Suppose I have such record type:
Record myRec : Type := {
myNat : nat;
myProof : myNat > 0
}.
How can I tell Coq to parse but not pretty-print (i.e., hide) the field myProof of a value of type myRec?
Can it be set while declaring the record type?
Or should it be done using the Notation command and the only printing syntax modifier?
The way I would do it is indeed with notations.
Record myRec : Type := myRecBuild {
myNat : nat ;
myProof : myNat > 0
}.
Notation "⟪ x ⟫" := (myRecBuild x _).
Now if you have a value of myRec it will only print the relevant part.
Lemma foo :
forall (x y : myRec),
x.(myNat) = y.(myNat) ->
x = x.
Proof.
intros [x hx] [y hy] e.
simpl in e. (* replaces myNat ⟪ x ⟫ with x *)
(* Goal is now ⟪ x ⟫ = ⟪ x ⟫, hiding hx and hy *)
Abort.
I would argue that you don't need the only printing option. In fact, this can be handy if you want to easily provide the relevant part and leave the rest to automation / tactics.
Lemma bar :
exists (x : myRec), x.(myNat) = 1.
Proof.
unshelve eexists ⟪ 1 ⟫.
- auto.
- reflexivity.
Qed.
This can become particularly handy when using Program or Equations.
Related
I am trying to prove a very limited form of Euler's criterion:
Variable F : finFieldType.
Hypothesis HF : (1 != -1 :> F).
Lemma euler (a : F) : a^+(#|F|.-1./2) = -1 -> forall x, x^+2 != a.
I have the bulk of the proof done already, but I am left with odd (#|F|.-1) = 0, that is, #|F|.-1 is even. (I'm not interested in characteristic 2). I can't seem to find useful facts in the math comp library about the cardinality of finFieldTypes. For example, I would expect a lemma saying there exists a p such that prime p and #|F| = p. Am I missing something here?
By the way, I could also have totally missed an already existing proof of Euler's criterion in the math comp library itself.
I am not aware of a proof of Euler's criterion, but I found two lemmas, in finfield, that give you the two results that you expected in order to finish your proof (the second one may not be presented as you expected):
First, you have the following lemma which gives you the prime p corresponding to the characteristic of your field F (as long as it is a finFieldType):
Lemma finCharP : {p | prime p & p \in [char F]}.
Then, another lemma gives you the cardinality argument :
Let n := logn p #|R|.
Lemma card_primeChar : #|R| = (p ^ n)%N.
The problem with the second lemma is that your field should be recognized as a PrimeCharType, which roughly corresponds to a ringType with an explicit characteristic. But given the first lemma, you are able to give such a structure to your field (which canonically has a ringType), on the fly. A possible proof could be the following
Lemma odd_card : ~~ odd (#|F|.-1).
Proof.
suff : odd (#|F|) by have /ltnW/prednK {1}<- /= := finRing_gt1 F.
have [p prime_p char_F] := (finCharP F); set F_pC := PrimeCharType p_char.
have H : #|F| = #|F_primeChar| by []; rewrite H card_primeChar -H odd_exp => {H F_pC}.
apply/orP; right; have := HF; apply: contraR=> /(prime_oddPn prime_p) p_eq2.
by move: char_F; rewrite p_eq2=> /oppr_char2 ->.
Qed.
I have a function Z -> Z -> whatever which I treat as a sort of a map from (Z, Z) to whatever, let's type it as FF.
With whatever being a simple sum constructible from nix or inj_whatever.
This map I initialize with some data, in the fashion of:
Definition i (x y : Z) (f : FF) : FF :=
fun x' y' =>
if andb (x =? x') (y =? y')
then inj_whatever
else f x y.
The =? represents boolean decidable equality on Z, from Coq's ZArith.
Now I would like to have equality on two of such FFs, I don't mind invoking functional_extensionality. What I would like to do now is to have Coq computationally decide equality of two FFs.
For example, suppose we do something along the lines of:
Definition empty : FF := fun x y => nix.
Now we add some arbitrary values to make foo and foo', those are equivalent under functional extensionality:
Definition foo := i 0 0 (i 0 (-42) (i 56 1 empty)).
Definition foo' := i 0 (-42) (i 56 1 (i 0 0 empty)).
What is a good way to automatically have Coq determine foo = foo'. Ltac level stuff? Actual terminating computation? Do I need domain restriction to a finite one?
The domain restriction is a bit of an intricate one. I manipulate the maps in a way f : FF -> FF, where f can extend the subset of Z x Z that the computation is defined on. As such, come to think of it, it can't be f : FF -> FF, but more like f : FF -> FF_1 where FF_1 is a subset of Z x Z that is extended by a small constant. As such, when one applies f n times, one ends up with FF_n which is equivalent to domain restriction of FF plus n * constant to the domain. So the function f slowly (by a constant factor) expands the domain FF is defined on.
