(** * Tactics: More Basic Tactics *)
(** This chapter introduces several additional proof strategies
and tactics that allow us to begin proving more interesting
properties of functional programs. We will see:
- how to use auxiliary lemmas in both "forward-style" and
"backward-style" proofs;
- how to reason about data constructors (in particular, how to use
the fact that they are injective and disjoint);
- how to strengthen an induction hypothesis (and when such
strengthening is required); and
- more details on how to reason by case analysis. *)
Set Warnings "-notation-overridden,-parsing".
From LF Require Export Poly.
(* ################################################################# *)
(** * The [apply] Tactic *)
(** We often encounter situations where the goal to be proved is
_exactly_ the same as some hypothesis in the context or some
previously proved lemma. *)
Theorem silly1 : forall (n m o p : nat),
n = m ->
[n;o] = [n;p] ->
[n;o] = [m;p].
Proof.
intros n m o p eq1 eq2.
rewrite <- eq1.
(** Here, we could finish with "[rewrite -> eq2. reflexivity.]" as we
have done several times before. We can finish this proof in
a single step by using the [apply] tactic instead: *)
apply eq2. Qed.
(** The [apply] tactic also works with _conditional_ hypotheses
and lemmas: if the statement being applied is an implication, then
the premises of this implication will be added to the list of
subgoals needing to be proved. *)
Theorem silly2 : forall (n m o p : nat),
n = m ->
(n = m -> [n;o] = [m;p]) ->
[n;o] = [m;p].
Proof.
intros n m o p eq1 eq2.
apply eq2. apply eq1. Qed.
(** Typically, when we use [apply H], the statement [H] will
begin with a [forall] that binds some _universal variables_. When
Coq matches the current goal against the conclusion of [H], it
will try to find appropriate values for these variables. For
example, when we do [apply eq2] in the following proof, the
universal variable [q] in [eq2] gets instantiated with [n] and [r]
gets instantiated with [m]. *)
Theorem silly2a : forall (n m : nat),
(n,n) = (m,m) ->
(forall (q r : nat), (q,q) = (r,r) -> [q] = [r]) ->
[n] = [m].
Proof.
intros n m eq1 eq2.
apply eq2. apply eq1. Qed.
(** **** Exercise: 2 stars, standard, optional (silly_ex)
Complete the following proof using only [intros] and [apply]. *)
Theorem silly_ex :
(forall n, evenb n = true -> oddb (S n) = true) ->
evenb 4 = true ->
oddb 3 = true.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** To use the [apply] tactic, the (conclusion of the) fact
being applied must match the goal exactly -- for example, [apply]
will not work if the left and right sides of the equality are
swapped. *)
Theorem silly3_firsttry : forall (n : nat),
true = (n =? 5) ->
(S (S n)) =? 7 = true.
Proof.
intros n H.
(** Here we cannot use [apply] directly, but we can use the [symmetry]
tactic, which switches the left and right sides of an equality in
the goal. *)
symmetry.
simpl. (** (This [simpl] is optional, since [apply] will perform
simplification first, if needed.) *)
apply H. Qed.
(** **** Exercise: 3 stars, standard (apply_exercise1)
(_Hint_: You can use [apply] with previously defined lemmas, not
just hypotheses in the context. Remember that [Search] is
your friend.) *)
Theorem rev_exercise1 : forall (l l' : list nat),
l = rev l' ->
l' = rev l.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, standard, optional (apply_rewrite)
Briefly explain the difference between the tactics [apply] and
[rewrite]. What are the situations where both can usefully be
applied? *)
(* FILL IN HERE
[] *)
(* ################################################################# *)
(** * The [apply with] Tactic *)
(** The following silly example uses two rewrites in a row to
get from [[a;b]] to [[e;f]]. *)
Example trans_eq_example : forall (a b c d e f : nat),
[a;b] = [c;d] ->
[c;d] = [e;f] ->
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
rewrite -> eq1. rewrite -> eq2. reflexivity. Qed.
(** Since this is a common pattern, we might like to pull it out
as a lemma recording, once and for all, the fact that equality is
transitive. *)
Theorem trans_eq : forall (X:Type) (n m o : X),
n = m -> m = o -> n = o.
Proof.
intros X n m o eq1 eq2. rewrite -> eq1. rewrite -> eq2.
reflexivity. Qed.
(** Now, we should be able to use [trans_eq] to prove the above
example. However, to do this we need a slight refinement of the
[apply] tactic. *)
Example trans_eq_example' : forall (a b c d e f : nat),
[a;b] = [c;d] ->
[c;d] = [e;f] ->
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
(** If we simply tell Coq [apply trans_eq] at this point, it can
tell (by matching the goal against the conclusion of the lemma)
that it should instantiate [X] with [[nat]], [n] with [[a,b]], and
[o] with [[e,f]]. However, the matching process doesn't determine
an instantiation for [m]: we have to supply one explicitly by
adding [with (m:=[c,d])] to the invocation of [apply]. *)
apply trans_eq with (m:=[c;d]).
apply eq1. apply eq2. Qed.
