(** * MoreInd: More on Induction *)
Require Export ProofObjects.
(* ##################################################### *)
(** * Induction Principles *)
(** This is a good point to pause and take a deeper look at induction
principles.
Every time we declare a new [Inductive] datatype, Coq
automatically generates and proves an _induction principle_
for this type.
The induction principle for a type [t] is called [t_ind]. Here is
the one for natural numbers: *)
Check nat_ind.
(* ===> nat_ind :
forall P : nat -> Prop,
P 0 ->
(forall n : nat, P n -> P (S n)) ->
forall n : nat, P n *)
(** The [induction] tactic is a straightforward wrapper that, at
its core, simply performs [apply t_ind]. To see this more
clearly, let's experiment a little with using [apply nat_ind]
directly, instead of the [induction] tactic, to carry out some
proofs. Here, for example, is an alternate proof of a theorem
that we saw in the [Basics] chapter. *)
Theorem mult_0_r' : forall n:nat,
n * 0 = 0.
Proof.
apply nat_ind.
Case "O". reflexivity.
Case "S". simpl. intros n IHn. rewrite -> IHn.
reflexivity. Qed.
(** This proof is basically the same as the earlier one, but a
few minor differences are worth noting. First, in the induction
step of the proof (the ["S"] case), we have to do a little
bookkeeping manually (the [intros]) that [induction] does
automatically.
Second, we do not introduce [n] into the context before applying
[nat_ind] -- the conclusion of [nat_ind] is a quantified formula,
and [apply] needs this conclusion to exactly match the shape of
the goal state, including the quantifier. The [induction] tactic
works either with a variable in the context or a quantified
variable in the goal.
Third, the [apply] tactic automatically chooses variable names for
us (in the second subgoal, here), whereas [induction] lets us
specify (with the [as...] clause) what names should be used. The
automatic choice is actually a little unfortunate, since it
re-uses the name [n] for a variable that is different from the [n]
in the original theorem. This is why the [Case] annotation is
just [S] -- if we tried to write it out in the more explicit form
that we've been using for most proofs, we'd have to write [n = S
n], which doesn't make a lot of sense! All of these conveniences
make [induction] nicer to use in practice than applying induction
principles like [nat_ind] directly. But it is important to
realize that, modulo this little bit of bookkeeping, applying
[nat_ind] is what we are really doing. *)
(** **** Exercise: 2 stars, optional (plus_one_r') *)
(** Complete this proof as we did [mult_0_r'] above, without using
the [induction] tactic. *)
Theorem plus_one_r' : forall n:nat,
n + 1 = S n.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** Coq generates induction principles for every datatype defined with
[Inductive], including those that aren't recursive. (Although
we don't need induction to prove properties of non-recursive
datatypes, the idea of an induction principle still makes sense
for them: it gives a way to prove that a property holds for all
values of the type.)
These generated principles follow a similar pattern. If we define a
type [t] with constructors [c1] ... [cn], Coq generates a theorem
with this shape:
t_ind :
forall P : t -> Prop,
... case for c1 ... ->
... case for c2 ... ->
...
... case for cn ... ->
forall n : t, P n
The specific shape of each case depends on the arguments to the
corresponding constructor. Before trying to write down a general
rule, let's look at some more examples. First, an example where
the constructors take no arguments: *)
Inductive yesno : Type :=
| yes : yesno
| no : yesno.
Check yesno_ind.
(* ===> yesno_ind : forall P : yesno -> Prop,
P yes ->
P no ->
forall y : yesno, P y *)
(** **** Exercise: 1 star, optional (rgb) *)
(** Write out the induction principle that Coq will generate for the
following datatype. Write down your answer on paper or type it
into a comment, and then compare it with what Coq prints. *)
Inductive rgb : Type :=
| red : rgb
| green : rgb
| blue : rgb.
Check rgb_ind.
(** [] *)
(** Here's another example, this time with one of the constructors
taking some arguments. *)
Inductive natlist : Type :=
| nnil : natlist
| ncons : nat -> natlist -> natlist.
Check natlist_ind.
(* ===> (modulo a little variable renaming for clarity)
natlist_ind :
forall P : natlist -> Prop,
P nnil ->
(forall (n : nat) (l : natlist), P l -> P (ncons n l)) ->
forall n : natlist, P n *)
(** **** Exercise: 1 star, optional (natlist1) *)
(** Suppose we had written the above definition a little
differently: *)
Inductive natlist1 : Type :=
| nnil1 : natlist1
| nsnoc1 : natlist1 -> nat -> natlist1.
(** Now what will the induction principle look like? *)
(** [] *)
(** From these examples, we can extract this general rule:
- The type declaration gives several constructors; each
corresponds to one clause of the induction principle.
- Each constructor [c] takes argument types [a1]...[an].
- Each [ai] can be either [t] (the datatype we are defining) or
some other type [s].