As I said in the comment more specifics are needed in order to elaborate a satisfactory answer. See the below example --- intended for a step by step description --- on how to play with equality on restricted function ranges using mathcomp:
From mathcomp Require Import all_ssreflect all_algebra.
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
(* We need this in order for the computation to work. *)
Section AllU.
Variable n : nat.
(* Bounded and unbounded fun *)
Definition FFb := {ffun 'I_n -> nat}.
Implicit Type (f : FFb).
Lemma FFP1 f1 f2 : reflect (f1 = f2) [forall x : 'I_n, f1 x == f2 x].
Proof. exact/(equivP eqfunP)/ffunP. Qed.
Lemma FFP2 f1 f2 :
[forall x : 'I_n, f1 x == f2 x] = all [fun x => f1 x == f2 x] (enum 'I_n).
Proof.
by apply/eqfunP/allP=> [eqf x he|eqf x]; apply/eqP/eqf; rewrite ?enumT.
Qed.
Definition f_inj (f : nat -> nat) : FFb := [ffun x => f (val x)].
Lemma FFP3 (f1 f2 : nat -> nat) :
all [fun x => f1 x == f2 x] (iota 0 n) -> f_inj f1 = f_inj f2.
Proof.
move/allP=> /= hb; apply/FFP1; rewrite FFP2; apply/allP=> x hx /=.
by rewrite !ffunE; apply/hb; rewrite mem_iota ?ltn_ord.
Qed.
(* Exercise, derive bounded eq from f_inj f1 = f_inj f2 *)
End AllU.
The final lemma should indeed allow you reduce equality of functions to a computational, fully runnable Gallina function.
A simpler version of the above, and likely more useful to you is:
Lemma FFP n (f1 f2 : nat -> nat) :
[forall x : 'I_n, f1 x == f2 x] = all [pred x | f1 x == f2 x] (iota 0 n).
Proof.
apply/eqfunP/allP=> eqf x; last by apply/eqP/eqf; rewrite mem_iota /=.
by rewrite mem_iota; case/andP=> ? hx; have /= -> := eqf (Ordinal hx).
Qed.
But it depends on how you (absent) condition on range restriction is specified.
After your edit, I think I should add a note on the more general topic of map equality, indeed you can define a more specific type of maps other than A -> B and then build a decision procedure.
Most typical map types [including the ones in the stdlib] will work, as long as they support the operation of "binding retrieval", so you can reduce equality to the check of finitely-many bound values.
In fact, the maps in Coq's standard library do already provide you such computational equality function.
Ok, this is a rather brutal solution which does not attempt to avoid doing the same case distinctions multiple times but it's fully automated.
We start with a tactic which inspects whether two integers are equal (using Z.eqb) and translates the results to a proposition which omega can deal with.
Ltac inspect_eq y x :=
let p := fresh "p" in
let q := fresh "q" in
let H := fresh "H" in
assert (p := proj1 (Z.eqb_eq x y));
assert (q := proj1 (Z.eqb_neq x y));
destruct (Z.eqb x y) eqn: H;
[apply (fun p => p eq_refl) in p; clear q|
apply (fun p => p eq_refl) in q; clear p].
We can then write a function which fires the first occurence of i it can find. This may introduce contradictory assumptions in the context e.g. if a previous match has revealed x = 0 but we now call inspect x 0, the second branch will have both x = 0 and x <> 0 in the context. It will be automatically dismissed by omega.
Ltac fire_i x y := match goal with
| [ |- context[i ?x' ?y' _ _] ] =>
unfold i at 1; inspect_eq x x'; inspect_eq y y'; (omega || simpl)
end.
We can then put everything together: call functional extensionality twice, repeat fire_i until there's nothing else to inspect and conclude by reflexivity (indeed all the branches with contradictions have been dismissed automatically!).
Ltac eqFF :=
let x := fresh "x" in
let y := fresh "y" in
intros;
apply functional_extensionality; intro x;
apply functional_extensionality; intro y;
repeat fire_i x y; reflexivity.
We can see that it discharges your lemma without any issue:
Lemma foo_eq : foo = foo'.
Proof.
unfold foo, foo'; eqFF.
Qed.
Here is a self-contained gist with all the imports and definitions.
I'm currently trying to write a tactic that instantiates an existential quantifier using a term that can be generated easily (in this specific example, from tauto). My first attempt:
Ltac mytac :=
match goal with
| |- (exists (_ : ?X), _) => cut X;
[ let t := fresh "t" in intro t ; exists t; firstorder
| tauto ]
end.
This tactic will work on a simple problem like
Lemma obv1(X : Set) : exists f : X -> X, f = f.
mytac.
Qed.
