I have this code
Parameter foo : nat -> option bool.
Definition foo_valid x :=
match foo x with
Some _ => True
| None => False
end.
Axiom foo_is_valid : forall x, foo x = Some true.
Lemma foo_some_is_true_for_real : forall {x : nat}, foo_valid x.
cbv.
intros.
At this point I have this goal
x : nat
========================= (1 / 1)
match foo x with
| Some _ => True
| None => False
end
Is it possible to use foo_is_valid axiom to eliminate the match and finish the proof? How can I do that?
foo_is_valid x is an equality so you can use it to rewrite.
The easiest way to do so is using the rewrite tactic.
rewrite foo_is_valid.
should replace foo x by Some true in your goal, and then simpl will make the match compute.
Is there an easy way to rename an existential variable in a hypothesis?
Sometimes the variable names are confusing, because the same names are reused in unrelated hypotheses.
For example, I want to change H1 : exists p : nat, n0 = p * 2 to H1 : exists pminus1 : nat, n0 = pminus1 * 2.
Here is a piece of code that does it:
match goal with
an_h : #ex _ (?f) |- _ =>
let new_f := eval lazy beta in (fun pminus_one => f pminus_one) in
assert (my_h : #ex _ new_f) by exact an_h; clear an_h
end.
I was curious about how the discriminate tactic works behind the curtain. Therefore I did some experiments.
First a simple Inductive definition:
Inductive AB:=A|B.
Then a simple lemma which can be proved by the discriminate tactic:
Lemma l1: A=B -> False.
intro.
discriminate.
Defined.
Let's see what the proof looks like:
Print l1.
l1 =
fun H : A = B =>
(fun H0 : False => False_ind False H0)
(eq_ind A
(fun e : AB => match e with
| A => True
| B => False
end) I B H)
: A = B -> False
This looks rather complicated and I do not understand what is happening here. Therefore I tried to prove the same lemma more explicitly:
Lemma l2: A=B -> False.
apply (fun e:(A=B) => match e with end).
Defined.
Let's again see what Coq has made with this:
Print l2.
l2 =
fun e : A = B =>
match
e as e0 in (_ = a)
return
(match a as x return (A = x -> Type) with
| A => fun _ : A = A => IDProp
| B => fun _ : A = B => False
end e0)
with
| eq_refl => idProp
end
: A = B -> False
Now I am totally confused. This is still more complicated.
Can anyone explain what is going on here?
Let's go over this l1 term and describe every part of it.
l1 : A = B -> False
l1 is an implication, hence by Curry-Howard correspondence it's an abstraction (function):
fun H : A = B =>
Now we need to construct the body of our abstraction, which must have type False. The discriminate tactic chooses to implement the body as an application f x, where f = fun H0 : False => False_ind False H0 and it's just a wrapper around the induction principle for False, which says that if you have a proof of False, you can get a proof of any proposition you want (False_ind : forall P : Prop, False -> P):
(fun H0 : False => False_ind False H0)
(eq_ind A
(fun e : AB => match e with
| A => True
| B => False
end) I B H)
If we perform one step of beta-reduction, we'll simplify the above into
False_ind False
(eq_ind A
(fun e : AB => match e with
| A => True
| B => False
end) I B H)
The first argument to False_ind is the type of the term we are building. If you were to prove A = B -> True, it would have been False_ind True (eq_ind A ...).
By the way, it's easy to see that we can simplify our body further - for False_ind to work it needs to be provided with a proof of False, but that's exactly what we are trying to construct here! Thus, we can get rid of False_ind completely, getting the following:
eq_ind A
(fun e : AB => match e with
| A => True
| B => False
end) I B H
eq_ind is the induction principle for equality, saying that equals can be substituted for equals:
eq_ind : forall (A : Type) (x : A) (P : A -> Prop),
P x -> forall y : A, x = y -> P y
In other words, if one has a proof of P x, then for all y equal to x, P y holds.