(** Actually, we usually don't have to include the name [m] in
the [with] clause; Coq is often smart enough to figure out which
instantiation we're giving. We could instead write: [apply
trans_eq with [c;d]]. *)
(** **** Exercise: 3 stars, standard, optional (apply_with_exercise) *)
Example trans_eq_exercise : forall (n m o p : nat),
m = (minustwo o) ->
(n + p) = m ->
(n + p) = (minustwo o).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################################# *)
(** * The [injection] and [discriminate] Tactics *)
(** Recall the definition of natural numbers:
Inductive nat : Type :=
| O
| S (n : nat).
It is obvious from this definition that every number has one of
two forms: either it is the constructor [O] or it is built by
applying the constructor [S] to another number. But there is more
here than meets the eye: implicit in the definition (and in our
informal understanding of how datatype declarations work in other
programming languages) are two more facts:
- The constructor [S] is _injective_ aka _one-to-one_. That is,
if [S n = S m], it must be the case that [n = m].
- The constructors [O] and [S] are _disjoint_. That is, [O] is not
equal to [S n] for any [n].
Similar principles apply to all inductively defined types: all
constructors are injective, and the values built from distinct
constructors are never equal. For lists, the [cons] constructor
is injective and [nil] is different from every non-empty list.
For booleans, [true] and [false] are different. (Since neither
[true] nor [false] take any arguments, their injectivity is not
interesting.) And so on. *)
(** For example, we can prove the injectivity of [S] by using the
[pred] function defined in [Basics.v]. *)
Theorem S_injective : forall (n m : nat),
S n = S m ->
n = m.
Proof.
intros n m H1.
assert (H2: n = pred (S n)). { reflexivity. }
rewrite H2. rewrite H1. reflexivity.
Qed.
(** This technique can be generalized to any constructor by
writing the equivalent of [pred] for that constructor -- i.e.,
writing a function that "undoes" one application of the
constructor. As a more convenient alternative, Coq provides a
tactic called [injection] that allows us to exploit the
injectivity of any constructor. Here is an alternate proof of the
above theorem using [injection]: *)
Theorem S_injective' : forall (n m : nat),
S n = S m ->
n = m.
Proof.
intros n m H.
(** By writing [injection H] at this point, we are asking Coq to
generate all equations that it can infer from [H] using the
injectivity of constructors. Each such equation is added as a
premise to the goal. In the present example, adds the premise
[n = m]. *)
injection H. intros Hnm. apply Hnm.
Qed.
(** Here's a more interesting example that shows how [injection] can
derive multiple equations at once. *)
Theorem injection_ex1 : forall (n m o : nat),
[n; m] = [o; o] ->
[n] = [m].
Proof.
intros n m o H.
injection H. intros H1 H2.
rewrite H1. rewrite H2. reflexivity.
Qed.
(** The "[as]" variant of [injection] permits us to choose names for
the equations and immediately introduce them as hypotheses, rather
than as premises. The original equality, in this case [H], is
also removed from the hypotheses with this variant. *)
Theorem injection_ex2 : forall (n m : nat),
[n] = [m] ->
n = m.
Proof.
intros n m H.
injection H as Hnm. rewrite Hnm.
reflexivity. Qed.
(** **** Exercise: 3 stars, standard (injection_ex3) *)
Example injection_ex3 : forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = z :: j ->
j = z :: l ->
x = y.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** So much for injectivity of constructors. What about disjointness?
The principle of disjointness says that two terms beginning with
different constructors (like [O] and [S], or [true] and [false])
can never be equal. This means that, any time we find ourselves
working in a context where we've _assumed_ that two such terms are
equal, we are justified in concluding anything we want to (because
the assumption is nonsensical).
The [discriminate] tactic embodies this principle: It is used on a
hypothesis involving an equality between different
constructors (e.g., [S n = O]), and it solves the current goal
immediately. For example: *)
Theorem eqb_0_l : forall n,
0 =? n = true -> n = 0.
Proof.
intros n.
(** We can proceed by case analysis on [n]. The first case is
trivial. *)
destruct n as [| n'] eqn:E.
- (* n = 0 *)
intros H. reflexivity.
(** However, the second one doesn't look so simple: assuming [0
=? (S n') = true], we must show [S n' = 0]! The way forward is to
observe that the assumption itself is nonsensical: *)
- (* n = S n' *)
simpl.
