(** * Prop: Propositions and Evidence *)
Require Export MoreCoq.
(** In previous chapters, we have seen many examples of factual
claims (_propositions_) and ways of presenting evidence of their
truth (_proofs_). In particular, we have worked extensively with
_equality propositions_ of the form [e1 = e2], with
implications ([P -> Q]), and with quantified propositions
([forall x, P]).
This chapter will take us on a first tour of the
propositional (logical) side of Coq.
In particular, we will expand our repertoire of primitive
propositions to include _user-defined_ propositions, not just
equality propositions (which are more-or-less "built in" to Coq).
*)
(* ##################################################### *)
(** * Inductively Defined Propositions *)
(** As a running example, let's
define a simple property of natural numbers -- we'll call it
"[beautiful]." *)
(** Informally, a number is [beautiful] if it is [0], [3], [5], or the
sum of two [beautiful] numbers.
More pedantically, we can define [beautiful] numbers by giving four
rules:
- Rule [b_0]: The number [0] is [beautiful].
- Rule [b_3]: The number [3] is [beautiful].
- Rule [b_5]: The number [5] is [beautiful].
- Rule [b_sum]: If [n] and [m] are both [beautiful], then so is
their sum. *)
(** We will see many definitions like this one during the rest
of the course, and for purposes of informal discussions, it is
helpful to have a lightweight notation that makes them easy to
read and write. _Inference rules_ are one such notation: *)
(**
----------- (b_0)
beautiful 0
------------ (b_3)
beautiful 3
------------ (b_5)
beautiful 5
beautiful n beautiful m
--------------------------- (b_sum)
beautiful (n+m)
*)
(** Each of the textual rules above is reformatted here as an
inference rule; the intended reading is that, if the _premises_
above the line all hold, then the _conclusion_ below the line
follows. For example, the rule [b_sum] says that, if [n] and [m]
are both [beautiful] numbers, then it follows that [n+m] is
[beautiful] too. If a rule has no premises above the line, then
its conclusion hold unconditionally.
These rules _define_ the property [beautiful]. That is, if we
want to convince someone that some particular number is [beautiful],
our argument must be based on these rules. For a simple example,
suppose we claim that the number [5] is [beautiful]. To support
this claim, we just need to point out that rule [b_5] says so.
Or, if we want to claim that [8] is [beautiful], we can support our
claim by first observing that [3] and [5] are both [beautiful] (by
rules [b_3] and [b_5]) and then pointing out that their sum, [8],
is therefore [beautiful] by rule [b_sum]. This argument can be
expressed graphically with the following _proof tree_: *)
(**
----------- (b_3) ----------- (b_5)
beautiful 3 beautiful 5
------------------------------- (b_sum)
beautiful 8
Of course, there are other ways of using these rules to argue that
[8] is [beautiful], for instance:
----------- (b_5) ----------- (b_3)
beautiful 5 beautiful 3
------------------------------- (b_sum)
beautiful 8
*)
(** **** Exercise: 1 star (varieties_of_beauty) *)
(** How many different ways are there to show that [8] is [beautiful]? *)
(* FILL IN HERE *)
(** [] *)
(** In Coq, we can express the definition of [beautiful] as
follows: *)
Inductive beautiful : nat -> Prop :=
b_0 : beautiful 0
| b_3 : beautiful 3
| b_5 : beautiful 5
| b_sum : forall n m, beautiful n -> beautiful m -> beautiful (n+m).
(** The first line declares that [beautiful] is a proposition -- or,
more formally, a family of propositions "indexed by" natural
numbers. (That is, for each number [n], the claim that "[n] is
[beautiful]" is a proposition.) Such a family of propositions is
often called a _property_ of numbers. Each of the remaining lines
embodies one of the rules for [beautiful] numbers.
The rules introduced this way have the same status as proven
theorems; that is, they are true axiomatically.
So we can use Coq's [apply] tactic with the rule names to prove
that particular numbers are [beautiful]. *)
Theorem three_is_beautiful: beautiful 3.
Proof.
(* This simply follows from the rule [b_3]. *)
apply b_3.
Qed.
Theorem eight_is_beautiful: beautiful 8.
Proof.
(* First we use the rule [b_sum], telling Coq how to
instantiate [n] and [m]. *)
apply b_sum with (n:=3) (m:=5).
(* To solve the subgoals generated by [b_sum], we must provide
evidence of [beautiful 3] and [beautiful 5]. Fortunately we
have rules for both. *)
apply b_3.
apply b_5.
Qed.
