I was trying to write a very simple program to sum nats in a list (copy pasted from here):
Fixpoint sum (l : list nat) : nat :=
match l with
| [] => 0
| x :: xs => x + sum xs
end.
but my local Coq and jsCoq complain:
Syntax error: 'end' expected after [branches] (in [term_match]).
Why is this? (note I didn't even implement this but my implementation looks the same pretty much)
I've implemented recursive functions before:
Inductive my_nat : Type :=
| O : my_nat
| S : my_nat -> my_nat.
Fixpoint add_left (n m : my_nat) : my_nat :=
match n with
| O => m
| S n' => S (add_left n' m)
end.
which doesn't complain...
I did see this question How to match a "match" expression? but it seems to address some special issue in ltac and I am not using ltac.
The location of the error is on the [], which suggests that Coq does not understand that notation. Upon finding undefined notation, the parser has no idea what to do and produces an essentially meaningless error message.
To have standard list notation be defined, you need to import it from the standard library via:
Require Import List.
Import ListNotations.
The stdlib module List contains the module ListNotations, which defines [] (and more generally [ x ; y ; .. ; z ]). List also defines the notation x :: xs.
when picking up excerpts from a development, you also have to find what are the syntax changing commands that have an effect on this exceprts: Module importation, scope opening, argument declarations (for implicits), Notations, and coercions". In the current case, the file is actually provided by the author of the exercises through this pointer.
Related
For example, if I define a function from nat to nat, it would be
Definition plusfive(a:nat): nat := a + 5.
However, I would like to define a function whose arguments are nats constructed using the "S" constructor (i.e. nonzero) is that possible to directly specify as a type? something like
Definition plusfive(a: nat.S): nat := a + 5.
(I know that for this case I could also add an argument with a proof that a is nonzero, but I am wondering if it is possible to directly name the type based on the 'S' constructor).
Functions have to be complete, so you will have to use some subtype instead of nat, or add an argument that reduces input space, for instance (H: a<>0)
Definition plusfive(a:nat) (H:a<>0) :=
match a as e return a=e -> _ with
| S _ => fun _ => a + 5
| _ => fun H0 => match (H H0) with end
end eq_refl.
However, these kinds of tricks have been discovered to be very cumbersome to work with in large developments, and often one instead uses complete functions on the base type that return dummy values for bad input values, and prove that the function is called with correct arguments separately from the function definition. See for example how division is defined in the standard library.
Require Import Nat.
Print div.
div =
fun x y : nat => match y with
| 0 => y
| S y' => fst (divmod x y' 0 y')
end
: nat -> nat -> nat
So Compute (div 1 0). gives you 0.
The nice thing is that you can use div in expressions directly, without having to interleave proofs that the denominator is non-zero. Proving that an expression is correct is then done after it has been defined, not at the same time.
I want to set Coq up, without redefining the : with a notation (and without a plugin, and without replacing the standard library or redefining the constants I'm using---no cheating like that), so that I have something like a partial coercion from option nat to nat, which is defined only on Some _. In particular, I want
Eval compute in Some 0 : nat.
to evaluate to 0, and I want
Check None : nat.
to raise an error.
The closest I've managed is the ability to do this with two :s:
Definition dummy {A} (x : option A) := A.
Definition inverted_option {A} (x : option A)
:= match x with Some _ => A | _ => True end.
Definition invert_Some {A} (x : option A) : inverted_option x
:= match x with Some v => v | None => I end.
Coercion invert_Some : option >-> inverted_option.
Notation nat' := (inverted_option (A:=nat) (Some _)).
Eval compute in (Some 0 : nat') : nat.
Check (None : nat') : nat.
(* The term "None" has type "option ?A" while it is expected to have type
"nat'". *)
However, this only works when nat' is a notation, and I can't define a coercion out of a notation. (And trying to define a coercion from inverted_option (Some _) to nat violated the uniform inheritance condition.) I thought I might be able to get around this issue by using canonical structures, but I haven't managed to figure out how to interleave canonical structure resolution with coercion insertion (see also Can canonical structure resolution be interleaved with coercion insertion?).
(I ran into this issue when attempting to answer Coq: Defining a subtype.)
Exercise 6.7 in Coq'Art, or the final exercise of the Logic chapter in Software Foundations: show that the following are equivalent.
Definition peirce := forall P Q:Prop, ((P->Q)->P)->P.
Definition classic := forall P:Prop, ~~P -> P.
Definition excluded_middle := forall P:Prop, P\/~P.
Definition de_morgan_not_and_not := forall P Q:Prop, ~(~P/\~Q)->P\/Q.
Definition implies_to_or := forall P Q:Prop, (P->Q)->(~P\/Q).
The solution set expresses this by a circular chain of implications, using five separate lemmas. But "TFAE" proofs are common enough in mathematics that I'd like to have an idiom to express them. Is there one in Coq?
This type of pattern is very easy to express in Coq, although setting up the infrastructure to do so might take some effort.
First, we define a proposition that expresses that all propositions in a list are equivalent:
Require Import Coq.Lists.List. Import ListNotations.
Definition all_equivalent (Ps : list Prop) : Prop :=
forall n m : nat, nth n Ps False -> nth m Ps True.
Next, we want to capture the standard pattern for proving this kind of result: if each proposition in the list implies the next one, and the last implies the first, we know they are all equivalent. (We could also have a more general pattern, where we replace a straight list of implications with a graph of implications between the propositions, whose transitive closure generates a complete graph. We'll avoid that in the interest of simplicity.) The premise of this pattern is easy to express; it is just a code transcription of the English explanation above.