- The corresponding case of the induction principle
says (in English):
- "for all values [x1]...[xn] of types [a1]...[an], if [P]
holds for each of the inductive arguments (each [xi] of
type [t]), then [P] holds for [c x1 ... xn]".
*)
(** **** Exercise: 1 star, optional (byntree_ind) *)
(** Write out the induction principle that Coq will generate for the
following datatype. Write down your answer on paper or type it
into a comment, and then compare it with what Coq prints. *)
Inductive byntree : Type :=
| bempty : byntree
| bleaf : yesno -> byntree
| nbranch : yesno -> byntree -> byntree -> byntree.
(** [] *)
(** **** Exercise: 1 star, optional (ex_set) *)
(** Here is an induction principle for an inductively defined
set.
ExSet_ind :
forall P : ExSet -> Prop,
(forall b : bool, P (con1 b)) ->
(forall (n : nat) (e : ExSet), P e -> P (con2 n e)) ->
forall e : ExSet, P e
Give an [Inductive] definition of [ExSet]: *)
Inductive ExSet : Type :=
(* FILL IN HERE *)
.
(** [] *)
(** What about polymorphic datatypes?
The inductive definition of polymorphic lists
Inductive list (X:Type) : Type :=
| nil : list X
| cons : X -> list X -> list X.
is very similar to that of [natlist]. The main difference is
that, here, the whole definition is _parameterized_ on a set [X]:
that is, we are defining a _family_ of inductive types [list X],
one for each [X]. (Note that, wherever [list] appears in the body
of the declaration, it is always applied to the parameter [X].)
The induction principle is likewise parameterized on [X]:
list_ind :
forall (X : Type) (P : list X -> Prop),
P [] ->
(forall (x : X) (l : list X), P l -> P (x :: l)) ->
forall l : list X, P l
Note the wording here (and, accordingly, the form of [list_ind]):
The _whole_ induction principle is parameterized on [X]. That is,
[list_ind] can be thought of as a polymorphic function that, when
applied to a type [X], gives us back an induction principle
specialized to the type [list X]. *)
(** **** Exercise: 1 star, optional (tree) *)
(** Write out the induction principle that Coq will generate for
the following datatype. Compare your answer with what Coq
prints. *)
Inductive tree (X:Type) : Type :=
| leaf : X -> tree X
| node : tree X -> tree X -> tree X.
Check tree_ind.
(** [] *)
(** **** Exercise: 1 star, optional (mytype) *)
(** Find an inductive definition that gives rise to the
following induction principle:
mytype_ind :
forall (X : Type) (P : mytype X -> Prop),
(forall x : X, P (constr1 X x)) ->
(forall n : nat, P (constr2 X n)) ->
(forall m : mytype X, P m ->
forall n : nat, P (constr3 X m n)) ->
forall m : mytype X, P m
*)
(** [] *)
(** **** Exercise: 1 star, optional (foo) *)
(** Find an inductive definition that gives rise to the
following induction principle:
foo_ind :
forall (X Y : Type) (P : foo X Y -> Prop),
(forall x : X, P (bar X Y x)) ->
(forall y : Y, P (baz X Y y)) ->
(forall f1 : nat -> foo X Y,
(forall n : nat, P (f1 n)) -> P (quux X Y f1)) ->
forall f2 : foo X Y, P f2
*)
(** [] *)
(** **** Exercise: 1 star, optional (foo') *)
(** Consider the following inductive definition: *)
Inductive foo' (X:Type) : Type :=
| C1 : list X -> foo' X -> foo' X
| C2 : foo' X.
(** What induction principle will Coq generate for [foo']? Fill
in the blanks, then check your answer with Coq.)
foo'_ind :
forall (X : Type) (P : foo' X -> Prop),
(forall (l : list X) (f : foo' X),
_______________________ ->
_______________________ ) ->
___________________________________________ ->
forall f : foo' X, ________________________
*)
(** [] *)
(* ##################################################### *)
(** ** Induction Hypotheses *)
(** Where does the phrase "induction hypothesis" fit into this story?
The induction principle for numbers
forall P : nat -> Prop,
P 0 ->
(forall n : nat, P n -> P (S n)) ->
forall n : nat, P n
is a generic statement that holds for all propositions
[P] (strictly speaking, for all families of propositions [P]
indexed by a number [n]). Each time we use this principle, we
are choosing [P] to be a particular expression of type
[nat->Prop].
We can make the proof more explicit by giving this expression a
name. For example, instead of stating the theorem [mult_0_r] as
"[forall n, n * 0 = 0]," we can write it as "[forall n, P_m0r
n]", where [P_m0r] is defined as... *)
Definition P_m0r (n:nat) : Prop :=
n * 0 = 0.
(** ... or equivalently... *)
Definition P_m0r' : nat->Prop :=
fun n => n * 0 = 0.
