I'm interested in trying my hand at constructing set theory using Coq. I would like to define a type sets without specifying what its members are, and a function mapping two sets to a Prop
Definition elem (s1 s1 : sets) : Prop.
I would then make the axioms of set theory hypotheses, and express theorems as (for example)
Theorem : ZFC -> (forall s : sets, ~ elem s s).
However, the syntax above doesn't work. Is this idea something that can be done in Coq? Is there a better way to accomplish this goal in Coq? I am very new to Coq, so I apologize if there is an obvious way of doing this that I don't know.
You need to give names to theorems. And for postulating things, use Parameter and Axiom (which technically mean the same thing but you can use to informally distinguish between concepts and facts).
Parameter set : Type.
Parameter elem : set -> set -> Prop.
Axiom set_extensionality : forall x y, (forall z, elem z x <-> elem z y) -> x = y.
(* etc. *)
In comparison, Definition and Theorem are used for defining and proving things. Having postulated the ZFC axioms, you can prove that a set is not an element of itself. The Theorem command takes a theorem name first (used to refer to it in the future):
Theorem no_self_elem : forall x, ~ elem x x.
Proof.
(* tactics here. *)
Qed.
Related
TLDR: I want to be able to compare two terms -- one with a hole and the other without the hole -- and extract the actual lambda term that complete the term. Either in Coq or in OCaml or a Coq plugin or in anyway really.
For example, as a toy example say I have the theorem:
Theorem add_easy_0'':
forall n:nat,
0 + n = n.
Proof.
The (lambda term) proof for this is:
fun n : nat => eq_refl : 0 + n = n
if you had a partial proof script say:
Theorem add_easy_0'':
forall n:nat,
0 + n = n.
Proof.
Show Proof.
intros.
Show Proof.
Inspected the proof you would get as your partial lambda proof as:
(fun n : nat => ?Goal)
but in fact you can close the proof and therefore implicitly complete the term with the ddt unification algorithm using apply:
Theorem add_easy_0'':
forall n:nat,
0 + n = n.
Proof.
Show Proof.
intros.
Show Proof.
apply (fun n : nat => eq_refl : 0 + n = n).
Show Proof.
Qed.
This closes the proof but goes not give you the solution for ?Goal -- though obviously Coq must have solved the CIC/ddt/Coq unification problem implicitly and closes the goals. I want to get the substitution solution from apply.
How does one do this from Coq's internals? Ideally while remaining in Coq but OCaml internals or coq plugin solutions or in fact any solution I am happy with.
Appendix 1: how did I realize apply must use some sort of "coq unification"
I knew that apply must be doing this because in the description of the apply tactic I know apply must be using unification due to it saying this:
This tactic applies to any goal. The argument term is a term well-formed in the local context. The tactic apply tries to match the current goal against the conclusion of the type of term. If it succeeds, then the tactic returns as many subgoals as the number of non-dependent premises of the type of term.
This is very very similar to what I once saw in a lecture for unification in Isabelle:
with some notes on what that means:
- You have/know rule [[A1; … ;An]] => A (*)
- that says: that given A1; …; An facts then you can conclude A
- or in backwards reasoning, if you want to conclude A, then you must give a proof of A1; …;An (or know Ai's are true)
- you want to close the proof of [[B1; …; Bm]] => C (**) (since thats your subgoal)
- so you already have the assumptions B1; …; Bm lying around for you, but you wish to be able to conclude C
- Say you want to transform subgoal (**) using rule (*). Then this is what’s going on:
- first you need to see if your subgoal (**) is a "special case" of your rule (*). You commence by checking if the conclusion (targets) of the rules are "equivalent". If the conclusions match then instead of showing C you can now show A instead. But for you to have (or show) A, you now need to show A1; … ;An using the substitution that made C and A match. The reason you need to show A1;...;An is because if you show them you get A automatically according to rule (*) -- which by the "match" (unification) shows the original goal you were after. The main catch is that you need to do this by using the substitution that made A and C match. So:
- first see if you can “match” A and C. The conclusions from both side must match. This matching is called unification and returns a substitution sig that makes the terms equal
- sig = Unify(A,C) s.t. sig(A) = sig(C)
- then because we transformed the subgoal (**) using rule (*), we must then proceed to prove the obligations from the rule (*) we used to match to conclusion of the subgoal (**). from the assumptions of the original subgoal in (**) (since those are still true) but using the substitution sig that makes the rules match.
