Given these two types,
Inductive unit : Set := tt.
Inductive otherUnit : Set := ttt.
Can Coq prove they are different, ie unit <> otherUnit ?
Both are singleton types in sort Set so it is not easy to find a Set -> Prop that differentiates them. For example, this does not even type check
Definition singletonTT (A : Set) : Prop := forall x : A, x = tt.
because tt has type unit instead of A.
It's actually admissible in Coq that these two types are equal; that is, you can neither prove they are equal or different, and it's consistent to assume either.
You can't express singletonTT in terms of tt because as you point out, it only type checks for the unit type. Instead you need to treat A opaquely; for example, you can express the property of being a singleton as A /\ forall (x y:A), x = y. Of course, both types are singletons so this doesn't distinguish them.
If you do assume Axiom unit_otherUnit : unit = otherUnit, then you can express something like singletonTT by casting tt to otherUnit: eq_rec _ (fun S => S) tt otherUnit ax = ttt.
When types have different cardinalities (which means they aren't isomorphic) you can use their cardinality to distinguish them and prove the types are distinct. Easy examples include False <> True and unit <> bool, and a more complicated example is nat <> (nat -> nat).
Related
I have the following two definitions that result in two different error messages.
The first definition is declined because of strict positivity and the second one because of a universe inconsistency.
(* non-strictly positive *)
Inductive SwitchNSP (A : Type) : Type :=
| switchNSP : SwitchNSP bool -> SwitchNSP A.
Fail Inductive UseSwitchNSP :=
| useSwitchNSP : SwitchNSP UseSwitchNSP -> UseSwitchNSP.
(* universe inconsistency *)
Inductive SwitchNSPI : Type -> Type :=
| switchNSPI : forall A, SwitchNSPI bool -> SwitchNSPI A.
Fail Inductive UseSwitchNSPI :=
| useSwitchNSPI : SwitchNSPI UseSwitchNSPI -> UseSwitchNSPI.
Chatting on gitter revealed that universe (in)consistencies are checked first, that is, the first definition adheres this check, but then fails because of a strict positivity issue.
As far as I understand the strict positivity restriction, if Coq allows non-strictly positivity data type definitions, I could construct non-terminating functions without using fix (which is pretty bad).
In order to make it even more confusing, the first definition is accepted in Agda and the second one gives a strict positivity error.
data Bool : Set where
True : Bool
False : Bool
data SwitchNSP (A : Set) : Set where
switchNSP : SwitchNSP Bool -> SwitchNSP A
data UseSwitchNSP : Set where
useSwitchNSP : SwitchNSP UseSwitchNSP -> UseSwitchNSP
data SwitchNSPI : Set -> Set where
switchNSPI : forall A -> SwitchNSPI Bool -> SwitchNSPI A
data UseSwitchNSPI : Set where
useSwitchNSP : SwitchNSPI UseSwitchNSPI -> UseSwitchNSPI
Now my question is two-folded: first, what is the "evil example" I could construct with the above definition? Second, which of the rules applies to the above definition?
Some notes:
To clarify, I think that I do understand why the second definition is not allowed type-checking-wise, but still feel that there is nothing "evil" happening here, when the definition is allowed.
I first thought that my example is an instance of this question, but enabling universe polymorphism does not help for the second definition.
Can I use some "trick" do adapt my definition such that it is accepted by Coq?
Unfortunately, there's nothing super deep about this example. As you noted Agda accepts it, and what trips Coq is the lack of uniformity in the parameters. For example, it accepts this:
Inductive SwitchNSPA (A : Type) : Type :=
| switchNSPA : SwitchNSPA A -> SwitchNSPA A.
Inductive UseSwitchNSPA :=
| useSwitchNSPA : SwitchNSPA UseSwitchNSPA -> UseSwitchNSPA.
Positivity criteria like the one used by Coq are not complete, so they will reject harmless types; the problem with supporting more types is that it often makes the positivity checker more complex, and that's already one of the most complex pieces of the kernel.
As for the concrete details of why it rejects it, well, I'm not 100% sure. Going by the rules in the manual, I think it should be accepted?
EDIT: The manual is being updated.
Specifically, using the following shorter names to simplify the following:
Inductive Inner (A : Type) : Type := inner : Inner bool -> Inner A.
Inductive Outer := outer : Inner Outer -> Outer.
Correctness rules
Positivity condition
Here,
X = Outer
T = forall x: Inner X, X
So we're in the second case with
U = Inner X
V = X
V is easy, so let's do that first:
V = (X) falls in the first case, with no t_i, hence is positive for X
For U: is U = Inner X strictly positive wrt X?
Here,
T = Inner X
Hence we're in the last case: T converts to (I a1) (no t_i) with
I = Inner
a1 = X
and X does not occur in the t_i, since there are no t_i.
Do the instantiated types of the constructors satisfy the nested positivity condition?