(** Now when we do the proof it is easier to see where [P_m0r]
appears. *)
Theorem mult_0_r'' : forall n:nat,
P_m0r n.
Proof.
apply nat_ind.
Case "n = O". reflexivity.
Case "n = S n'".
(* Note the proof state at this point! *)
intros n IHn.
unfold P_m0r in IHn. unfold P_m0r. simpl. apply IHn. Qed.
(** This extra naming step isn't something that we'll do in
normal proofs, but it is useful to do it explicitly for an example
or two, because it allows us to see exactly what the induction
hypothesis is. If we prove [forall n, P_m0r n] by induction on
[n] (using either [induction] or [apply nat_ind]), we see that the
first subgoal requires us to prove [P_m0r 0] ("[P] holds for
zero"), while the second subgoal requires us to prove [forall n',
P_m0r n' -> P_m0r n' (S n')] (that is "[P] holds of [S n'] if it
holds of [n']" or, more elegantly, "[P] is preserved by [S]").
The _induction hypothesis_ is the premise of this latter
implication -- the assumption that [P] holds of [n'], which we are
allowed to use in proving that [P] holds for [S n']. *)
(* ##################################################### *)
(** ** More on the [induction] Tactic *)
(** The [induction] tactic actually does even more low-level
bookkeeping for us than we discussed above.
Recall the informal statement of the induction principle for
natural numbers:
- If [P n] is some proposition involving a natural number n, and
we want to show that P holds for _all_ numbers n, we can
reason like this:
- show that [P O] holds
- show that, if [P n'] holds, then so does [P (S n')]
- conclude that [P n] holds for all n.
So, when we begin a proof with [intros n] and then [induction n],
we are first telling Coq to consider a _particular_ [n] (by
introducing it into the context) and then telling it to prove
something about _all_ numbers (by using induction).
What Coq actually does in this situation, internally, is to
"re-generalize" the variable we perform induction on. For
example, in our original proof that [plus] is associative...
*)
Theorem plus_assoc' : forall n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
(* ...we first introduce all 3 variables into the context,
which amounts to saying "Consider an arbitrary [n], [m], and
[p]..." *)
intros n m p.
(* ...We now use the [induction] tactic to prove [P n] (that
is, [n + (m + p) = (n + m) + p]) for _all_ [n],
and hence also for the particular [n] that is in the context
at the moment. *)
induction n as [| n'].
Case "n = O". reflexivity.
Case "n = S n'".
(* In the second subgoal generated by [induction] -- the
"inductive step" -- we must prove that [P n'] implies
[P (S n')] for all [n']. The [induction] tactic
automatically introduces [n'] and [P n'] into the context
for us, leaving just [P (S n')] as the goal. *)
simpl. rewrite -> IHn'. reflexivity. Qed.
(** It also works to apply [induction] to a variable that is
quantified in the goal. *)
Theorem plus_comm' : forall n m : nat,
n + m = m + n.
Proof.
induction n as [| n'].
Case "n = O". intros m. rewrite -> plus_0_r. reflexivity.
Case "n = S n'". intros m. simpl. rewrite -> IHn'.
rewrite <- plus_n_Sm. reflexivity. Qed.
(** Note that [induction n] leaves [m] still bound in the goal --
i.e., what we are proving inductively is a statement beginning
with [forall m].
If we do [induction] on a variable that is quantified in the goal
_after_ some other quantifiers, the [induction] tactic will
automatically introduce the variables bound by these quantifiers
into the context. *)
Theorem plus_comm'' : forall n m : nat,
n + m = m + n.
Proof.
(* Let's do induction on [m] this time, instead of [n]... *)
induction m as [| m'].
Case "m = O". simpl. rewrite -> plus_0_r. reflexivity.
Case "m = S m'". simpl. rewrite <- IHm'.
rewrite <- plus_n_Sm. reflexivity. Qed.
(** **** Exercise: 1 star, optional (plus_explicit_prop) *)
(** Rewrite both [plus_assoc'] and [plus_comm'] and their proofs in
the same style as [mult_0_r''] above -- that is, for each theorem,
give an explicit [Definition] of the proposition being proved by
induction, and state the theorem and proof in terms of this
defined proposition. *)
(* FILL IN HERE *)
(** [] *)
(** ** Generalizing Inductions. *)
(** One potentially confusing feature of the [induction] tactic is
that it happily lets you try to set up an induction over a term
that isn't sufficiently general. The net effect of this will be
do lose information (much as [destruct] can do), and leave
you unable to complete the proof. Here's an example: *)
Lemma one_not_beautiful_FAILED: ~ beautiful 1.
Proof.
intro H.
(* Just doing an [inversion] on [H] won't get us very far in the [b_sum]
case. (Try it!). So we'll need induction. A naive first attempt: *)
induction H.