(** If we use [discriminate] on this hypothesis, Coq confirms
that the subgoal we are working on is impossible and removes it
from further consideration. *)
intros H. discriminate H.
Qed.
(** This is an instance of a logical principle known as the _principle
of explosion_, which asserts that a contradictory hypothesis
entails anything, even false things! *)
Theorem discriminate_ex1 : forall (n : nat),
S n = O ->
2 + 2 = 5.
Proof.
intros n contra. discriminate contra. Qed.
Theorem discriminate_ex2 : forall (n m : nat),
false = true ->
[n] = [m].
Proof.
intros n m contra. discriminate contra. Qed.
(** If you find the principle of explosion confusing, remember
that these proofs are _not_ showing that the conclusion of the
statement holds. Rather, they are showing that, if the
nonsensical situation described by the premise did somehow arise,
then the nonsensical conclusion would follow. We'll explore the
principle of explosion in more detail in the next chapter. *)
(** **** Exercise: 1 star, standard (discriminate_ex3) *)
Example discriminate_ex3 :
forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = [] ->
x = z.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** The injectivity of constructors allows us to reason that
[forall (n m : nat), S n = S m -> n = m]. The converse of this
implication is an instance of a more general fact about both
constructors and functions, which we will find convenient in a few
places below: *)
Theorem f_equal : forall (A B : Type) (f: A -> B) (x y: A),
x = y -> f x = f y.
Proof. intros A B f x y eq. rewrite eq. reflexivity. Qed.
Theorem eq_implies_succ_equal : forall (n m : nat),
n = m -> S n = S m.
Proof. intros n m H. apply f_equal. apply H. Qed.
(** There is also a tactic named `f_equal` that can prove such
theorems. Given a goal [f a1 ... an = g b1 ... bn], tactic [f_equal]
will produce subgoals [f = g], [a1 = b1], ..., [an = bn]. Any of those
subgoals that are simple enough (e.g., immediately provable by
[reflexivity] will be automatically discharged by [f_equal]. *)
Theorem eq_implies_succ_equal' : forall (n m : nat),
n = m -> S n = S m.
Proof. intros n m H. f_equal. apply H. Qed.
(* ################################################################# *)
(** * Using Tactics on Hypotheses *)
(** By default, most tactics work on the goal formula and leave
the context unchanged. However, most tactics also have a variant
that performs a similar operation on a statement in the context.
For example, the tactic [simpl in H] performs simplification in
the hypothesis named [H] in the context. *)
Theorem S_inj : forall (n m : nat) (b : bool),
(S n) =? (S m) = b ->
n =? m = b.
Proof.
intros n m b H. simpl in H. apply H. Qed.
(** Similarly, [apply L in H] matches some conditional statement
[L] (of the form [X -> Y], say) against a hypothesis [H] in the
context. However, unlike ordinary [apply] (which rewrites a goal
matching [Y] into a subgoal [X]), [apply L in H] matches [H]
against [X] and, if successful, replaces it with [Y].
In other words, [apply L in H] gives us a form of "forward
reasoning": from [X -> Y] and a hypothesis matching [X], it
produces a hypothesis matching [X]. By contrast, [apply L] is
"backward reasoning": it says that if we know [X -> Y] and we
are trying to prove [Y], it suffices to prove [X].
Here is a variant of a proof from above, using forward reasoning
throughout instead of backward reasoning. *)
Theorem silly3' : forall (n : nat),
(n =? 5 = true -> (S (S n)) =? 7 = true) ->
true = (n =? 5) ->
true = ((S (S n)) =? 7).
Proof.
intros n eq H.
symmetry in H. apply eq in H. symmetry in H.
apply H. Qed.
(** Forward reasoning starts from what is _given_ (premises,
previously proven theorems) and iteratively draws conclusions from
them until the goal is reached. Backward reasoning starts from
the _goal_, and iteratively reasons about what would imply the
goal, until premises or previously proven theorems are reached.
If you've seen informal proofs before (for example, in a math or
computer science class), they probably used forward reasoning. In
general, idiomatic use of Coq tends to favor backward reasoning,
but in some situations the forward style can be easier to think
about. *)
(* ################################################################# *)
(** * Varying the Induction Hypothesis *)
(** Sometimes it is important to control the exact form of the
induction hypothesis when carrying out inductive proofs in Coq.
In particular, we need to be careful about which of the
assumptions we move (using [intros]) from the goal to the context
before invoking the [induction] tactic. For example, suppose
we want to show that [double] is injective -- i.e., that it maps
different arguments to different results:
Theorem double_injective: forall n m,
double n = double m -> n = m.
The way we _start_ this proof is a bit delicate: if we begin with
intros n. induction n.
all is well. But if we begin it with
intros n m. induction n.
we get stuck in the middle of the inductive case... *)
Theorem double_injective_FAILED : forall n m,
double n = double m ->
n = m.