(** As you would expect, we can also prove theorems that have
hypotheses about [beautiful]. *)
Theorem beautiful_plus_eight: forall n, beautiful n -> beautiful (8+n).
Proof.
intros n B.
apply b_sum with (n:=8) (m:=n).
apply eight_is_beautiful.
apply B.
Qed.
(** **** Exercise: 2 stars (b_timesm) *)
Theorem b_timesm: forall n m, beautiful n -> beautiful (m*n).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** ** Induction Over Evidence *)
(** Besides _constructing_ evidence that numbers are beautiful, we can
also _reason about_ such evidence. *)
(** The fact that we introduced [beautiful] with an [Inductive]
declaration tells Coq not only that the constructors [b_0], [b_3],
[b_5] and [b_sum] are ways to build evidence, but also that these
two constructors are the _only_ ways to build evidence that
numbers are beautiful. *)
(** In other words, if someone gives us evidence [E] for the assertion
[beautiful n], then we know that [E] must have one of four shapes:
- [E] is [b_0] (and [n] is [O]),
- [E] is [b_3] (and [n] is [3]),
- [E] is [b_5] (and [n] is [5]), or
- [E] is [b_sum n1 n2 E1 E2] (and [n] is [n1+n2], where [E1] is
evidence that [n1] is beautiful and [E2] is evidence that [n2]
is beautiful). *)
(** This permits us to _analyze_ any hypothesis of the form [beautiful
n] to see how it was constructed, using the tactics we already
know. In particular, we can use the [induction] tactic that we
have already seen for reasoning about inductively defined _data_
to reason about inductively defined _evidence_.
To illustrate this, let's define another property of numbers: *)
Inductive gorgeous : nat -> Prop :=
g_0 : gorgeous 0
| g_plus3 : forall n, gorgeous n -> gorgeous (3+n)
| g_plus5 : forall n, gorgeous n -> gorgeous (5+n).
(** **** Exercise: 1 star (gorgeous_tree) *)
(** Write out the definition of [gorgeous] numbers using inference rule
notation.
(* FILL IN HERE *)
[]
*)
(** **** Exercise: 1 star (gorgeous_plus13) *)
Theorem gorgeous_plus13: forall n,
gorgeous n -> gorgeous (13+n).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** It seems intuitively obvious that, although [gorgeous] and
[beautiful] are presented using slightly different rules, they are
actually the same property in the sense that they are true of the
same numbers. Indeed, we can prove this. *)
Theorem gorgeous__beautiful : forall n,
gorgeous n -> beautiful n.
Proof.
intros n H.
induction H as [|n'|n'].
Case "g_0".
apply b_0.
Case "g_plus3".
apply b_sum. apply b_3.
apply IHgorgeous.
Case "g_plus5".
apply b_sum. apply b_5. apply IHgorgeous.
Qed.
(** Notice that the argument proceeds by induction on the _evidence_ [H]! *)
(** Let's see what happens if we try to prove this by induction on [n]
instead of induction on the evidence [H]. *)
Theorem gorgeous__beautiful_FAILED : forall n,
gorgeous n -> beautiful n.
Proof.
intros. induction n as [| n'].
Case "n = 0". apply b_0.
Case "n = S n'". (* We are stuck! *)
Abort.
(** The problem here is that doing induction on [n] doesn't yield a
useful induction hypothesis. Knowing how the property we are
interested in behaves on the predecessor of [n] doesn't help us
prove that it holds for [n]. Instead, we would like to be able to
have induction hypotheses that mention other numbers, such as [n -
3] and [n - 5]. This is given precisely by the shape of the
constructors for [gorgeous]. *)
(** **** Exercise: 2 stars (gorgeous_sum) *)
Theorem gorgeous_sum : forall n m,
gorgeous n -> gorgeous m -> gorgeous (n + m).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, advanced (beautiful__gorgeous) *)
Theorem beautiful__gorgeous : forall n, beautiful n -> gorgeous n.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, optional (g_times2) *)
(** Prove the [g_times2] theorem below without using [gorgeous__beautiful].
You might find the following helper lemma useful. *)
Lemma helper_g_times2 : forall x y z, x + (z + y)= z + x + y.
Proof.
(* FILL IN HERE *) Admitted.
Theorem g_times2: forall n, gorgeous n -> gorgeous (2*n).
Proof.
intros n H. simpl.
induction H.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** ** From Boolean Functions to Propositions *)
(** In chapter [Basics] we defined a _function_ [evenb] that tests a
number for evenness, yielding [true] if so. We can use this
function to define the _proposition_ that some number [n] is
even: *)
Definition even (n:nat) : Prop :=
evenb n = true.