(* But now, although we get four cases, as we would expect from
the definition of [beautiful], we lose all information about [H] ! *)
Abort.
(** The problem is that [induction] over a Prop only works properly over
completely general instances of the Prop, i.e. one in which all
the arguments are free (unconstrained) variables.
In this respect it behaves more
like [destruct] than like [inversion].
When you're tempted to do use [induction] like this, it is generally
an indication that you need to be proving something more general.
But in some cases, it suffices to pull out any concrete arguments
into separate equations, like this: *)
Lemma one_not_beautiful: forall n, n = 1 -> ~ beautiful n.
Proof.
intros n E H.
induction H as [| | | p q Hp IHp Hq IHq].
Case "b_0".
inversion E.
Case "b_3".
inversion E.
Case "b_5".
inversion E.
Case "b_sum".
(* the rest is a tedious case analysis *)
destruct p as [|p'].
SCase "p = 0".
destruct q as [|q'].
SSCase "q = 0".
inversion E.
SSCase "q = S q'".
apply IHq. apply E.
SCase "p = S p'".
destruct q as [|q'].
SSCase "q = 0".
apply IHp. rewrite plus_0_r in E. apply E.
SSCase "q = S q'".
simpl in E. inversion E. destruct p'. inversion H0. inversion H0.
Qed.
(** There's a handy [remember] tactic that can generate the second
proof state out of the original one. *)
Lemma one_not_beautiful': ~ beautiful 1.
Proof.
intros H.
remember 1 as n eqn:E.
(* now carry on as above *)
induction H.
Admitted.
(* ####################################################### *)
(** * Informal Proofs (Advanced) *)
(** Q: What is the relation between a formal proof of a proposition
[P] and an informal proof of the same proposition [P]?
A: The latter should _teach_ the reader how to produce the
former.
Q: How much detail is needed??
Unfortunately, There is no single right answer; rather, there is a
range of choices.
At one end of the spectrum, we can essentially give the reader the
whole formal proof (i.e., the informal proof amounts to just
transcribing the formal one into words). This gives the reader
the _ability_ to reproduce the formal one for themselves, but it
doesn't _teach_ them anything.
At the other end of the spectrum, we can say "The theorem is true
and you can figure out why for yourself if you think about it hard
enough." This is also not a good teaching strategy, because
usually writing the proof requires some deep insights into the
thing we're proving, and most readers will give up before they
rediscover all the same insights as we did.
In the middle is the golden mean -- a proof that includes all of
the essential insights (saving the reader the hard part of work
that we went through to find the proof in the first place) and
clear high-level suggestions for the more routine parts to save the
reader from spending too much time reconstructing these
parts (e.g., what the IH says and what must be shown in each case
of an inductive proof), but not so much detail that the main ideas
are obscured.
Another key point: if we're comparing a formal proof of a
proposition [P] and an informal proof of [P], the proposition [P]
doesn't change. That is, formal and informal proofs are _talking
about the same world_ and they _must play by the same rules_. *)
(** ** Informal Proofs by Induction *)
(** Since we've spent much of this chapter looking "under the hood" at
formal proofs by induction, now is a good moment to talk a little
about _informal_ proofs by induction.
In the real world of mathematical communication, written proofs
range from extremely longwinded and pedantic to extremely brief
and telegraphic. The ideal is somewhere in between, of course,
but while you are getting used to the style it is better to start
out at the pedantic end. Also, during the learning phase, it is
probably helpful to have a clear standard to compare against.
With this in mind, we offer two templates below -- one for proofs
by induction over _data_ (i.e., where the thing we're doing
induction on lives in [Type]) and one for proofs by induction over
_evidence_ (i.e., where the inductively defined thing lives in
[Prop]). In the rest of this course, please follow one of the two
for _all_ of your inductive proofs. *)
(** *** Induction Over an Inductively Defined Set *)
(** _Template_:
- _Theorem_:
_Proof_: By induction on [n].
- Suppose [n = c a1 ... ak], where <...and here we state
the IH for each of the [a]'s that has type [S], if any>.
We must show <...and here we restate [P(c a1 ... ak)]>.
- []
_Example_:
- _Theorem_: For all sets [X], lists [l : list X], and numbers
[n], if [length l = n] then [index (S n) l = None].
_Proof_: By induction on [l].
- Suppose [l = []]. We must show, for all numbers [n],
that, if length [[] = n], then [index (S n) [] =
None].
This follows immediately from the definition of index.
- Suppose [l = x :: l'] for some [x] and [l'], where
[length l' = n'] implies [index (S n') l' = None], for
any number [n']. We must show, for all [n], that, if
[length (x::l') = n] then [index (S n) (x::l') =
None].
Let [n] be a number with [length l = n]. Since
length l = length (x::l') = S (length l'),
it suffices to show that
index (S (length l')) l' = None.
