that the tactic [pclearbot] simplifies all hypotheses of the form [upaco{n} gf bot{n}] to [paco{n} gf bot{n}]. *)
(** We also provide two tactics [pdestruct] and [pinversion]. They are simply defined as follows:  - [pdestruct H] := [punfold H; destruct H; pclearbot] - [pinversion H] := [punfold H; inversion H; pclearbot]  Using this the proof of the above theorem [seq'_cons] can be simplified as [intros; pinversion SEQ; auto.] *)(* *)
(** ** Example: infinite tree equality  The second example involves infinite binary trees of natural numbers and an equality thereon.  We prove two particular equalities via [cofix] in an incremental fashion, and then show how these proofs can be done just as easily via [pcofix]. We then show how, using [pcofix], the proofs can additionally be modularly decomposed.  *)
(** As before, we first define the coinductive type and the unfolding trick. *)
CoInductive inftree :=
| node : nat -> inftree -> inftree -> inftree.
Definition tunf t : inftree :=
match t with node n tl tr => node n tl tr end.
(** In order to state our example equalities, we define the following four trees. *)
CoFixpoint one : inftree := node 1 one two
with       two : inftree := node 2 one two.
CoFixpoint eins : inftree := node 1 eins (node 2 eins zwei)
with       zwei : inftree := node 2 eins zwei.
(** *** Incremental proofs using [cofix]  As before, we define the tree equality as the greatest fixed point of its generating function. *)
Inductive teq_gen teq : inftree -> inftree -> Prop :=
| _teq_gen : forall n t1l t1r t2l t2r (Rl : teq t1l t2l : Prop) (Rr : teq t1r t2r),
teq_gen teq (node n t1l t1r) (node n t2l t2r).
Hint Constructors teq_gen.
CoInductive teq t1 t2 : Prop :=
| teq_fold (IN : teq_gen teq t1 t2).
(** We now prove, using [cofix], that [one] equals [eins] and, separately, that [two] equals [zwei].  Note the nested use of [cofix] in either proof script, which lets us strengthen the coinductive hypothesis in the middle of the proof.  For instance, in the proof [teq_one], we start out with the coinductive hypothesis that [one] and [eins] are equal ([CIH]).  Later on, we use [cofix] again to additionally assume that [two] and [zwei] are equal ([CIH']).  This is a prime example of _incremental_ proof: we start with no coinductive assumptions, and we progressively add more coinductive assumptions as the proof progresses. *)
(** *** Incremental proofs using our [pcofix]  As before, we can easily turn the above [cofix]-proofs into [pcofix]-proofs. *)
Definition teq' t1 t2 := paco2 teq_gen bot2 t1 t2.
Hint Unfold teq'.
Hint Resolve teq_gen_mon : paco.
(** *** Pseudo-compositional proofs using [cofix]  It is easy to see that the proofs of [teq_one] and [teq_two] are essentially the same.  We can avoid duplicating some work by decomposing the proofs as follows.  First we prove that [teq two zwei] holds under the coinductive assumption [teq one eins].  This is basically the second part of the proof of [teq_one] (and the first part of the proof of [teq_two]).  Then we prove the converse, i.e., that [teq one eins] holds under the coinductive assumption [teq two zwei]. This is basically the first part of the proof of [teq_one] (and the second part of the proof of [teq_two]).  The issue is that there seems to be no way to express these two properties.  The best we can do is prove [teq two zwei -> teq one eins] and [teq one eins -> teq two zwei], and make sure that in these proofs any use of the assumption is syntactically guarded. For instance, we are not allowed to prove [teq two zwei -> teq one eins] by inspecting [teq two zwei] and extracting [teq one eins] from that (see the proof of [teq_two_one_bad] below). Although a valid proof of that goal by itself, it could not be composed later to yield a proof of [teq one eins] or [teq two zwei] (see the aborted theorem [teq_eins_bad] below).  Moreover, both lemmas must then be made transparent by using [Defined] instead of [Qed] at the end, such that, when composing them to prove [teq one eins] or [teq two zwei], guardedness checking can inspect their proof terms.  In other words, [cofix] is not properly compositional. *)
(** The following proof would fail guardedness checking at [Qed]-time, because in the proof of the lemma [teq_two_one_bad] the assumption is used "too early". *)
(** We now compose them with the help of the lemma [paco2_mult]: - [paco2_mult f : paco2 f (paco2 f r) <2= paco2 f r]  The tactic [pmult] applies [paco{n}_mult] to the conclusion for an appropriate [n]. *)
(** ** Example: mutual coinduction  The third and last example shows that [paco] also works for mutual coinduction.  Here, the mutuality involves two predicates. *)(* *)
(** We define two generating functions (using the [inftree] type from before). *)
Inductive eqone_gen eqone eqtwo : inftree -> Prop :=
| _eqone_gen : forall tl tr (EQL : eqone tl : Prop) (EQR : eqtwo tr : Prop),
eqone_gen eqone eqtwo (node 1 tl tr).
Inductive eqtwo_gen eqone eqtwo : inftree -> Prop :=
| _eqtwo_gen : forall tl tr (EQL : eqone tl : Prop) (EQR : eqtwo tr : Prop),
eqtwo_gen eqone eqtwo (node 2 tl tr).
Hint Constructors eqone_gen eqtwo_gen.
(** *** A proof via [cofix]  Using these, we now define two mutually coinductive predicates [eqone] and [eqtwo].  It is easy to see intuitively that they contain all trees that are equal to [one] and [two] from above, respectively.  We then prove that [eqone] contains [eins]. *)
CoInductive eqone (t : inftree) : Prop :=
| eqone_fold (EQ : eqone_gen eqone eqtwo t)
with eqtwo (t : inftree) : Prop :=
| eqtwo_fold (EQ : eqtwo_gen eqone eqtwo t).
(** *** A proof via [pcofix]  To define the [paco] versions of [eqone] and [eqtwo], we apply the two constructors [paco1_2_0] and [paco1_2_1], respectively ("1" because we are dealing with unary predicates).  Again, the translation of the lemma and of its proof is almost trivial. *)
Definition eqone' t := paco1_2_0 eqone_gen eqtwo_gen bot1 bot1 t.
Definition eqtwo' t := paco1_2_1 eqone_gen eqtwo_gen bot1 bot1 t.
Hint Unfold eqone' eqtwo'.
Hint Resolve eqone_gen_mon eqtwo_gen_mon : paco.

Lemma eqone'_eins: eqone' eins.