(** That is, we can define "[n] is even" to mean "the function [evenb]
returns [true] when applied to [n]."
Note that here we have given a name
to a proposition using a [Definition], just as we have
given names to expressions of other sorts. This isn't a fundamentally
new kind of proposition; it is still just an equality. *)
(** Another alternative is to define the concept of evenness
directly. Instead of going via the [evenb] function ("a number is
even if a certain computation yields [true]"), we can say what the
concept of evenness means by giving two different ways of
presenting _evidence_ that a number is even. *)
Inductive ev : nat -> Prop :=
| ev_0 : ev O
| ev_SS : forall n:nat, ev n -> ev (S (S n)).
(** This definition says that there are two ways to give
evidence that a number [m] is even. First, [0] is even, and
[ev_0] is evidence for this. Second, if [m = S (S n)] for some
[n] and we can give evidence [e] that [n] is even, then [m] is
also even, and [ev_SS n e] is the evidence. *)
(** **** Exercise: 1 star (double_even) *)
Theorem double_even : forall n,
ev (double n).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** *** Discussion: Computational vs. Inductive Definitions *)
(** We have seen that the proposition "[n] is even" can be
phrased in two different ways -- indirectly, via a boolean testing
function [evenb], or directly, by inductively describing what
constitutes evidence for evenness. These two ways of defining
evenness are about equally easy to state and work with. Which we
choose is basically a question of taste.
However, for many other properties of interest, the direct
inductive definition is preferable, since writing a testing
function may be awkward or even impossible.
One such property is [beautiful]. This is a perfectly sensible
definition of a set of numbers, but we cannot translate its
definition directly into a Coq Fixpoint (or into a recursive
function in any other common programming language). We might be
able to find a clever way of testing this property using a
[Fixpoint] (indeed, it is not too hard to find one in this case),
but in general this could require arbitrarily deep thinking. In
fact, if the property we are interested in is uncomputable, then
we cannot define it as a [Fixpoint] no matter how hard we try,
because Coq requires that all [Fixpoint]s correspond to
terminating computations.
On the other hand, writing an inductive definition of what it
means to give evidence for the property [beautiful] is
straightforward. *)
(** **** Exercise: 1 star (ev__even) *)
(** Here is a proof that the inductive definition of evenness implies
the computational one. *)
Theorem ev__even : forall n,
ev n -> even n.
Proof.
intros n E. induction E as [| n' E'].
Case "E = ev_0".
unfold even. reflexivity.
Case "E = ev_SS n' E'".
unfold even. apply IHE'.
Qed.
(** Could this proof also be carried out by induction on [n] instead
of [E]? If not, why not? *)
(* FILL IN HERE *)
(** [] *)
(** The induction principle for inductively defined propositions does
not follow quite the same form as that of inductively defined
sets. For now, you can take the intuitive view that induction on
evidence [ev n] is similar to induction on [n], but restricts our
attention to only those numbers for which evidence [ev n] could be
generated. We'll look at the induction principle of [ev] in more
depth below, to explain what's really going on. *)
(** **** Exercise: 1 star (l_fails) *)
(** The following proof attempt will not succeed.
Theorem l : forall n,
ev n.
Proof.
intros n. induction n.
Case "O". simpl. apply ev_0.
Case "S".
...
Intuitively, we expect the proof to fail because not every
number is even. However, what exactly causes the proof to fail?
(* FILL IN HERE *)
*)
(** [] *)
(** **** Exercise: 2 stars (ev_sum) *)
(** Here's another exercise requiring induction. *)
Theorem ev_sum : forall n m,
ev n -> ev m -> ev (n+m).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** ** [Inversion] on Evidence *)
(** Another situation where we want to analyze evidence for evenness
is when proving that, if [n] is even, then [pred (pred n))] is
too. In this case, we don't need to do an inductive proof. The
right tactic turns out to be [inversion]. *)
Theorem ev_minus2: forall n,
ev n -> ev (pred (pred n)).
Proof.
intros n E.
inversion E as [| n' E'].
Case "E = ev_0". simpl. apply ev_0.
Case "E = ev_SS n' E'". simpl. apply E'. Qed.
(** **** Exercise: 1 star, optional (ev_minus2_n) *)
(** What happens if we try to use [destruct] on [n] instead of [inversion] on [E]? *)
(* FILL IN HERE *)
(** [] *)
(** Another example, in which [inversion] helps narrow down to
the relevant cases. *)
Theorem SSev__even : forall n,
ev (S (S n)) -> ev n.
Proof.
intros n E.
inversion E as [| n' E'].
apply E'. Qed.