]]
But this follows directly from the induction hypothesis,
picking [n'] to be length [l']. [] *)
(** *** Induction Over an Inductively Defined Proposition *)
(** Since inductively defined proof objects are often called
"derivation trees," this form of proof is also known as _induction
on derivations_.
_Template_:
- _Theorem_: P]," where [Q] is
some inductively defined proposition (more generally,
"For all [x] [y] [z], [Q x y z -> P x y z]")>
_Proof_: By induction on a derivation of [Q].
- Suppose the final rule used to show [Q] is [c]. Then
<...and here we state the types of all of the [a]'s
together with any equalities that follow from the
definition of the constructor and the IH for each of
the [a]'s that has type [Q], if there are any>. We must
show <...and here we restate [P]>.
- []
_Example_
- _Theorem_: The [<=] relation is transitive -- i.e., for all
numbers [n], [m], and [o], if [n <= m] and [m <= o], then
[n <= o].
_Proof_: By induction on a derivation of [m <= o].
- Suppose the final rule used to show [m <= o] is
[le_n]. Then [m = o] and we must show that [n <= m],
which is immediate by hypothesis.
- Suppose the final rule used to show [m <= o] is
[le_S]. Then [o = S o'] for some [o'] with [m <= o'].
We must show that [n <= S o'].
By induction hypothesis, [n <= o'].
But then, by [le_S], [n <= S o']. [] *)
(* ##################################################### *)
(** * Optional Material *)
(** The remainder of this chapter offers some additional details on
how induction works in Coq, the process of building proof
trees, and the "trusted computing base" that underlies
Coq proofs. It can safely be skimmed on a first reading. (We
recommend skimming rather than skipping over it outright: it
answers some questions that occur to many Coq users at some point,
so it is useful to have a rough idea of what's here.) *)
(* ##################################################### *)
(** ** Induction Principles in [Prop] *)
(** Earlier, we looked in detail at the induction principles that Coq
generates for inductively defined _sets_. The induction
principles for inductively defined _propositions_ like [gorgeous]
are a tiny bit more complicated. As with all induction
principles, we want to use the induction principle on [gorgeous]
to prove things by inductively considering the possible shapes
that something in [gorgeous] can have -- either it is evidence
that [0] is gorgeous, or it is evidence that, for some [n], [3+n]
is gorgeous, or it is evidence that, for some [n], [5+n] is
gorgeous and it includes evidence that [n] itself is. Intuitively
speaking, however, what we want to prove are not statements about
_evidence_ but statements about _numbers_. So we want an
induction principle that lets us prove properties of numbers by
induction on evidence.
For example, from what we've said so far, you might expect the
inductive definition of [gorgeous]...
Inductive gorgeous : nat -> Prop :=
g_0 : gorgeous 0
| g_plus3 : forall n, gorgeous n -> gorgeous (3+m)
| g_plus5 : forall n, gorgeous n -> gorgeous (5+m).
...to give rise to an induction principle that looks like this...
gorgeous_ind_max :
forall P : (forall n : nat, gorgeous n -> Prop),
P O g_0 ->
(forall (m : nat) (e : gorgeous m),
P m e -> P (3+m) (g_plus3 m e) ->
(forall (m : nat) (e : gorgeous m),
P m e -> P (5+m) (g_plus5 m e) ->
forall (n : nat) (e : gorgeous n), P n e
... because:
- Since [gorgeous] is indexed by a number [n] (every [gorgeous]
object [e] is a piece of evidence that some particular number
[n] is gorgeous), the proposition [P] is parameterized by both
[n] and [e] -- that is, the induction principle can be used to
prove assertions involving both a gorgeous number and the
evidence that it is gorgeous.
- Since there are three ways of giving evidence of gorgeousness
([gorgeous] has three constructors), applying the induction
principle generates three subgoals:
- We must prove that [P] holds for [O] and [b_0].
- We must prove that, whenever [n] is a gorgeous
number and [e] is an evidence of its gorgeousness,
if [P] holds of [n] and [e],
then it also holds of [3+m] and [g_plus3 n e].
- We must prove that, whenever [n] is a gorgeous
number and [e] is an evidence of its gorgeousness,
if [P] holds of [n] and [e],
then it also holds of [5+m] and [g_plus5 n e].
- If these subgoals can be proved, then the induction principle
tells us that [P] is true for _all_ gorgeous numbers [n] and
evidence [e] of their gorgeousness.
But this is a little more flexibility than we actually need or
want: it is giving us a way to prove logical assertions where the
assertion involves properties of some piece of _evidence_ of
gorgeousness, while all we really care about is proving
properties of _numbers_ that are gorgeous -- we are interested in
assertions about numbers, not about evidence. It would therefore
be more convenient to have an induction principle for proving
propositions [P] that are parameterized just by [n] and whose
conclusion establishes [P] for all gorgeous numbers [n]:
forall P : nat -> Prop,
... ->
forall n : nat, gorgeous n -> P n
For this reason, Coq actually generates the following simplified
induction principle for [gorgeous]: *)
Check gorgeous_ind.