Now, let's create step-by-step a proof of False using eq_ind (in the end we should obtain the eq_ind A (fun e : AB ...) term).
We start, of course, with eq_ind, then we apply it to some x - let's use A for that purpose. Next, we need the predicate P. One important thing to keep in mind while writing P down is that we must be able to prove P x. This goal is easy to achieve - we are going to use the True proposition, which has a trivial proof. Another thing to remember is the proposition we are trying to prove (False) - we should be returning it if the input parameter is not A.
With all the above the predicate almost writes itself:
fun x : AB => match x with
| A => True
| B => False
end
We have the first two arguments for eq_ind and we need three more: the proof for the branch where x is A, which is the proof of True, i.e. I. Some y, which will lead us to the proposition we want to get proof of, i.e. B, and a proof that A = B, which is called H at the very beginning of this answer. Stacking these upon each other we get
eq_ind A
(fun x : AB => match x with
| A => True
| B => False
end)
I
B
H
And this is exactly what discriminate gave us (modulo some wrapping).
Another answer focuses on the discriminate part, I will focus on the manual proof. You tried:
Lemma l2: A=B -> False.
apply (fun e:(A=B) => match e with end).
Defined.
What should be noted and makes me often uncomfortable using Coq is that Coq accepts ill-defined definitions that it internally rewrites into well-typed terms. This allows to be less verbose, since Coq adds itself some parts. But on the other hand, Coq manipulates a different term than the one we entered.
This is the case for your proof. Naturally, the pattern-matching on e should involve the constructor eq_refl which is the single constructor of the eq type. Here, Coq detects that the equality is not inhabited and thus understands how to modify your code, but what you entered is not a proper pattern-matching.
Two ingredients can help understand what is going on here:
the definition of eq
the full pattern-matching syntax, with as, in and return terms
First, we can look at the definition of eq.
Inductive eq {A : Type} (x : A) : A -> Prop := eq_refl : x = x.
Note that this definition is different from the one that seems more natural (in any case, more symmetric).
Inductive eq {A : Type} : A -> A -> Prop := eq_refl : forall (x:A), x = x.
This is really important that eq is defined with the first definition and not the second. In particular, for our problem, what is important is that, in x = y, x is a parameter while y is an index. That is to say, x is constant across all the constructors while y can be different in each constructor. You have the same difference with the type Vector.t. The type of the elements of a vector will not change if you add an element, that's why it is implemented as a parameter. Its size, however, can change, that's why it is implemented as an index.
Now, let us look at the extended pattern-matching syntax. I give here a very brief explanation of what I have understood. Do not hesitate to look at the reference manual for safer information. The return clause can help specify a return type that will be different for each branch. That clause can use the variables defined in the as and in clauses of the pattern-matching, which binds respectively the matched term and the type indices. The return clause will both be interpreted in the context of each branch, substituting the variables of as and in using this context, to type-check the branches one by one, and be used to type the match from an external point of view.
Here is a contrived example with an as clause:
Definition test n :=
match n as n0 return (match n0 with | 0 => nat | S _ => bool end) with
| 0 => 17
| _ => true
end.
Depending on the value of n, we are not returning the same type. The type of test is forall n : nat, match n with | 0 => nat | S _ => bool end. But when Coq can decide in which case of the match we are, it can simplify the type. For example:
Definition test2 n : bool := test (S n).
Here, Coq knows that, whatever is n, S n given to test will result as something of type bool.
For equality, we can do something similar, this time using the in clause.
Definition test3 (e:A=B) : False :=
match e in (_ = c) return (match c with | B => False | _ => True end) with
| eq_refl => I
end.
What's going on here ? Essentially, Coq type-checks separately the branches of the match and the match itself. In the only branch eq_refl, c is equal to A (because of the definition of eq_refl which instantiates the index with the same value as the parameter), therefore we claimed we returned some value of type True, here I. But when seen from an external point of view, c is equal to B (because e is of type A=B), and this time the return clause claims that the match returns some value of type False. We use here the capability of Coq to simplify pattern-matching in types that we have just seen with test2. Note that we used True in the other cases than B, but we don't need True in particular. We only need some inhabited type, such that we can return something in the eq_refl branch.