- so the new subgoals if we match the current subgoal (*) with rule (**) is:
- [[sig(B1); … ; sig(Bm) ]] => sigm(A1)
- ...
- [[sig(B1); … ; sig(Bm) ]] => sigm(An)
- Completing/closing the proof above (i.e. proving it) shows/proves:
- [[sig(B1); …;sig(Bm) ]] => sig(C)
- Command: apply (rule <(*)>) where (*) stands for the rule names
Appendix2: why not exact?
Note that initially I thought exact was the Coq tactic I wanted to intercept but I was wrong I believe. My notes on exact:
- exact p. (assuming p has type U).
- closes a proof if the goal term T (i.e. a Type) matches the type of the given term p.
- succeeds iff T and U are convertible (basically, intuitively if they unify https://coq.inria.fr/refman/language/core/conversion.html#conversion-rules since are saying if T is convertible to U)
conversion seems to be equality check not really unification i.e. it doesn't try to solve a system of symbolic equations.
Appendix 3: Recall unification
brief notes:
- unification https://en.wikipedia.org/wiki/Unification_(computer_science)
- an algorithm that solves a system of equations between symbolic expressions/terms
- i.e. you want
- cons2( cons1( x, y, ...,) ..., cons3(a, b, c), ... ) = cons1(x, nil)
- x = y
- basically a bunch of term LHS term RHS and want to know if you can make them all equal given the terms/values and variables in them...
- term1 = term2, term3 = term4 ? with some variables perhaps.
- the solution is the substitution of the variables that satisfies all the equations
bounty
I'm genuinely curious about intercepting the apply tactic or call its unification algorithm.
apply indeed solve a unification, according to the document.
The tactic apply relies on first-order unification with dependent types unless the conclusion of the type of term is of the form P (t1 ... tn) with P to be instantiated.
Note that generally, the apply will turn one "hole" to several "hole"s that each cooresponds to a subgoal generated by it.
I have no idea how to access the internal progress of apply and get the substitution it uses.
However, You can call unify t u to do unification maully. you can refer to the official document. As far as I am concerned, the unicoq plugin provides another unification algorithm, and you can use munify t u to find unification between two items, see the Unicoq official repo.
An example of using unify and mutify:
From Unicoq Require Import Unicoq.
Theorem add_easy_0'':
forall n:nat,
0 + n = n.
Proof.
Show Proof.
intros.
Show Proof.
refine ?[my_goal].
Show my_goal.
munify (fun t : nat => eq_refl : 0 + t = t) (fun n : nat => ?my_goal).
(* unify (fun t : nat => eq_refl : 0 + t = t) (fun n : nat => ?my_goal). *)
Qed.
However, I wonder whether I have understand your question correctly.
Do you want to name the goal?
If you want to "extract the actual lambda term that complete the (parial) term". The so-called "lamda term" is the goal at that time, isn't it? If so, why to you want to "extract" it? It is just over there! Do you want to store the current subgoal and name it? If so, the abstract tactic perhaps helps, as mentioned in How to save the current goal / subgoal as an `assert` lemma
For example:
Theorem add_easy_0'':
forall n:nat,
0 + n = n.
Proof.
Show Proof.
intros.
Show Proof.
abstract apply eq_refl using my_name.
Check my_name.
(*my_name : forall n : nat, 0 + n = n*)
Show Proof.
(*(fun n : nat => my_name n)*)
Qed.
Do you want to get the substituion?
Are you asking a substituion that make the goal term and the conclusion of the theorem applied match? For example:
Require Import Arith.
Lemma example4 : 3 <= 3.
Proof.
Show Proof.
Check le_n.
(* le_n : forall n : nat, n <= n *)
apply le_n.
Are you looking forward to get something like n=3? If you want to get such a "substitution", I am afraid the two tactics mentioned above will not help. Writing OCaml codes should be needed.