(* ===> gorgeous_ind
: forall P : nat -> Prop,
P 0 ->
(forall n : nat, gorgeous n -> P n -> P (3 + n)) ->
(forall n : nat, gorgeous n -> P n -> P (5 + n)) ->
forall n : nat, gorgeous n -> P n *)
(** In particular, Coq has dropped the evidence term [e] as a
parameter of the the proposition [P], and consequently has
rewritten the assumption [forall (n : nat) (e: gorgeous n), ...]
to be [forall (n : nat), gorgeous n -> ...]; i.e., we no longer
require explicit evidence of the provability of [gorgeous n]. *)
(** In English, [gorgeous_ind] says:
- Suppose, [P] is a property of natural numbers (that is, [P n] is
a [Prop] for every [n]). To show that [P n] holds whenever [n]
is gorgeous, it suffices to show:
- [P] holds for [0],
- for any [n], if [n] is gorgeous and [P] holds for
[n], then [P] holds for [3+n],
- for any [n], if [n] is gorgeous and [P] holds for
[n], then [P] holds for [5+n]. *)
(** As expected, we can apply [gorgeous_ind] directly instead of using [induction]. *)
Theorem gorgeous__beautiful' : forall n, gorgeous n -> beautiful n.
Proof.
intros.
apply gorgeous_ind.
Case "g_0".
apply b_0.
Case "g_plus3".
intros.
apply b_sum. apply b_3.
apply H1.
Case "g_plus5".
intros.
apply b_sum. apply b_5.
apply H1.
apply H.
Qed.
(** The precise form of an Inductive definition can affect the
induction principle Coq generates.
For example, in [Logic], we have defined [<=] as: *)
(* Inductive le : nat -> nat -> Prop :=
| le_n : forall n, le n n
| le_S : forall n m, (le n m) -> (le n (S m)). *)
(** This definition can be streamlined a little by observing that the
left-hand argument [n] is the same everywhere in the definition,
so we can actually make it a "general parameter" to the whole
definition, rather than an argument to each constructor. *)
Inductive le (n:nat) : nat -> Prop :=
| le_n : le n n
| le_S : forall m, (le n m) -> (le n (S m)).
Notation "m <= n" := (le m n).
(** The second one is better, even though it looks less symmetric.
Why? Because it gives us a simpler induction principle. *)
Check le_ind.
(* ===> forall (n : nat) (P : nat -> Prop),
P n ->
(forall m : nat, n <= m -> P m -> P (S m)) ->
forall n0 : nat, n <= n0 -> P n0 *)
(** By contrast, the induction principle that Coq calculates for the
first definition has a lot of extra quantifiers, which makes it
messier to work with when proving things by induction. Here is
the induction principle for the first [le]: *)
(* le_ind :
forall P : nat -> nat -> Prop,
(forall n : nat, P n n) ->
(forall n m : nat, le n m -> P n m -> P n (S m)) ->
forall n n0 : nat, le n n0 -> P n n0 *)
(* ##################################################### *)
(** * Additional Exercises *)
(** **** Exercise: 2 stars, optional (foo_ind_principle) *)
(** Suppose we make the following inductive definition:
Inductive foo (X : Set) (Y : Set) : Set :=
| foo1 : X -> foo X Y
| foo2 : Y -> foo X Y
| foo3 : foo X Y -> foo X Y.
Fill in the blanks to complete the induction principle that will be
generated by Coq.
foo_ind
: forall (X Y : Set) (P : foo X Y -> Prop),
(forall x : X, __________________________________) ->
(forall y : Y, __________________________________) ->
(________________________________________________) ->
________________________________________________
*)
(** [] *)
(** **** Exercise: 2 stars, optional (bar_ind_principle) *)
(** Consider the following induction principle:
bar_ind
: forall P : bar -> Prop,
(forall n : nat, P (bar1 n)) ->
(forall b : bar, P b -> P (bar2 b)) ->
(forall (b : bool) (b0 : bar), P b0 -> P (bar3 b b0)) ->
forall b : bar, P b
Write out the corresponding inductive set definition.
Inductive bar : Set :=
| bar1 : ________________________________________
| bar2 : ________________________________________
| bar3 : ________________________________________.