Going back to the strange term produced by Coq, the method used by Coq does something similar, but on this example, certainly more complicated. In particular, Coq often uses types IDProp inhabited by idProp when it needs useless types and terms. They correspond to True and I used just above.
Finally, I give the link of a discussion on coq-club that really helped me understand how extended pattern-matching is typed in Coq.
I have come by this problem many times: I have a proof state in Coq that includes matches on both sides of an equality that are the same.
Is there a standard way to rewrite multiple matches into one?
Eg.
match expression_evaling_to_Z with
Zarith.Z0 => something
Zartih.Pos _ => something_else
Zarith.Neg _ => something_else
end = yet_another_thing.
And if I destruct on expresion_evaling_to_Z I get two identical goals. I would like to find a way to get only one of the goals.
A standard solution is to define "a view" of your datatype using a type family that will introduce the proper conditions and cases when destructed. For your particular case, you could do:
Require Import Coq.ZArith.ZArith.
Inductive zero_view_spec : Z -> Type :=
| Z_zero : zero_view_spec Z0
| Z_zeroN : forall z, z <> Z0 -> zero_view_spec z.
Lemma zero_viewP z : zero_view_spec z.
Proof. now destruct z; [constructor|constructor 2|constructor 2]. Qed.
Lemma U z : match z with
Z0 => 0
| Zpos _ | Zneg _ => 1
end = 0.
Proof.
destruct (zero_viewP z).
Abort.
This is a common idiom in some libraries like math-comp, which provides special support for instantiating the z argument of the type family.
You can write the match expression more succinctly:
match expression_evaling_to_Z with
| Z0 => something
| Zpos _ | Zneg _ => something_else
end = yet_another_thing.
But that will give you 3 subgoals when using destruct.
In this particular case we may use the fact that you actually need to distinguish the zero and non-zero cases, and it looks like a job for the Z.abs_nat : Z -> nat function.
Require Import Coq.ZArith.BinIntDef.
match Z.abs_nat (expression_evaling_to_Z) with
| O => something
| S _ => something_else
end = yet_another_thing.
This will get you only two subcases, but you need to destruct on Z.abs_nat (expression_evaling_to_Z) or introduce a new variable. If you choose the 1st variant, then you'll probably need destruct (...) eqn:Heq. to put the equation into context.
Basically this approach is about finding a new datatype (or defining one) and a suitable function to map from the old type to the new one.
If you don't mind typing you can use replace to replace the RHS with the LHS of your goal, which makes it trivial to solve, and then you just have to prove once that the rewrite is indeed ok.
Open Scope Z.
Lemma L a b :
match a + b with
Z0 => a + b
| Zpos _ => b + a
| Zneg _ => b + a
end = a + b.
replace (b+a) with (a+b). (* 1. replace the RHS with something trivially true *)
destruct (a+b); auto. (* 2. solve the branches in one fell swoop *)
apply Z.add_comm. (* 3. solve only once what is required for the two brances *)
Qed.
Perhaps you can use some Ltac-fu or other lemma to not have to type in the RHS manually too.
I have the following function in Coq. Now I want to define an instance as below. In my case, equivalence [==] on type [nat] is defined but not on [StringMap.t String.t]. Please if you solve the Instace SC_Proper. Thanks.
Definition SC (u: nat) (zm: StringMap.t String.t):
StringMap.t String.t :=
match u with
| S p => match p with
| 2 => zm
| _ =>
match StringMap.find "S" zm with
| Some k => StringMap.empty
| _ => zm
end
end
| O => zm
end.
Instance SC_Proper
: Proper (equiv ==> equiv ==> equiv) SC.
Proof with o.
repeat red. intros u u' Hu zm1 zm2 Hzm.
Admitted.