(** These uses of [inversion] may seem a bit mysterious at first.
Until now, we've only used [inversion] on equality
propositions, to utilize injectivity of constructors or to
discriminate between different constructors. But we see here
that [inversion] can also be applied to analyzing evidence
for inductively defined propositions.
(You might also expect that [destruct] would be a more suitable
tactic to use here. Indeed, it is possible to use [destruct], but
it often throws away useful information, and the [eqn:] qualifier
doesn't help much in this case.)
Here's how [inversion] works in general. Suppose the name
[I] refers to an assumption [P] in the current context, where
[P] has been defined by an [Inductive] declaration. Then,
for each of the constructors of [P], [inversion I] generates
a subgoal in which [I] has been replaced by the exact,
specific conditions under which this constructor could have
been used to prove [P]. Some of these subgoals will be
self-contradictory; [inversion] throws these away. The ones
that are left represent the cases that must be proved to
establish the original goal.
In this particular case, the [inversion] analyzed the construction
[ev (S (S n))], determined that this could only have been
constructed using [ev_SS], and generated a new subgoal with the
arguments of that constructor as new hypotheses. (It also
produced an auxiliary equality, which happens to be useless here.)
We'll begin exploring this more general behavior of inversion in
what follows. *)
(** **** Exercise: 1 star (inversion_practice) *)
Theorem SSSSev__even : forall n,
ev (S (S (S (S n)))) -> ev n.
Proof.
(* FILL IN HERE *) Admitted.
(** The [inversion] tactic can also be used to derive goals by showing
the absurdity of a hypothesis. *)
Theorem even5_nonsense :
ev 5 -> 2 + 2 = 9.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, advanced (ev_ev__ev) *)
(** Finding the appropriate thing to do induction on is a
bit tricky here: *)
Theorem ev_ev__ev : forall n m,
ev (n+m) -> ev n -> ev m.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, optional (ev_plus_plus) *)
(** Here's an exercise that just requires applying existing lemmas. No
induction or even case analysis is needed, but some of the rewriting
may be tedious. *)
Theorem ev_plus_plus : forall n m p,
ev (n+m) -> ev (n+p) -> ev (m+p).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** * Additional Exercises *)
(** **** Exercise: 4 stars (palindromes) *)
(** A palindrome is a sequence that reads the same backwards as
forwards.
- Define an inductive proposition [pal] on [list X] that
captures what it means to be a palindrome. (Hint: You'll need
three cases. Your definition should be based on the structure
of the list; just having a single constructor
c : forall l, l = rev l -> pal l
may seem obvious, but will not work very well.)
- Prove that
forall l, pal (l ++ rev l).
- Prove that
forall l, pal l -> l = rev l.
*)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 5 stars, optional (palindrome_converse) *)
(** Using your definition of [pal] from the previous exercise, prove
that
forall l, l = rev l -> pal l.
*)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 4 stars, advanced (subsequence) *)
(** A list is a _subsequence_ of another list if all of the elements
in the first list occur in the same order in the second list,
possibly with some extra elements in between. For example,
[1,2,3]
is a subsequence of each of the lists
[1,2,3]
[1,1,1,2,2,3]
[1,2,7,3]
[5,6,1,9,9,2,7,3,8]
but it is _not_ a subsequence of any of the lists
[1,2]
[1,3]
[5,6,2,1,7,3,8]
- Define an inductive proposition [subseq] on [list nat] that
captures what it means to be a subsequence. (Hint: You'll need
three cases.)
- Prove that subsequence is reflexive, that is, any list is a
subsequence of itself.
- Prove that for any lists [l1], [l2], and [l3], if [l1] is a
subsequence of [l2], then [l1] is also a subsequence of [l2 ++
l3].
- (Optional, harder) Prove that subsequence is transitive -- that
is, if [l1] is a subsequence of [l2] and [l2] is a subsequence
of [l3], then [l1] is a subsequence of [l3]. Hint: choose your
induction carefully!
*)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 2 stars, optional (R_provability) *)
(** Suppose we give Coq the following definition:
Inductive R : nat -> list nat -> Prop :=
| c1 : R 0 []
| c2 : forall n l, R n l -> R (S n) (n :: l)
| c3 : forall n l, R (S n) l -> R n l.
Which of the following propositions are provable?
- [R 2 [1,0]]
- [R 1 [1,2,1,0]]
- [R 6 [3,2,1,0]]
*)
(** [] *)
(* $Date: 2013-07-01 18:48:47 -0400 (Mon, 01 Jul 2013) $ *)