Fixpoint all_equivalent'_aux
(first current : Prop) (rest : list Prop) : Prop :=
match rest with
| [] => current -> first
| P :: rest' => (current -> P) /\ all_equivalent'_aux first P rest'
end.
Definition all_equivalent' (Ps : list Prop) : Prop :=
match Ps with
| first :: second :: rest =>
(first -> second) /\ all_equivalent' first second rest
| _ => True
end.
The difficult part is showing that this premise implies the conclusion we want:
Lemma all_equivalentP Ps : all_equivalent' Ps -> all_equivalent Ps.
Showing that this lemma holds probably requires some ingenuity to find a strong enough inductive generalization. I can't quite prove it right now, but might add a solution later to the answer if you want.
I'm trying to write a function that takes a list of natural numbers and returns as output the amount of different elements in it. For example, if I have the list [1,2,2,4,1], my function DifElem should output "3". I've tried many things, the closest I've gotten is this:
Fixpoint DifElem (l : list nat) : nat :=
match l with
| [] => 0
| m::tm =>
let n := listWidth tm in
if (~ In m tm) then S n else n
end.
My logic is this: if m is not in the tail of the list then add one to the counter. If it is, do not add to the counter, so I'll only be counting once: when it's the last time it appears. I get the error:
Error: The term "~ In m tm" has type "Prop"
which is not a (co-)inductive type.
In is part of Coq's list standard library Coq.Lists.List. It is defined there as:
Fixpoint In (a:A) (l:list A) : Prop :=
match l with
| [] => False
| b :: m => b = a \/ In a m
end.
I think I don't understand well enough how to use If then statements in definitions, Coq's documentation was not helpful enough.
I also tried this definition with nodup from the same library:
Definition Width (A : list nat ) := length (nodup ( A ) ).
In this case what I get as error is:
The term "A" has type "list nat" while it is expected to have
type "forall x y : ?A0, {x = y} + {x <> y}".
And I'm quiet confused as to what's going on here. I'd appreciate your help to solve this issue.
You seem to be confusing propositions (Prop) and booleans (bool). I'll try to explain in simple terms: a proposition is something you prove (according to Martin-Lof's interpretation it is a set of proofs), and a boolean is a datatype which can hold only 2 values (true / false). Booleans can be useful in computations, when there are only two possible outcomes and no addition information is not needed. You can find more on this topic in this answer by #Ptival or a thorough section on this in the Software Foundations book by B.C. Pierce et al. (see Propositions and Booleans section).
Actually, nodup is the way to go here, but Coq wants you to provide a way of deciding on equality of the elements of the input list. If you take a look at the definition of nodup:
Hypothesis decA: forall x y : A, {x = y} + {x <> y}.
Fixpoint nodup (l : list A) : list A :=
match l with
| [] => []
| x::xs => if in_dec decA x xs then nodup xs else x::(nodup xs)
end.
you'll notice a hypothesis decA, which becomes an additional argument to the nodup function, so you need to pass eq_nat_dec (decidable equality fot nats), for example, like this: nodup eq_nat_dec l.
So, here is a possible solution:
Require Import Coq.Arith.Arith.
Require Import Coq.Lists.List.
Import ListNotations.
Definition count_uniques (l : list nat) : nat :=
length (nodup eq_nat_dec l).
Eval compute in count_uniques [1; 2; 2; 4; 1].
(* = 3 : nat *)
Note: The nodup function works since Coq v8.5.
In addition to Anton's solution using the standard library I'd like to remark that mathcomp provides specially good support for this use case along with a quite complete theory on count and uniq. Your function becomes:
From mathcomp Require Import ssreflect ssrfun ssrbool eqtype ssrnat seq.
Definition count_uniques (T : eqType) (s : seq T) := size (undup s).
In fact, I think the count_uniques name is redundant, I'd prefer to directly use size (undup s) where needed.
Using sets:
Require Import MSets.
Require List. Import ListNotations.
Module NatSet := Make Nat_as_OT.
Definition no_dup l := List.fold_left (fun s x => NatSet.add x s) l NatSet.empty.
Definition count_uniques l := NatSet.cardinal (no_dup l).
Eval compute in count_uniques [1; 2; 2; 4; 1].
I'm trying to use the list remove function in Coq standard library but it asks for a bizarre typing and I don't know how to solve.
The function I'm implementing is to make a list of free variables in a lambda term as follows:
Fixpoint fv (t : trm) : vars :=
match t with
| Var v => [v]
| App t1 t2 => (fv t1) ++ (fv t2)
| Abs x t' => remove x (fv t')
end.
And it gives me the following error:
Error: In environment
fv : trm -> vars
t : trm
x : nat
t' : trm
The term "x" has type "nat" while it is expected to have type
"forall x0 y : ?171, {x0 = y} + {x0 <> y}".
I'm pretty sure there is something to do with that hyphotesis thing in the definition of the remove function. I have no idea how to deal with it though, any helps?
remove is defined in a context containing:
Hypothesis eq_dec : forall x y : A, {x = y}+{x <> y}.
The function removes takes this as a first argument (which you can see by doing Print remove.)
This hypothesis is a function deciding equality of elements of the type in your list. In your case, you will have to provide a function to decide equality of var (which seems to be nat, so there is probably such a function in the standard library too).
If you do not know the "{p} + {q}" notation, you can look it up here:
http://coq.inria.fr/library/Coq.Init.Specif.html#sumbool