Proof.
intros n m. induction n as [| n' IHn'].
- (* n = O *) simpl. intros eq. destruct m as [| m'] eqn:E.
+ (* m = O *) reflexivity.
+ (* m = S m' *) discriminate eq.
- (* n = S n' *) intros eq. destruct m as [| m'] eqn:E.
+ (* m = O *) discriminate eq.
+ (* m = S m' *) apply f_equal.
(** At this point, the induction hypothesis, [IHn'], does _not_ give us
[n' = m'] -- there is an extra [S] in the way -- so the goal is
not provable. *)
Abort.
(** What went wrong? *)
(** The problem is that, at the point we invoke the induction
hypothesis, we have already introduced [m] into the context --
intuitively, we have told Coq, "Let's consider some particular [n]
and [m]..." and we now have to prove that, if [double n = double
m] for _these particular_ [n] and [m], then [n = m].
The next tactic, [induction n] says to Coq: We are going to show
the goal by induction on [n]. That is, we are going to prove, for
_all_ [n], that the proposition
- [P n] = "if [double n = double m], then [n = m]"
holds, by showing
- [P O]
(i.e., "if [double O = double m] then [O = m]") and
- [P n -> P (S n)]
(i.e., "if [double n = double m] then [n = m]" implies "if
[double (S n) = double m] then [S n = m]").
If we look closely at the second statement, it is saying something
rather strange: it says that, for a _particular_ [m], if we know
- "if [double n = double m] then [n = m]"
then we can prove
- "if [double (S n) = double m] then [S n = m]".
To see why this is strange, let's think of a particular [m] --
say, [5]. The statement is then saying that, if we know
- [Q] = "if [double n = 10] then [n = 5]"
then we can prove
- [R] = "if [double (S n) = 10] then [S n = 5]".
But knowing [Q] doesn't give us any help at all with proving
[R]! (If we tried to prove [R] from [Q], we would start with
something like "Suppose [double (S n) = 10]..." but then we'd be
stuck: knowing that [double (S n)] is [10] tells us nothing about
whether [double n] is [10], so [Q] is useless.) *)
(** Trying to carry out this proof by induction on [n] when [m] is
already in the context doesn't work because we are then trying to
prove a statement involving _every_ [n] but just a _single_ [m]. *)
(** The successful proof of [double_injective] leaves [m] in the goal
statement at the point where the [induction] tactic is invoked on
[n]: *)
Theorem double_injective : forall n m,
double n = double m ->
n = m.
Proof.
intros n. induction n as [| n' IHn'].
- (* n = O *) simpl. intros m eq. destruct m as [| m'] eqn:E.
+ (* m = O *) reflexivity.
+ (* m = S m' *) discriminate eq.
- (* n = S n' *) simpl.
(** Notice that both the goal and the induction hypothesis are
different this time: the goal asks us to prove something more
general (i.e., to prove the statement for _every_ [m]), but the IH
is correspondingly more flexible, allowing us to choose any [m] we
like when we apply the IH. *)
intros m eq.
(** Now we've chosen a particular [m] and introduced the assumption
that [double n = double m]. Since we are doing a case analysis on
[n], we also need a case analysis on [m] to keep the two "in sync." *)
destruct m as [| m'] eqn:E.
+ (* m = O *) simpl.
(** The 0 case is trivial: *)
discriminate eq.
+ (* m = S m' *)
apply f_equal.
(** At this point, since we are in the second branch of the [destruct
m], the [m'] mentioned in the context is the predecessor of the
[m] we started out talking about. Since we are also in the [S]
branch of the induction, this is perfect: if we instantiate the
generic [m] in the IH with the current [m'] (this instantiation is
performed automatically by the [apply] in the next step), then
[IHn'] gives us exactly what we need to finish the proof. *)
apply IHn'. injection eq as goal. apply goal. Qed.
(** What you should take away from all this is that we need to be
careful, when using induction, that we are not trying to prove
something too specific: To prove a property of [n] and [m] by
induction on [n], it is sometimes important to leave [m]
generic. *)
(** The following exercise requires the same pattern. *)
(** **** Exercise: 2 stars, standard (eqb_true) *)
Theorem eqb_true : forall n m,
n =? m = true -> n = m.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, advanced (eqb_true_informal)
Give a careful informal proof of [eqb_true], being as explicit
as possible about quantifiers. *)
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_informal_proof : option (nat*string) := None.
(** [] *)
(** **** Exercise: 3 stars, standard, recommended (plus_n_n_injective)
In addition to being careful about how you use [intros], practice using "in"
variants in this proof. (Hint: use [plus_n_Sm].) *)
Theorem plus_n_n_injective : forall n m,
n + n = m + m ->
n = m.
