After proving tens of lemmas in propositional and predicate calculus (some more challenging than others but generally still provable on an intro-apply-destruct autopilot) I hit one starting w/ ~forall and was immediately snagged. Clearly, my understanding and knowledge of Coq was lacking. So, I'm asking for a low-level Coq technique for proving statements of the general form
~forall A [B].., C -> D.
exists A [B].., ~(C -> D).
In words, I'm hoping for a general Coq recipy for setting up and firing counterexamples. (The main reason for quantifying over functions above is that it's a (or the) primitive connective in Coq.) If you want examples, I suggest e.g.
~forall P Q: Prop, P -> Q.
~forall P: Prop, P -> ~P.
There is a related question which neither posed nor answered mine, so I suppose it's not a duplicate.
Recall that ~ P is defined as P -> False. In other words, to show such a statement, it suffices to assume P and derive a contradiction. The crucial point is that you are allowed to use P as a hypothesis in any way you like. In the particular case of universally quantified statements, the specialize tactic might come in handy. This tactic allows us to instantiate a universally quantified variable with a particular value. For instance,
Goal ~ forall P Q, P -> Q.
Proof.
intros contra.
specialize (contra True False). (* replace the hypothesis
by [True -> False] *)
apply contra. (* At this point the goal becomes [True] *)
trivial.
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.
I am trying to understand the apparent paradox of the logical framework of theorem provers like Coq not including LEM yet also being able to construct proofs by contradiction. Specifically the intuitionistic type theory that these theorem provers are based on does not allow for any logical construction of the form ￢(￢P)⇒P, and so what is required in order to artificially construct this in a language like Coq? And how is the constructive character of the system preserved if this is allowed?
I think you are mixing up two related uses of contradiction in logic. One is the technique of proof by contradiction, which says that you can prove P by proving ~ (~ P) -- that is, by showing that ~ P entails a contradiction. This technique is actually not available in general in constructive logics like Coq, unless one of the following applies.
You add the excluded middle forall P, P \/ ~ P as an axiom. Coq supports this, but this addition means that you are not working in a constructive logic anymore.
The proposition P is known to be decidable (i.e., P \/ ~ P holds). This is the case, for example, for the equality of two natural numbers n and m, which we can prove by induction.
The proposition P is of the form ~ Q. Since Q -> ~ (~ Q) holds constructively, by the law of contrapositives (which is also valid constructively), we obtain ~ (~ (~ Q)) -> (~ Q).
The other use of contradiction is the principle of explosion, which says that anything follows once you assume a contradiction (i.e., False in Coq). Unlike proof by contradiction, the principle of explosion is always valid in constructive logic, so there is no paradox here.
In constructive logic, by definition, a contradiction is an inhabitant of the empty type 0, and, also by definition, the negation ￢P of a proposition P is a function of type: P -> 0 that gives an inhabitant of the empty type 0 from an inhabitant (a proof) of P.
If you assume an inhabitant (proof) of P, and derive constructively an inhabitant of 0, you have defined a function inhabiting the type P -> 0, i.e. a proof of ￢P. This is a constructive sort of proof by contradiction: assume P, derive a contradiction, conclude ￢P.
Now if you assume ￢P and derive a contradiction, you have a constructive proof of ￢￢P, but cannot conclude constructively that you have a proof of P: for this you need the LEM axiom.
I am interested in the probably false lemma :
Lemma decideOr : forall (P Q : Prop),
(P \/ Q) -> {P} + {Q}.
that asserts we can algorithmically decide any proof of an or in sort Prop. Of course, Coq does not let us destruct the input to extract it in sort Set. However, a proof of P \/ Q is a lambda-term that Coq accepts to print, so external tools can process it.
First question : can this lambda-term be decided outside of Coq (assuming the term uses no axioms, only plain Coq) ? It might be, because the rules of constructive logic demand that all disjunctions be explicitely chosen, without cheating by a proof by contradiction. So can we code a parser of Coq proof terms, and try to decide whether the first or the second operand of the or was proved ? If the term starts with or_introl or or_intror we are done. So I guess the problems are when the term is a lambda-application. But then Coq terms are strongly normalizing, so we reduce it to a normal form and it seems it will start with either or_introl or or_intror.
Second question : if this problem can be decided outside of Coq, what prevents us from internalizing it within Coq, ie proving lemma decideOr above ?
First question
Yes, you can write a program that takes as input a Coq proof of A \/ B and outputs true or false depending on which side was used to prove the disjunction. Indeed, if you write
Compute P.
in Coq, where P : A \/ B, Coq will normalize the proof P and print which constructor was used. This will not work if P uses proofs that end in Qed (because those are not unfolded by the evaluator), but in principle it is possible to replace Qed by Defined everywhere and make it work.
Second question
What prevents us from proving decideOr is that the designers of Coq wanted to have a type of propositions that supports the excluded middle (using an axiom) while allowing programs to execute. If decideOr were a theorem and we wanted to use the excluded middle (classical : forall A : Prop, A \/ ~ A), it would not be possible to execute programs that branch on the result of decideOr (classical A). This does not mean that decideOr is false: it is perfectly possible to admit it as an axiom. There is a difference between not being provable ("there does not exist a proof of A") and being refutable ("there exists a proof of ~ A").