*)
(** [] *)
(** **** Exercise: 2 stars, optional (no_longer_than_ind) *)
(** Given the following inductively defined proposition:
Inductive no_longer_than (X : Set) : (list X) -> nat -> Prop :=
| nlt_nil : forall n, no_longer_than X [] n
| nlt_cons : forall x l n, no_longer_than X l n ->
no_longer_than X (x::l) (S n)
| nlt_succ : forall l n, no_longer_than X l n ->
no_longer_than X l (S n).
write the induction principle generated by Coq.
no_longer_than_ind
: forall (X : Set) (P : list X -> nat -> Prop),
(forall n : nat, ____________________) ->
(forall (x : X) (l : list X) (n : nat),
no_longer_than X l n -> ____________________ ->
_____________________________ ->
(forall (l : list X) (n : nat),
no_longer_than X l n -> ____________________ ->
_____________________________ ->
forall (l : list X) (n : nat), no_longer_than X l n ->
____________________
*)
(** [] *)
(* ##################################################### *)
(** ** Induction Principles for other Logical Propositions *)
(** Similarly, in [Logic] we have defined [eq] as: *)
(* Inductive eq (X:Type) : X -> X -> Prop :=
refl_equal : forall x, eq X x x. *)
(** In the Coq standard library, the definition of equality is
slightly different: *)
Inductive eq' (X:Type) (x:X) : X -> Prop :=
refl_equal' : eq' X x x.
(** The advantage of this definition is that the induction
principle that Coq derives for it is precisely the familiar
principle of _Leibniz equality_: what we mean when we say "[x] and
[y] are equal" is that every property on [P] that is true of [x]
is also true of [y]. *)
Check eq'_ind.
(* ===>
forall (X : Type) (x : X) (P : X -> Prop),
P x -> forall y : X, x =' y -> P y
===> (i.e., after a little reorganization)
forall (X : Type) (x : X) forall y : X,
x =' y ->
forall P : X -> Prop, P x -> P y *)
(** The induction principles for conjunction and disjunction are a
good illustration of Coq's way of generating simplified induction
principles for [Inductive]ly defined propositions, which we
discussed above. You try first: *)
(** **** Exercise: 1 star, optional (and_ind_principle) *)
(** See if you can predict the induction principle for conjunction. *)
(* Check and_ind. *)
(** [] *)
(** **** Exercise: 1 star, optional (or_ind_principle) *)
(** See if you can predict the induction principle for disjunction. *)
(* Check or_ind. *)
(** [] *)
Check and_ind.
(** From the inductive definition of the proposition [and P Q]
Inductive and (P Q : Prop) : Prop :=
conj : P -> Q -> (and P Q).
we might expect Coq to generate this induction principle
and_ind_max :
forall (P Q : Prop) (P0 : P /\ Q -> Prop),
(forall (a : P) (b : Q), P0 (conj P Q a b)) ->
forall a : P /\ Q, P0 a
but actually it generates this simpler and more useful one:
and_ind :
forall P Q P0 : Prop,
(P -> Q -> P0) ->
P /\ Q -> P0
In the same way, when given the inductive definition of [or P Q]
Inductive or (P Q : Prop) : Prop :=
| or_introl : P -> or P Q
| or_intror : Q -> or P Q.
instead of the "maximal induction principle"
or_ind_max :
forall (P Q : Prop) (P0 : P \/ Q -> Prop),
(forall a : P, P0 (or_introl P Q a)) ->
(forall b : Q, P0 (or_intror P Q b)) ->
forall o : P \/ Q, P0 o
what Coq actually generates is this:
or_ind :
forall P Q P0 : Prop,
(P -> P0) ->
(Q -> P0) ->
P \/ Q -> P0
]]
*)
(** **** Exercise: 1 star, optional (False_ind_principle) *)
(** Can you predict the induction principle for falsehood? *)
(* Check False_ind. *)
(** [] *)
(** Here's the induction principle that Coq generates for existentials: *)
Check ex_ind.
(* ===> forall (X:Type) (P: X->Prop) (Q: Prop),
(forall witness:X, P witness -> Q) ->
ex X P ->
Q *)
(** This induction principle can be understood as follows: If we have
a function [f] that can construct evidence for [Q] given _any_
witness of type [X] together with evidence that this witness has
property [P], then from a proof of [ex X P] we can extract the
witness and evidence that must have been supplied to the
constructor, give these to [f], and thus obtain a proof of [Q]. *)
(* ######################################################### *)
(** ** Explicit Proof Objects for Induction *)
(** Although tactic-based proofs are normally much easier to
work with, the ability to write a proof term directly is sometimes
very handy, particularly when we want Coq to do something slightly
non-standard. *)
(** Recall the induction principle on naturals that Coq generates for
us automatically from the Inductive declation for [nat]. *)
Check nat_ind.
(* ===>
nat_ind : forall P : nat -> Prop,
P 0 ->
(forall n : nat, P n -> P (S n)) ->
forall n : nat, P n *)
(** There's nothing magic about this induction lemma: it's just
another Coq lemma that requires a proof. Coq generates the proof
automatically too... *)
Print nat_ind.
Print nat_rect.