Proof.
intros n. induction n as [| n'].
(* FILL IN HERE *) Admitted.
(** [] *)
(** The strategy of doing fewer [intros] before an [induction] to
obtain a more general IH doesn't always work by itself; sometimes
some _rearrangement_ of quantified variables is needed. Suppose,
for example, that we wanted to prove [double_injective] by
induction on [m] instead of [n]. *)
Theorem double_injective_take2_FAILED : forall n m,
double n = double m ->
n = m.
Proof.
intros n m. induction m as [| m' IHm'].
- (* m = O *) simpl. intros eq. destruct n as [| n'] eqn:E.
+ (* n = O *) reflexivity.
+ (* n = S n' *) discriminate eq.
- (* m = S m' *) intros eq. destruct n as [| n'] eqn:E.
+ (* n = O *) discriminate eq.
+ (* n = S n' *) apply f_equal.
(* Stuck again here, just like before. *)
Abort.
(** The problem is that, to do induction on [m], we must first
introduce [n]. (If we simply say [induction m] without
introducing anything first, Coq will automatically introduce [n]
for us!) *)
(** What can we do about this? One possibility is to rewrite the
statement of the lemma so that [m] is quantified before [n]. This
works, but it's not nice: We don't want to have to twist the
statements of lemmas to fit the needs of a particular strategy for
proving them! Rather we want to state them in the clearest and
most natural way. *)
(** What we can do instead is to first introduce all the quantified
variables and then _re-generalize_ one or more of them,
selectively taking variables out of the context and putting them
back at the beginning of the goal. The [generalize dependent]
tactic does this. *)
Theorem double_injective_take2 : forall n m,
double n = double m ->
n = m.
Proof.
intros n m.
(* [n] and [m] are both in the context *)
generalize dependent n.
(* Now [n] is back in the goal and we can do induction on
[m] and get a sufficiently general IH. *)
induction m as [| m' IHm'].
- (* m = O *) simpl. intros n eq. destruct n as [| n'] eqn:E.
+ (* n = O *) reflexivity.
+ (* n = S n' *) discriminate eq.
- (* m = S m' *) intros n eq. destruct n as [| n'] eqn:E.
+ (* n = O *) discriminate eq.
+ (* n = S n' *) apply f_equal.
apply IHm'. injection eq as goal. apply goal. Qed.
(** Let's look at an informal proof of this theorem. Note that
the proposition we prove by induction leaves [n] quantified,
corresponding to the use of generalize dependent in our formal
proof.
_Theorem_: For any nats [n] and [m], if [double n = double m], then
[n = m].
_Proof_: Let [m] be a [nat]. We prove by induction on [m] that, for
any [n], if [double n = double m] then [n = m].
- First, suppose [m = 0], and suppose [n] is a number such
that [double n = double m]. We must show that [n = 0].
Since [m = 0], by the definition of [double] we have [double n =
0]. There are two cases to consider for [n]. If [n = 0] we are
done, since [m = 0 = n], as required. Otherwise, if [n = S n']
for some [n'], we derive a contradiction: by the definition of
[double], we can calculate [double n = S (S (double n'))], but
this contradicts the assumption that [double n = 0].
- Second, suppose [m = S m'] and that [n] is again a number such
that [double n = double m]. We must show that [n = S m'], with
the induction hypothesis that for every number [s], if [double s =
double m'] then [s = m'].
By the fact that [m = S m'] and the definition of [double], we
have [double n = S (S (double m'))]. There are two cases to
consider for [n].
If [n = 0], then by definition [double n = 0], a contradiction.
Thus, we may assume that [n = S n'] for some [n'], and again by
the definition of [double] we have [S (S (double n')) =
S (S (double m'))], which implies by injectivity that [double n' =
double m']. Instantiating the induction hypothesis with [n'] thus
allows us to conclude that [n' = m'], and it follows immediately
that [S n' = S m']. Since [S n' = n] and [S m' = m], this is just
what we wanted to show. [] *)
(** Before we close this section and move on to some exercises,
let's digress briefly and use [eqb_true] to prove a similar
property of identifiers that we'll need in later chapters: *)
Theorem eqb_id_true : forall x y,
eqb_id x y = true -> x = y.
Proof.
intros [m] [n]. simpl. intros H.
assert (H' : m = n). { apply eqb_true. apply H. }
rewrite H'. reflexivity.
Qed.
(** **** Exercise: 3 stars, standard, recommended (gen_dep_practice)
Prove this by induction on [l]. *)
Theorem nth_error_after_last: forall (n : nat) (X : Type) (l : list X),
length l = n ->
nth_error l n = None.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################################# *)
(** * Unfolding Definitions *)
(** It sometimes happens that we need to manually unfold a name that
has been introduced by a [Definition] so that we can manipulate
its right-hand side. For example, if we define... *)
Definition square n := n * n.