There is only one constructor:
inner : Inner bool -> Inner X.
Does this satisfy the nested positivity condition?
Here,
T = forall x: Inner bool, Inner X.
So we're in the second case with
U = Inner bool
V = Inner X
X does not occur in U, so X is strictly positive in U.
Does V satisfy the nested positivity condition for X?
Here,
T = Inner X
Hence we're in the first case: T converts to (I b1) (no u_i) with
I = Inner
b1 = X
There are no u_i, so V satisfies the nested positivity condition.
I have opened a bug report. The manual is being fixed.
Two more small things:
I can't resist pointing that your type is empty:
Theorem Inner_empty: forall A, Inner A -> False.
Proof. induction 1; assumption. Qed.
You wrote:
if Coq allows non-strictly positivity data type definitions, I could construct non-terminating functions without using fix (which is pretty bad).
That's almost correct, but not exactly: if Coq didn't enforce strict positivity, you could construct non-terminating functions period, which is bad. It doesn't matter whether they use fix or not: having non-termination in the logic basically makes it unsound (and hence Coq prevents you from writing fixpoints that do not terminate by lazy reduction).
In a constructive setting such as Coq's, I expect the proof of a disjunction A \/ B to be either a proof of A, or a proof of B. If I reformulate this on subsets of a type X, it says that if I have a proof that x is in A union B, then I either have a proof that x is in A, or a proof that x is in B. So I want to define the characteristic function of a union by case analysis,
Definition characteristicUnion (X : Type) (A B : X -> Prop)
(x : X) (un : A x \/ B x) : nat.
It will be equal to 1 when x is in A, and to 0 when x is in B. However Coq does not let me destruct un, because "Case analysis on sort Set is not allowed for inductive definition or".
Is there another way in Coq to model subsets of type X, that would allow me to construct those characteristic functions on unions ? I do not need to extract programs, so I guess simply disabling the previous error on case analysis would work for me.
Mind that I do not want to model subsets as A : X -> bool. That would be unecessarily stronger : I do not need laws of excluded middle such as "either x is in A or x is not in A".
As pointed out by #András Kovács, Coq prevents you from "extracting" computationally relevant information from types in Prop in order to allow some more advanced features to be used. There has been a lot of research on this topic, including recently Univalent Foundations / HoTT, but that would go beyond the scope of this question.
In your case you want indeed to use the type { A } + { B } which will allow you to do what you want.
I think the union of subsets should be a subset as well. We can do this by defining union as pointwise disjunction:
Definition subset (X : Type) : Type := X -> Prop.
Definition union {X : Type}(A B : subset X) : subset X := fun x => A x \/ B x.
Given a type (like List) in Coq, how do I figure out what the equality symbol "=" mean in that type? What commands should I type to figure out the definition?
The equality symbol is just special infix syntax for the eq predicate. Perhaps surprisingly, it is defined the same way for every type, and we can even ask Coq to print it for us:
Print eq.
(* Answer: *)
Inductive eq (A : Type) (x : A) : Prop :=
| eq_refl : eq x x.
This definition is so minimal that it might be hard to understand what is going on. Roughly speaking, it says that the most basic way to show that two expressions are equal is by reflexivity -- that is, when they are exactly the same. For instance, we can use eq_refl to prove that 5 = 5 or [4] = [4]:
Check eq_refl : 5 = 5.
Check eq_refl : [4] = [4].
There is more to this definition than meets the eye. First, Coq considers any two expressions that are equalivalent up to simplification to be equal. In these cases, we can use eq_refl to show that they are equal as well. For instance:
Check eq_refl : 2 + 2 = 4.
This works because Coq knows the definition of addition on the natural numbers and is able to mechanically simplify the expression 2 + 2 until it arrives at 4.
Furthermore, the above definition tells us how to use an equality to prove other facts. Because of the way inductive types work in Coq, we can show the following result:
eq_elim :
forall (A : Type) (x y : A),
x = y ->
forall (P : A -> Prop), P x -> P y
Paraphrasing, when two things are equal, any fact that holds of the first one also holds of the second one. This principle is roughly what Coq uses under the hood when you invoke the rewrite tactic.
Finally, equality interacts with other types in interesting ways. You asked what the definition of equality for list was. We can show that the following lemmas are valid:
forall A (x1 x2 : A) (l1 l2 : list A),
x1 :: l1 = x2 :: l2 -> x1 = x2 /\ l1 = l2
forall A (x : A) (l : list A),
x :: l <> nil.
In words:
if two nonempty lists are equal, then their heads and tails are equal;
a nonempty list is different from nil.
More generally, if T is an inductive type, we can show that:
if two expressions starting with the same constructor are equal, then their arguments are equal (that is, constructors are injective); and
two expressions starting with different constructors are always different (that is, different constructors are disjoint).