Do you want store the prove of current goal?
Or are you looking forward to store the proof of the current goal? Perhaps you can try assert, as mentioned in Using a proven subgoal in another subgoal in Coq.
Suppose I want to define a type of Monomials in Coq. These would be finite maps from some ordered set of variables to nat where, say, x²y³ is represented by the map that sends x to 2, y to 3 and where everything else gets the default value, zero.
The basic definitions don't seem so hard:
Require Import
Coq.FSets.FMapFacts
Coq.FSets.FMapList
Coq.Structures.OrderedType.
Module Monomial (K : OrderedType).
Module M := FMapList.Make(K).
Module P := WProperties_fun K M.
Module F := P.F.
Definition Var : Type := M.key.
Definition Monomial : Type := M.t nat.
Definition mon_one : Monomial := M.empty _.
Definition add_at (a : option nat) (b : option nat) : option nat :=
match a, b with
| Some aa, Some bb => Some (aa + bb)
| Some aa, None => Some aa
| None, Some bb => Some bb
| None, None => None
end.
Definition mon_times (M : Monomial) (M' : Monomial) : Monomial :=
M.map2 add_at M M'.
End Monomial.
At this point, I'd like to prove something like:
Lemma mon_times_comm : forall M M', mon_times M M' = mon_times M' M.
I can see how to prove that the two maps are Equal using the lemma Equal_mapsto_iff, but I'd really like to say that my type really represents monomials and that multiplication is genuinely commutative (and the maps are eq).
I'm pretty new to Coq: is this a reasonable thing to try to prove?
Also, I realise that this might depend on the finite map implementation: if FMapList was the wrong choice and another implementation makes this easier, please point me at that!
I can see how to prove that the two maps are Equal using the lemma Equal_mapsto_iff, but I'd really like to say that my type really represents monomials and that multiplication is genuinely commutative (and the maps are eq).
I'm pretty new to Coq: is this a reasonable thing to try to prove?
Also, I realise that this might depend on the finite map implementation: if FMapList was the wrong choice and another implementation makes this easier, please point me at that!
Indeed, you are on the right track. The set type you are using doesn't have the property that two sets with the same elements are definitionally equal in Coq. As such sets are implemented as binary trees you may have Node(A, Node(B,C)) <> Node(Node(A,B),C).
In particular, having a good "set type" is an extremely challenging task in Coq due to several issues, see the anwser How to define set in coq without defining set as a list of elements for a bit more of discussion.
Doing proper algebra does indeed require a lot of complex infrastructure, #ErikMD's pointer is the right one, you should have a look at math-comp and related papers to get an understanding on the state of the art. Of course, keep experimenting!
Regarding the formalization of monomials and multivariate polynomials in Coq, you could consider using the multinomials library. It is available on OPAM:
$ opam install coq-mathcomp-multinomials
and it naturally proves a similar result to your mon_times_comm lemma:
From mathcomp Require Import ssreflect ssrfun ssrbool eqtype ssrnat seq.
From mathcomp Require Import choice finfun tuple fintype ssralg bigop.
From SsrMultinomials Require Import freeg mpoly.
Lemma test1 (n : nat) (m1 m2 : 'X_{1..n}) : (m1 + m2 = m2 + m1)%MM.
Proof.
move=> *.
by rewrite addmC.
Qed.
Lemma test2 (n : nat) (R : comRingType) (p q : {mpoly R[n]}) :
(p * q = q * p)%R.
Proof.
move=> *.
by rewrite mpoly_mulC.
Qed.
Note that the multinomials library is built upon the MathComp library that is strongly related to the SSReflect extension of the Coq proof language.
Finally, note that this library is very convenient to develop Coq proofs involving multinomial polynomials, but doesn't directly allow computing with these Coq datatypes (Eval vm_compute in ...). If you are also interested in that aspect, you may also want to take a look at the CoqEAL library (and in particular its multipoly.v theory that relies on FMaps).
Can we assign cardinals in Coq to Prop, Set and each Type_i ? I only see the definition of finite cardinals in Coq's library, so maybe we need that of big cardinals to begin with.