(** ... and try to prove a simple fact about [square]... *)
Lemma square_mult : forall n m, square (n * m) = square n * square m.
Proof.
intros n m.
simpl.
(** ... we appear to be stuck: [simpl] doesn't simplify anything at
this point, and since we haven't proved any other facts about
[square], there is nothing we can [apply] or [rewrite] with.
To make progress, we can manually [unfold] the definition of
[square]: *)
unfold square.
(** Now we have plenty to work with: both sides of the equality are
expressions involving multiplication, and we have lots of facts
about multiplication at our disposal. In particular, we know that
it is commutative and associative, and from these it is not hard
to finish the proof. *)
rewrite mult_assoc.
assert (H : n * m * n = n * n * m).
{ rewrite mult_comm. apply mult_assoc. }
rewrite H. rewrite mult_assoc. reflexivity.
Qed.
(** At this point, some discussion of unfolding and simplification is
in order.
You may already have observed that tactics like [simpl],
[reflexivity], and [apply] will often unfold the definitions of
functions automatically when this allows them to make progress.
For example, if we define [foo m] to be the constant [5]... *)
Definition foo (x: nat) := 5.
(** .... then the [simpl] in the following proof (or the
[reflexivity], if we omit the [simpl]) will unfold [foo m] to
[(fun x => 5) m] and then further simplify this expression to just
[5]. *)
Fact silly_fact_1 : forall m, foo m + 1 = foo (m + 1) + 1.
Proof.
intros m.
simpl.
reflexivity.
Qed.
(** However, this automatic unfolding is somewhat conservative. For
example, if we define a slightly more complicated function
involving a pattern match... *)
Definition bar x :=
match x with
| O => 5
| S _ => 5
end.
(** ...then the analogous proof will get stuck: *)
Fact silly_fact_2_FAILED : forall m, bar m + 1 = bar (m + 1) + 1.
Proof.
intros m.
simpl. (* Does nothing! *)
Abort.
(** The reason that [simpl] doesn't make progress here is that it
notices that, after tentatively unfolding [bar m], it is left with
a match whose scrutinee, [m], is a variable, so the [match] cannot
be simplified further. It is not smart enough to notice that the
two branches of the [match] are identical, so it gives up on
unfolding [bar m] and leaves it alone. Similarly, tentatively
unfolding [bar (m+1)] leaves a [match] whose scrutinee is a
function application (that cannot itself be simplified, even
after unfolding the definition of [+]), so [simpl] leaves it
alone. *)
(** At this point, there are two ways to make progress. One is to use
[destruct m] to break the proof into two cases, each focusing on a
more concrete choice of [m] ([O] vs [S _]). In each case, the
[match] inside of [bar] can now make progress, and the proof is
easy to complete. *)
Fact silly_fact_2 : forall m, bar m + 1 = bar (m + 1) + 1.
Proof.
intros m.
destruct m eqn:E.
- simpl. reflexivity.
- simpl. reflexivity.
Qed.
(** This approach works, but it depends on our recognizing that the
[match] hidden inside [bar] is what was preventing us from making
progress. *)
(** A more straightforward way to make progress is to explicitly tell
Coq to unfold [bar]. *)
Fact silly_fact_2' : forall m, bar m + 1 = bar (m + 1) + 1.
Proof.
intros m.
unfold bar.
(** Now it is apparent that we are stuck on the [match] expressions on
both sides of the [=], and we can use [destruct] to finish the
proof without thinking too hard. *)
destruct m eqn:E.
- reflexivity.
- reflexivity.
Qed.
(* ################################################################# *)
(** * Using [destruct] on Compound Expressions *)
(** We have seen many examples where [destruct] is used to
perform case analysis of the value of some variable. But
sometimes we need to reason by cases on the result of some
_expression_. We can also do this with [destruct].
Here are some examples: *)
Definition sillyfun (n : nat) : bool :=
if n =? 3 then false
else if n =? 5 then false
else false.
Theorem sillyfun_false : forall (n : nat),
sillyfun n = false.
Proof.
intros n. unfold sillyfun.
destruct (n =? 3) eqn:E1.
- (* n =? 3 = true *) reflexivity.
- (* n =? 3 = false *) destruct (n =? 5) eqn:E2.
+ (* n =? 5 = true *) reflexivity.
+ (* n =? 5 = false *) reflexivity. Qed.
(** After unfolding [sillyfun] in the above proof, we find that
we are stuck on [if (n =? 3) then ... else ...]. But either
[n] is equal to [3] or it isn't, so we can use [destruct (eqb
n 3)] to let us reason about the two cases.