These facts are not, strictly speaking, part of the definition of equality, but rather consequences of the way inductive types work in Coq. Unfortunately, it doesn't work as well for other kinds of types in Coq; in particular, the notion of equality for functions in Coq is not very useful, unless you are willing to add extra axioms into the theory.
Coq would let me define this :
Definition teenagers : Set := { x : nat | x >= 13 /\ x <= 19 }.
and also :
Variable Julia:teenagers.
but not :
Example minus_20 : forall x:teenagers, x<20.
or :
Example Julia_fact1 : Julia > 12.
This is because Julia (of type teenagers) cannot be compared to 12 (nat).
Q: How should I inform Coq that Julia's support type is nat so that I can write anything useful about her ?
Q': My definition of teenagers seems like a dead end ; it is more declarative than constructive and I seem to have lost inductive properties of nat. How can I show up its inhabitants ? If there is no way, I can still stick to nat and work with Prop and functions. (newbie here, less than one week self learning with Pierce's SF).
The pattern you are using in teenagers is an instance of the "subType" pattern. As you have noted, a { x : nat | P x } is different from nat. Currently, Coq provides little support to handle these kind of types effectively, but if you restrict to "well-behaved" classes of P, you can actually work in a reasonable way. [This should really become a Coq FAQ BTW]
In the long term, you may want to use special support for this pattern. A good example of such support is provided by the math-comp library subType interface.
Describing this interface is beyond your original question, so I will end with a few comments:
In your minus_20 example, you want to use the first projection of your teenagers datatype. Try forall x : teenagers, proj1_sig x < 20. Coq can try to insert such projection automatically if you declare the projection as a Coercion:
Require Import Omega.
Definition teenagers : Set :=
{ x : nat | x >= 13 /\ x <= 19 }.
Coercion teen_to_nat := fun x : teenagers => proj1_sig x.
Implicit Type t : teenagers.
Lemma u t : t < 20.
Proof. now destruct t; simpl; omega. Qed.
As you have correctly observed, { x : T | P x } is not the same in Coq than x. In principle, you cannot transfer reasoning from objects of type T to objects of type { x : T | P x } as you must also reason in addition about objects of type P x. But for a wide class of P, you can show that the teen_to_nat projection is injective, that is:
forall t1 t2, teen_to_nat t1 = teen_to_nat t2 -> t1 = t2.
Then, reasoning over the base type can be transferred to the subtype. See also: Inductive subset of an inductive set in Coq
[edit]: I've added a couple typical examples of subTypes in math-comp, as I think they illustrate well the concept:
Sized lists or n.-tuples. A list of length n is represented in math-comp as a pair of a single list plus a proof of size, that is to say n.-tuple T = { s : seq T | size s == n}. Thanks to injectivity and coertions, you can use all the regular list functions over tuples and they work fine.
Bounded natural or ordinals: similarly, the type 'I_n = { x : nat | x < n } works almost the same than a natural number, but with a bound.
How to define isomorphism classes in Coq?
Suppose I have a record ToyRec:
Record ToyRec {Labels : Set} := {
X:Set;
r:X->Labels
}.
And a definition of isomorphisms between two objects of type ToyRec, stating that two
objects T1 and T2 are isomorphic if there exists a bijection f:T1.(X)->T2.(X) which preserves the label of mapped elements.
Definition Isomorphic{Labels:Set} (T1 T2 : #ToyRec Labels) : Prop :=
exists f:T1.(X)->T2.(X), (forall x1 x2:T1.(X), f x1 <> f x2) /\
(forall x2:T2.(X), exists x1:T1.(X), f x1 = f x2) /\
(forall x1:T1.(X) T1.(r) x1 = T2.(r) (f x1)).
Now I would like to define a function that takes an object T1 and returns a set
containing all objects that are isomorphic to T1.
g(T1) = {T2 | Isomorphic T1 T2}
How one does such a thing in coq? I know that I might be reasoning too set theoretically
here, but what would be the right type theoretic notion of isomorphism class? Or even more basically, how one would define a set (or type) of all elemenets satisfying a given property?
It really depends on what you want to do with it. In Coq, there is a comprehension type {x : T | P x} which is the type of all elements x in type T that satisfy property P. However, it is a type, meaning that it is used to classify other terms, and not a data-structure you can compute with in the traditional sense. Thus, you can use it, for instance, to write a function on T that only works on elements that satisfy P (in which case the type of the function would be {x : T | P x} -> Y, where Y is its result type), but you can't use it to, say, write a function that computes how many elements of T satisfy P.
If you want to compute with this set, things become a bit more complicated. Let's suppose P is a decidable property so that things become a bit easier. If T is a finite type, then you can a set data-structure that has a comprehension operator (have a look at the Ssreflect library, for instance). However, this breaks when T is infinite, which is the case of your ToyRec type. As Vinz said, there's no generic way of constructively building this set as a data-structure.
Perhaps it would be easier to have a complete answer if you explained what you want to do with this type exactly.