(* ===> (after some manual inlining and tidying)
nat_ind =
fun (P : nat -> Prop)
(f : P 0)
(f0 : forall n : nat, P n -> P (S n)) =>
fix F (n : nat) : P n :=
match n with
| 0 => f
| S n0 => f0 n0 (F n0)
end.
*)
(** We can read this as follows:
Suppose we have evidence [f] that [P] holds on 0, and
evidence [f0] that [forall n:nat, P n -> P (S n)].
Then we can prove that [P] holds of an arbitrary nat [n] via
a recursive function [F] (here defined using the expression
form [Fix] rather than by a top-level [Fixpoint]
declaration). [F] pattern matches on [n]:
- If it finds 0, [F] uses [f] to show that [P n] holds.
- If it finds [S n0], [F] applies itself recursively on [n0]
to obtain evidence that [P n0] holds; then it applies [f0]
on that evidence to show that [P (S n)] holds.
[F] is just an ordinary recursive function that happens to
operate on evidence in [Prop] rather than on terms in [Set].
*)
(** We can adapt this approach to proving [nat_ind] to help prove
_non-standard_ induction principles too. Recall our desire to
prove that
[forall n : nat, even n -> ev n].
Attempts to do this by standard induction on [n] fail, because the
induction principle only lets us proceed when we can prove that
[even n -> even (S n)] -- which is of course never provable. What
we did in [Logic] was a bit of a hack:
[Theorem even__ev : forall n : nat,
(even n -> ev n) /\ (even (S n) -> ev (S n))].
We can make a much better proof by defining and proving a
non-standard induction principle that goes "by twos":
*)
Definition nat_ind2 :
forall (P : nat -> Prop),
P 0 ->
P 1 ->
(forall n : nat, P n -> P (S(S n))) ->
forall n : nat , P n :=
fun P => fun P0 => fun P1 => fun PSS =>
fix f (n:nat) := match n with
0 => P0
| 1 => P1
| S (S n') => PSS n' (f n')
end.
(** Once you get the hang of it, it is entirely straightforward to
give an explicit proof term for induction principles like this.
Proving this as a lemma using tactics is much less intuitive (try
it!).
The [induction ... using] tactic variant gives a convenient way to
specify a non-standard induction principle like this. *)
Lemma even__ev' : forall n, even n -> ev n.
Proof.
intros.
induction n as [ | |n'] using nat_ind2.
Case "even 0".
apply ev_0.
Case "even 1".
inversion H.
Case "even (S(S n'))".
apply ev_SS.
apply IHn'. unfold even. unfold even in H. simpl in H. apply H.
Qed.
(* ######################################################### *)
(** ** The Coq Trusted Computing Base *)
(** One issue that arises with any automated proof assistant is "why
trust it?": what if there is a bug in the implementation that
renders all its reasoning suspect?
While it is impossible to allay such concerns completely, the fact
that Coq is based on the Curry-Howard correspondence gives it a
strong foundation. Because propositions are just types and proofs
are just terms, checking that an alleged proof of a proposition is
valid just amounts to _type-checking_ the term. Type checkers are
relatively small and straightforward programs, so the "trusted
computing base" for Coq -- the part of the code that we have to
believe is operating correctly -- is small too.
What must a typechecker do? Its primary job is to make sure that
in each function application the expected and actual argument
types match, that the arms of a [match] expression are constructor
patterns belonging to the inductive type being matched over and
all arms of the [match] return the same type, and so on.
There are a few additional wrinkles:
- Since Coq types can themselves be expressions, the checker must
normalize these (by using the computation rules) before
comparing them.
- The checker must make sure that [match] expressions are
_exhaustive_. That is, there must be an arm for every possible
constructor. To see why, consider the following alleged proof
object:
Definition or_bogus : forall P Q, P \/ Q -> P :=
fun (P Q : Prop) (A : P \/ Q) =>
match A with
| or_introl H => H
end.
All the types here match correctly, but the [match] only
considers one of the possible constructors for [or]. Coq's
exhaustiveness check will reject this definition.
- The checker must make sure that each [fix] expression
terminates. It does this using a syntactic check to make sure
that each recursive call is on a subexpression of the original
argument. To see why this is essential, consider this alleged
proof:
Definition nat_false : forall (n:nat), False :=
fix f (n:nat) : False := f n.
Again, this is perfectly well-typed, but (fortunately) Coq will
reject it. *)
(** Note that the soundness of Coq depends only on the correctness of
this typechecking engine, not on the tactic machinery. If there
is a bug in a tactic implementation (and this certainly does
happen!), that tactic might construct an invalid proof term. But
when you type [Qed], Coq checks the term for validity from
scratch. Only lemmas whose proofs pass the type-checker can be
used in further proof developments. *)
(* $Date: 2013-07-17 16:19:11 -0400 (Wed, 17 Jul 2013) $ *)