According to the proof-irrelevance semantics, for example exposed here, Set and the Type_i form an increasing sequence of inaccessible cardinals. Can this be proven in Coq ?
Prop seems more complex because of impredicativity. Proof-irrelevance means we identify all proofs of the same P : Prop, and interpret Prop itself as the pair {false, true}. So the cardinal of Prop would be 2. However, for any two proofs p1 p2 : P, Coq does not accept eq_refl p1 as a proof of p1 = p2. So Coq does not completely identify p1 and p2. And on the other hand, impredicativity means that for any A : Type and P : Prop, A -> P is of type Prop. That makes a lot more inhabitants than in Set.
If this is too hard, can Coq prove that Prop and Set are uncountable ? By Cantor's theorem, Coq easily proves that there is not surjection nat -> (nat -> Prop). This does not seem very far from proving there is not surjection nat -> Prop. But then we need the filter Prop -> (nat -> Prop), that isolates which Prop's have a free nat variable. Can we define this filter in Coq, since we cannot pattern-match on Prop ?
It is not possible to show that Prop is uncountable inside Coq. The ClassicalFacts module in the standard library shows that the axiom of propositional degeneracy, forall A : Prop, A = True \/ A = False, is equivalent to the presence of the excluded middle and propositional extensionality. Since Coq's set theoretic model validates these two axioms, Coq cannot refute degeneracy.
It is certainly possible to show that Set and Type are infinite, since they contain all types Fin n of natural numbers bounded by n, and these types are provably different from one another since they have different cardinality. I suspect that it is possible to show that they are uncountable by adapting the usual diagonalization argument -- that is, assume that some invertible count function e : nat -> Set, and try to encode something like the type of all natural numbers that "do not contain themselves". I do not know how you would go about proving that these types are inaccessible cardinals.
I am trying to define a recursive predicate using well-founded fixpoints with the obligation to show F_ext when rewriting with Fix_eq. The CPDT says that most such obligations are dischargeable with straightforward proof automation, but unhappily this does not appear to be so for my predicate.
I have reduced the problem to the following lemma (from Proper (pointwise_relation A eq ==> eq) (#all A)). Is it provable in Coq without additional axioms?
Lemma ext_fa:
forall (A : Type) (f g : A -> Prop),
(forall x, f x = g x) ->
(forall x, f x) = (forall x, g x).
It can be shown with extensionality of predicates or functions, but since the conclusion is weaker than the usual one (f = g) I naively thought it would be possible to produce a proof without using additional axioms. After all, both sides of the equality only involve applications of f and g; how could any intensional differences be discerned?
Have I missed a simple proof or is the lemma unprovable?
You might be interested in this code I wrote a while ago, which includes variants of Fix_eq for various numbers of arguments, and don't depend on function extensionality. Note that you don't need to change Fix_F, and can instead just prove variants of Fix_eq.
To answer the question you asked, rather than solve your context, the lemma you state is called "forall extensionality".
It is present in Coq.Logic.FunctionalExtensionality, where the axiom of function extensionality is used to prove it. The fact that the standard library version uses an axiom to prove this lemma is, at the very least, strong evidence that it is not provable without axioms in Coq.
Here is a proof sketch of that fact. Since Coq is strongly normalizing*, every proof of x = y in the empty context is judgmentally equal to eq_refl. That is, if you can prove x = y in the empty context, then x and y are convertible. Let f x := inhabited (Vector.t (x + 1)) and let g x := inhabited (Vector.t (1 + x)). It is straightforward to prove forall x, f x = g x by induction on x. Therefore, if your lemma were true without axioms, we could get a proof of
(forall x, inhabited (Vector.t (x + 1))) = (forall x, inhabited (Vector.t (1 + x)))
in the empty context, and hence eq_refl ought to prove this statement. We can easily check and see that eq_refl does not prove this statement. So your lemma ext_fa is not provable without axioms.