Let's assume we have two proofs for a simple lemma.
Lemma l1: exists x:nat, x <> 0.
exists 1.
intro.
discriminate.
Defined.
Lemma l2: exists x:nat, x <> 0.
exists 2.
discriminate.
Defined.
Intuitively, I would say that those are two different proofs.
So, can I prove the following lemma?
Lemma l3: l1 <> l2
I suppose this is undecidable.
What happens if we introduce the Univalence Axiom ?
First, a small note on terminology. There is another sense of "undecidable" often used in theoretical computer science to refer to problems of deciding, given an arbitrary element of some set, whether a fixed property holds or not of that element. If there is an algorithm computable by, say, a Turing machine, that correctly answers the question for any possible input, we say the problem is decidable; otherwise, it is undecidable. Your notion of "undecidable" is often referred to as "independence" (The two issues are of course, related. The problem of whether an arbitrary Coq proposition is provable or not is undecidable.)
Now, back to your question. I believe (although I am not entirely sure) that your lemma l3 cannot be proved or refuted in Coq even if you incorporate the univalence axiom. The reason is that the univalence axiom only violates proof irrelevance for a particular kind of proposition: equality assertions. And there is nothing about it that has any obvious consequences for existential quantification. Perhaps some intuition here can help. There is a computationally relevant analog of existential quantification (that is, something that lives in Type) that allows you to prove your principle, independently of assuming univalence:
Lemma l1: { x:nat | x <> 0 }.
exists 1.
intro.
discriminate.
Defined.
Lemma l2: { x:nat | x <> 0 }.
exists 2.
discriminate.
Defined.
Lemma l3: l1 <> l2.
Proof.
intros H. inversion H.
Qed.
However, even if this is possible for this type, it is still safe to assume irrelevance for existential quantification, because Coq's logic prevents us from manipulating its proofs in a way that allow us to extract which witnesses were used.
When refineing a program, I tried to end proof by inversion on a False hypothesis when the goal was a Type. Here is a reduced version of the proof I tried to do.
Lemma strange1: forall T:Type, 0>0 -> T.
intros T H.
inversion H. (* Coq refuses inversion on 'H : 0 > 0' *)
Coq complained
Error: Inversion would require case analysis on sort
Type which is not allowed for inductive definition le
However, since I do nothing with T, it shouldn't matter, ... or ?
I got rid of the T like this, and the proof went through:
Lemma ex_falso: forall T:Type, False -> T.
inversion 1.
Qed.
Lemma strange2: forall T:Type, 0>0 -> T.
intros T H.
apply ex_falso. (* this changes the goal to 'False' *)
inversion H.
Qed.
What is the reason Coq complained? Is it just a deficiency in inversion, destruct, etc. ?
I had never seen this issue before, but it makes sense, although one could probably argue that it is a bug in inversion.
This problem is due to the fact that inversion is implemented by case analysis. In Coq's logic, one cannot in general perform case analysis on a logical hypothesis (i.e., something whose type is a Prop) if the result is something of computational nature (i.e., if the sort of the type of the thing being returned is a Type). One reason for this is that the designers of Coq wanted to make it possible to erase proof arguments from programs when extracting them into code in a sound way: thus, one is only allowed to do case analysis on a hypothesis to produce something computational if the thing being destructed cannot alter the result. This includes:
Propositions with no constructors, such as False.
Propositions with only one constructor, as long as that constructor takes no arguments of computational nature. This includes True, Acc (the accessibility predicated used for doing well-founded recursion), but excludes the existential quantifier ex.
As you noticed, however, it is possible to circumvent that rule by converting some proposition you want to use for producing your result to another one you can do case analysis on directly. Thus, if you have a contradictory assumption, like in your case, you can first use it to prove False (which is allowed, since False is a Prop), and then eliminating False to produce your result (which is allowed by the above rules).
In your example, inversion is being too conservative by giving up just because it cannot do case analysis on something of type 0 < 0 in that context. It is true that it can't do case analysis on it directly by the rules of the logic, as explained above; however, one could think of making a slightly smarter implementation of inversion that recognizes that we are eliminating a contradictory hypothesis and adds False as an intermediate step, just like you did. Unfortunately, it seems that we need to do this trick by hand to make it work.
In addition to Arthur's answer, there is a workaround using constructive_definite_description axiom. Using this axiom in a function would not allow to perform calculations and extract code from it, but it still could be used in other proofs:
From Coq Require Import Description.
Definition strange1: forall T:Type, 0>0 -> T.
intros T H.
assert (exists! t:T, True) as H0 by inversion H.
apply constructive_definite_description in H0.
destruct H0 as [x ?].
exact x.
Defined.
Or same function without proof editing mode:
Definition strange2 (T: Type) (H: 0 > 0): T :=
proj1_sig (constructive_definite_description (fun _ => True) ltac: (inversion H)).
Also there's a stronger axiom constructive_indefinite_description that converts a proposition exists x:T, P x (without uniqueness) into a corresponding sigma-type {x:T | P x}.