In general, the [destruct] tactic can be used to perform case
analysis of the results of arbitrary computations. If [e] is an
expression whose type is some inductively defined type [T], then,
for each constructor [c] of [T], [destruct e] generates a subgoal
in which all occurrences of [e] (in the goal and in the context)
are replaced by [c]. *)
(** **** Exercise: 3 stars, standard, optional (combine_split)
Here is an implementation of the [split] function mentioned in
chapter [Poly]: *)
Fixpoint split {X Y : Type} (l : list (X*Y))
: (list X) * (list Y) :=
match l with
| [] => ([], [])
| (x, y) :: t =>
match split t with
| (lx, ly) => (x :: lx, y :: ly)
end
end.
(** Prove that [split] and [combine] are inverses in the following
sense: *)
Theorem combine_split : forall X Y (l : list (X * Y)) l1 l2,
split l = (l1, l2) ->
combine l1 l2 = l.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** The [eqn:] part of the [destruct] tactic is optional: We've chosen
to include it most of the time, just for the sake of
documentation, but many Coq proofs omit it.
When [destruct]ing compound expressions, however, the information
recorded by the [eqn:] can actually be critical: if we leave it
out, then [destruct] can sometimes erase information we need to
complete a proof.
For example, suppose we define a function [sillyfun1] like
this: *)
Definition sillyfun1 (n : nat) : bool :=
if n =? 3 then true
else if n =? 5 then true
else false.
(** Now suppose that we want to convince Coq of the (rather
obvious) fact that [sillyfun1 n] yields [true] only when [n] is
odd. If we start the proof like this (with no [eqn:] on the
destruct)... *)
Theorem sillyfun1_odd_FAILED : forall (n : nat),
sillyfun1 n = true ->
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (n =? 3).
(* stuck... *)
Abort.
(** ... then we are stuck at this point because the context does
not contain enough information to prove the goal! The problem is
that the substitution performed by [destruct] is quite brutal --
in this case, it thows away every occurrence of [n =? 3], but we
need to keep some memory of this expression and how it was
destructed, because we need to be able to reason that, since [n =?
3 = true] in this branch of the case analysis, it must be that [n
= 3], from which it follows that [n] is odd.
What we want here is to substitute away all existing occurences of
[n =? 3], but at the same time add an equation to the context that
records which case we are in. This is precisely what the [eqn:]
qualifier does. *)
Theorem sillyfun1_odd : forall (n : nat),
sillyfun1 n = true ->
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (n =? 3) eqn:Heqe3.
(* Now we have the same state as at the point where we got
stuck above, except that the context contains an extra
equality assumption, which is exactly what we need to
make progress. *)
- (* e3 = true *) apply eqb_true in Heqe3.
rewrite -> Heqe3. reflexivity.
- (* e3 = false *)
(* When we come to the second equality test in the body
of the function we are reasoning about, we can use
[eqn:] again in the same way, allowing us to finish the
proof. *)
destruct (n =? 5) eqn:Heqe5.
+ (* e5 = true *)
apply eqb_true in Heqe5.
rewrite -> Heqe5. reflexivity.
+ (* e5 = false *) discriminate eq. Qed.
(** **** Exercise: 2 stars, standard (destruct_eqn_practice) *)
Theorem bool_fn_applied_thrice :
forall (f : bool -> bool) (b : bool),
f (f (f b)) = f b.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################################# *)
(** * Review *)
(** We've now seen many of Coq's most fundamental tactics. We'll
introduce a few more in the coming chapters, and later on we'll
see some more powerful _automation_ tactics that make Coq help us
with low-level details. But basically we've got what we need to
get work done.
Here are the ones we've seen:
- [intros]: move hypotheses/variables from goal to context
- [reflexivity]: finish the proof (when the goal looks like [e =
e])
- [apply]: prove goal using a hypothesis, lemma, or constructor
- [apply... in H]: apply a hypothesis, lemma, or constructor to
a hypothesis in the context (forward reasoning)
- [apply... with...]: explicitly specify values for variables
that cannot be determined by pattern matching
- [simpl]: simplify computations in the goal
- [simpl in H]: ... or a hypothesis
- [rewrite]: use an equality hypothesis (or lemma) to rewrite
the goal
- [rewrite ... in H]: ... or a hypothesis
- [symmetry]: changes a goal of the form [t=u] into [u=t]
- [symmetry in H]: changes a hypothesis of the form [t=u] into
[u=t]
- [unfold]: replace a defined constant by its right-hand side in
the goal
- [unfold... in H]: ... or a hypothesis
- [destruct... as...]: case analysis on values of inductively
defined types
- [destruct... eqn:...]: specify the name of an equation to be
added to the context, recording the result of the case
analysis
- [induction... as...]: induction on values of inductively
defined types
- [injection]: reason by injectivity on equalities
between values of inductively defined types
- [discriminate]: reason by disjointness of constructors on
equalities between values of inductively defined types
- [assert (H: e)] (or [assert (e) as H]): introduce a "local
lemma" [e] and call it [H]
- [generalize dependent x]: move the variable [x] (and anything
else that depends on it) from the context back to an explicit
hypothesis in the goal formula *)
(* ################################################################# *)
(** * Additional Exercises *)
(** **** Exercise: 3 stars, standard (eqb_sym) *)
Theorem eqb_sym : forall (n m : nat),
(n =? m) = (m =? n).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, advanced, optional (eqb_sym_informal)
Give an informal proof of this lemma that corresponds to your
formal proof above:
Theorem: For any [nat]s [n] [m], [(n =? m) = (m =? n)].