Note that equality for functions and equality for types are severely under-specified in Coq. Essentially, the only types (or functions) that you can prove equal in Coq are the ones that are judgmentally equal (or, more precisely, the ones that are expressible as two judgmentally equal lambdas applied to provably-equal closed terms). The only types that you can prove not equal are the ones which are provably not isomorphic. The only functions that you can prove not equal are the ones which provably differ on some concrete element of the domain that you provide. There's a lot of space between the equalities that you can prove, and the inequalities you can prove, and you don't get to say anything about things in this space without axioms.
*Coq isn't actually strongly normalizing because there are some issues with coinductives. But modulo that, it's strongly normalizing.
I am having trouble with formalizing definitions of the following form: define an integer such that some property holds.
Let's say that I formalized the definition of the property:
Definition IsGood (x : Z) : Prop := ...
Now I need a definition of the form:
Definition Good : Z := ...
assuming that I proved that an integer with the property exists and is unique:
Lemma Lemma_GoodExistsUnique : exists! (x : Z), IsGood x.
Is there an easy way of defining Good using IsGood and Lemma_GoodExistsUnique?
Since, the property is defined on integer numbers, it seems that no additional axioms should be necessary. In any event, I don't see how adding something like the axiom of choice can help with the definition.
Also, I am having trouble with formalizing definitions of the following form (I suspect this is related to the problem I described above, but please indicate if that is not the case): for every x, there exists y, and these y are different for different x. Like, for example, how to define that there are N distinct good integer numbers using IsGood:
Definition ThereAreNGoodIntegers (N : Z) (IsGood : Z -> Prop) := ...?
In real-world mathematics, definitions like that occur every now and again, so this should not be difficult to formalize if Coq is intended to be suitable for practical mathematics.
The short answer to your first question is: in general, it is not possible, but in your particular case, yes.
In Coq's theory, propositions (i.e., Props) and their proofs have a very special status. In particular, it is in general not possible to write a choice operator that extracts the witness of an existence proof. This is done to make the theory compatible with certain axioms and principles, such as proof irrelevance, which says that all proofs of a given proposition are equal to each other. If you want to be able to do this, you need to add this choice operator as an additional axiom to your theory, as in the standard library.
However, in certain particular cases, it is possible to extract a witness out of an abstract existence proof without recurring to any additional axioms. In particular, it is possible to do this for countable types (such as Z) when the property in question is decidable. You can for instance use the choiceType interface in the Ssreflect library to get exactly what you want (look for the xchoose function).
That being said, I would usually advice against doing things in this style, because it leads to unnecessary complexity. It is probably easier to define Good directly, without resorting to the existence proof, and then prove separately that Good has the sought property.
Definition Good : Z := (* ... *)
Definition IsGood (z : Z) : Prop := (* ... *)
Lemma GoodIsGood : IsGood Good.
Proof. (* ... *) Qed.
Lemma GoodUnique : forall z : Z, IsGood z -> z = Good.
If you absolutely want to define Good with an existence proof, you can also change the proof of Lemma_GoodExistsUnique to use a connective in Type instead of Prop, since it allows you to extract the witness directly using the proj1_sig function:
Lemma Lemma_GoodExistsUnique : {z : Z | Good z /\ forall z', Good z' -> z' = z}.
Proof. (* ... *) Qed.
As for your second question, yes, it is a bit related to the first point. Once again, I would recommend that you write down a function y_from_x with type Z -> Z that will compute y given x, and then prove separately that this function relates inputs and outputs in a particular way. Then, you can say that the ys are different for different xs by proving that y_from_x is injective.
On the other hand, I'm not sure how your last example relates to this second question. If I understand what you want to do correctly, you can write something like
Definition ThereAreNGoodIntegers (N : Z) (IsGood : Z -> Prop) :=
exists zs : list Z,
Z.of_nat (length zs) = N
/\ NoDup zs
/\ Forall IsGood zs.
Here, Z.of_nat : nat -> Z is the canonical injection from naturals to integers, NoDup is a predicate asserting that a list doesn't contain repeated elements, and Forall is a higher-order predicate asserting that a given predicate (in this case, IsGood) holds of all elements of a list.
As a final note, I would advice against using Z for things that can only involve natural numbers. In your example, your using an integer to talk about the cardinality of a set, and this number is always a natural number.