However it won't work on a goal like
Lemma obv2(X : Set) : exists f : X -> X, forall x, f x = x.
mytac. (* goal becomes t x = x for arbitrary t,x *)
Here I would like to use this tactic, trusting that the f which tauto finds will be just fun x => x, thus subbing in the specific proof (which should be the identity function) and not just the generic t from my current script. How might I go about writing such a tactic?
It's much more common to create an existential variable and let some tactic (eauto or tauto for example) instantiate the variable by unification.
On the other hand, you can also literally use a tactic to provide the witness using tactics in terms:
Ltac mytac :=
match goal with
| [ |- exists (_:?T), _ ] =>
exists (ltac:(tauto) : T)
end.
Lemma obv1(X : Set) : exists f : X -> X, f = f.
Proof.
mytac.
auto.
Qed.
You need the type ascription : T so that the tactic-in-term ltac:(tauto) has the right goal (the type the exists expects).
I'm not sure this is all that useful (usually the type of the witness isn't very informative and you want to use the rest of the goal to choose it), but it's cool that you can do this nonetheless.
You can use eexists to introduce an existential variable, and let tauto instantiates it.
This give the following simple code.
Lemma obv2(X : Set) : exists f : X -> X, forall x, f x = x.
eexists; tauto.
Qed.
If I have the following:
H : some complicated expression = some other complicated expression
and I want to grab
u := some other complicated expression
without hardcoding it into my proof (i.e., using pose)
Is there a clean way to do this in LTac?
I am sure there are other ltac ways to do it, in my case I prefer to use SSReflect's contextual pattern language to do it. (You'll need to install the plugin or use Coq >= 8.7 which includes SSReflect):
(* ce_i = complicated expression i *)
Lemma example T (ce_1 ce_2 : T) (H : ce_1 = ce_2) : False.
set u := (X in _ = X) in H.
resulting goal:
T : Type
ce_1, ce_2 : T
u := ce_2 : T
H : ce_1 = u
============================
False
Usually you can refine the pattern more and more until you get a pretty stable match.
Note that this happens to be the first example of the section 8.3 "Contextual patterns" in the SSReflect manual.
Here is another version, which uses Ltac and its ability to pattern-match on types of terms:
Tactic Notation "assign" "rhs" "of" ident(H) "to" ident(u) "in" ident(H') :=
match type of H with _ = ?rhs => set (u := rhs) in H' end.
Tactic Notation "assign" "rhs" "of" ident(H) "to" ident(u) "in" "*" :=
match type of H with _ = ?rhs => set (u := rhs) in * end.
We can create more variants of the above (see e.g. here). Here is how to use it:
Lemma example {T} (ce1 ce2 ce3 : T) (H1 : ce1 = ce2) (H2 : ce2 = ce3) : ce1 = ce3.
Proof.
assign rhs of H1 to u in *.
Proof state:
u := ce2 : T
H1 : ce1 = u
H2 : u = ce3
============================
ce1 = ce3
One more time:
Undo.
assign rhs of H1 to u in H1.
Proof state:
u := ce2 : T
H1 : ce1 = u
H2 : ce2 = ce3
============================
ce1 = ce3
Suppose I have a premise like this:
H2: ~ a b c <> a b c
And I wish to change it to:
a b c = a b c
Where
a is Term -> Term -> Term
b and c are both Term
How can I do it? Thanks!
If you unfold the definitions of ~ and <>, you hypothesis has the following type:
H2: (a b c = a b c -> False) -> False
Therefore, what you wish to achieve is what logicians usually call "double negation elimination". It is not an intuitionistically-provable theorem, and is therefore defined in the Classical module of Coq (see http://coq.inria.fr/distrib/V8.4/stdlib/Coq.Logic.Classical_Prop.html for details):
Classical.NNPP : forall (p : Prop), ~ ~ p -> p
I assume your actual problem is more involved than a b c = a b c, but for the sake of mentioning it, if you really care about obtaining that particular hypothesis, you can safely prove it without even looking at H2:
assert (abc_refl : a b c = a b c) by reflexivity.
If your actual example is not immediately reflexive and the equality is actually false, maybe you want to turn your goal into showing that H2 is absurd. You can do so by eliminating H2 (elim H2., which is basically doing a cut on the False type), and you will end up in the context:
H2 : ~ a b c <> a b c
EQ : a b c = a b c
=====================
False
I'm not sure whether all of this helps, but you might have oversimplified your problem so that I cannot provide more insight on what your real problem is.
Just a little thought to add to Ptival's answer - if your desired goal was not trivially solved by reflexivity, you could still make progress provided you had decidable equality on your Term, for example by applying this little lemma:
Section S.
Parameter T : Type.
Parameter T_eq_dec : forall (x y : T), {x = y} + {x <> y}.
Lemma not_ne : forall (x y : T), ~ (x <> y) -> x = y.
Proof.
intros.
destruct (T_eq_dec x y); auto.
unfold not in *.
assert False.
apply (H n).
contradiction.
Qed.
End S.