Proof: *)
(* FILL IN HERE
[] *)
(** **** Exercise: 3 stars, standard, optional (eqb_trans) *)
Theorem eqb_trans : forall n m p,
n =? m = true ->
m =? p = true ->
n =? p = true.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, advanced (split_combine)
We proved, in an exercise above, that for all lists of pairs,
[combine] is the inverse of [split]. How would you formalize the
statement that [split] is the inverse of [combine]? When is this
property true?
Complete the definition of [split_combine_statement] below with a
property that states that [split] is the inverse of
[combine]. Then, prove that the property holds. (Be sure to leave
your induction hypothesis general by not doing [intros] on more
things than necessary. Hint: what property do you need of [l1]
and [l2] for [split (combine l1 l2) = (l1,l2)] to be true?) *)
Definition split_combine_statement : Prop
(* ("[: Prop]" means that we are giving a name to a
logical proposition here.) *)
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Theorem split_combine : split_combine_statement.
Proof.
(* FILL IN HERE *) Admitted.
(* Do not modify the following line: *)
Definition manual_grade_for_split_combine : option (nat*string) := None.
(** [] *)
(** **** Exercise: 3 stars, advanced (filter_exercise)
This one is a bit challenging. Pay attention to the form of your
induction hypothesis. *)
Theorem filter_exercise : forall (X : Type) (test : X -> bool)
(x : X) (l lf : list X),
filter test l = x :: lf ->
test x = true.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 4 stars, advanced, recommended (forall_exists_challenge)
Define two recursive [Fixpoints], [forallb] and [existsb]. The
first checks whether every element in a list satisfies a given
predicate:
forallb oddb [1;3;5;7;9] = true
forallb negb [false;false] = true
forallb evenb [0;2;4;5] = false
forallb (eqb 5) [] = true
The second checks whether there exists an element in the list that
satisfies a given predicate:
existsb (eqb 5) [0;2;3;6] = false
existsb (andb true) [true;true;false] = true
existsb oddb [1;0;0;0;0;3] = true
existsb evenb [] = false
Next, define a _nonrecursive_ version of [existsb] -- call it
[existsb'] -- using [forallb] and [negb].
Finally, prove a theorem [existsb_existsb'] stating that
[existsb'] and [existsb] have the same behavior.
*)
Fixpoint forallb {X : Type} (test : X -> bool) (l : list X) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_forallb_1 : forallb oddb [1;3;5;7;9] = true.
Proof. (* FILL IN HERE *) Admitted.
Example test_forallb_2 : forallb negb [false;false] = true.
Proof. (* FILL IN HERE *) Admitted.
Example test_forallb_3 : forallb evenb [0;2;4;5] = false.
Proof. (* FILL IN HERE *) Admitted.
Example test_forallb_4 : forallb (eqb 5) [] = true.
Proof. (* FILL IN HERE *) Admitted.
Fixpoint existsb {X : Type} (test : X -> bool) (l : list X) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_existsb_1 : existsb (eqb 5) [0;2;3;6] = false.
Proof. (* FILL IN HERE *) Admitted.
Example test_existsb_2 : existsb (andb true) [true;true;false] = true.
Proof. (* FILL IN HERE *) Admitted.
Example test_existsb_3 : existsb oddb [1;0;0;0;0;3] = true.
Proof. (* FILL IN HERE *) Admitted.
Example test_existsb_4 : existsb evenb [] = false.
Proof. (* FILL IN HERE *) Admitted.
Definition existsb' {X : Type} (test : X -> bool) (l : list X) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Theorem existsb_existsb' : forall (X : Type) (test : X -> bool) (l : list X),
existsb test l = existsb' test l.
Proof. (* FILL IN HERE *) Admitted.
(** [] *)
(* Fri 30 Aug 2019 02:44:45 PM CEST *)