I'm trying out Coq, but I'm not completely sure what I'm doing. Is:
Theorem new_theorem : forall x, P:Prop /\ Q:Prop
Equivalent to:
∀x ( P(x) and Q(x) )
Edit: I think they are.
Are you having problems with the syntax?
$ coqtop
Welcome to Coq 8.1pl3 (Dec. 2007)
Coq < Section Test.
Coq < Variable X:Set.
X is assumed
Coq < Variables P Q:X -> Prop.
P is assumed
Q is assumed
Coq < Theorem forall_test: forall x:X, P(x) /\ Q(x).
1 subgoal
X : Set
P : X -> Prop
Q : X -> Prop
============================
forall x : X, P x /\ Q x
forall_test <
Well, to answer your question:
Section test.
Variable A : Type. (* assume some universe A *)
Variable P Q : A -> Prop. (* and two predicates over A, P and Q *)
Goal forall x, P x /\ Q x. (* Ax, ( P(x) and Q(x) ) *)
End test.
Related
I have started playing around with Cubical Agda. Last thing I tried doing was building the type of integers (assuming the type of naturals is already defined) in a way similar to how it is done in classical mathematics (see the construction of integers on wikipedia). This is
data dInt : Set where
_⊝_ : ℕ → ℕ → dInt
canc : ∀ a b c d → a + d ≡ b + c → a ⊝ b ≡ c ⊝ d
trunc : isSet (dInt)
After doing that, I wanted to define addition
_++_ : dInt → dInt → dInt
(x ⊝ z) ++ (u ⊝ v) = (x + u) ⊝ (z + v)
(x ⊝ z) ++ canc a b c d u i = canc (x + a) (z + b) (x + c) (z + d) {! !} i
...
I am now stuck on the part between the two braces. A term of type x + a + (z + d) ≡ z + b + (x + c) is asked. Not wanting to prove this by hand, I wanted to use the ring solver made in Cubical Agda. But I could never manage to make it work, even trying to set it up for simple ring equalities like x + x + x ≡ 3 * x.
How can I make it work ? Is there a minimal example to make it work for naturals ? There is a file NatExamples.agda in the library, but it makes you have to rewrite your equalities in a convoluted way.
You can see how the solver for natural numbers is supposed to be used in this file in the cubical library:
Cubical/Tactics/NatSolver/Examples.agda
Note that this solver is different from the solver for commutative rings, which is designed for proving equations in abstract rings and is explained here:
Cubical/Tactics/CommRingSolver/Examples.agda
However, if I read your problem correctly, the equality you want to prove requires the use of other propositional equalities in Nat. This is not supported by any solver in the cubical library (as far as I know, also the standard library doesn't support it). But of course, you can use the solver for all the steps that don't use other equalities.
Just in case you didn't spot this: here is a definition of the integers in math-style using the SetQuotients of the cubical library. SetQuotients help you to avoid the work related to your third constructor trunc. This means you basically just need to show some constructions are well defined as you would in 'normal' math.
I've successfully used the ring solver for exactly the same problem: defining Int as a quotient of ℕ ⨯ ℕ. You can find the complete file here, the relevant parts are the following:
Non-cubical propositional equality to path equality:
open import Cubical.Core.Prelude renaming (_+_ to _+̂_)
open import Relation.Binary.PropositionalEquality renaming (refl to prefl; _≡_ to _=̂_) using ()
fromPropEq : ∀ {ℓ A} {x y : A} → _=̂_ {ℓ} {A} x y → x ≡ y
fromPropEq prefl = refl
An example of using the ring solver:
open import Function using (_$_)
import Data.Nat.Solver
open Data.Nat.Solver.+-*-Solver
using (prove; solve; _:=_; con; var; _:+_; _:*_; :-_; _:-_)
reorder : ∀ x y a b → (x +̂ a) +̂ (y +̂ b) ≡ (x +̂ y) +̂ (a +̂ b)
reorder x y a b = fromPropEq $ solve 4 (λ x y a b → (x :+ a) :+ (y :+ b) := (x :+ y) :+ (a :+ b)) prefl x y a b
So here, even though the ring solver gives us a proof of _=̂_, we can use _=̂_'s K and _≡_'s reflexivity to turn that into a path equality which can be used further downstream to e.g. prove that Int addition is representative-invariant.
I'm trying to wrap my head around type classes in Coq (I've dabbled with it in the past, but I'm a far cry from being an experienced user). As an exercise, I am trying to write a group theory library. This is what I've come up with:
Class Group {S : Type} {op : S → S → S} := {
id : S;
inverse : S → S;
id_left {x} : (op id x) = x;
id_right {x} : (op x id) = x;
assoc {x y z} : (op (op x y) z) = (op x (op y z));
right_inv {x} : (op x (inverse x)) = id;
}.
I am particularly fond of the implicit S and op parameters (assuming I understand them correctly).
Making some notation for inverses is easy:
Notation "- x" := (#inverse _ _ _ x)
(at level 35, right associativity) : group_scope.
Now, I would like to make x * y a shorthand for (op x y). When working with sections, this is straightforward enough:
Section Group.
Context {S} {op} { G : #Group S op }.
(* Reserved at top of file *)
Notation "x * y" := (op x y) : group_scope.
(* ... *)
End Group.
However, since this is declared within a section, the notation is inaccessible elsewhere. I would like to declare the notation globally if possible. The problem I am running into (as opposed to inverse) is that, since op is an implicit parameter to Group, it doesn't actually exist anywhere in the global scope (so I cannot refer to it by (#op _ _ _ x y)). This problem indicates to me that I am either using type classes wrong or don't understand how to integrate notation with implicit variables. Would someone be able to point me in the right direction?
Answer (25 Jan 2018)
Based on Anton Trunov's response, I was able to write the following, which works:
Reserved Notation "x * y" (at level 40, left associativity).
Class alg_group_binop (S : Type) := alg_group_op : S → S → S.
Delimit Scope group_scope with group.
Infix "*" := alg_group_op: group_scope.
Open Scope group_scope.
Class Group {S : Type} {op : alg_group_binop S} : Type := {
id : S;
inverse : S → S;
id_left {x} : id * x = x;
id_right {x} : x * id = x;
assoc {x y z} : (x * y) * z = x * (y * z);
right_inv {x} : x * (inverse x) = id;
}.
Here is how Pierre Castéran and Matthieu Sozeau solve this problem in A Gentle Introduction to Type Classes and Relations in Coq (§3.9.2):
A solution from ibid. consists in declaring a singleton type class for representing binary operators:
Class monoid_binop (A:Type) := monoid_op : A -> A -> A.
Nota: Unlike multi-field class types, monoid_op is not a constructor, but a transparent constant such that monoid_op f can be δβ-reduced into f.
It is now possible to declare an infix notation:
Delimit Scope M_scope with M.
Infix "*" := monoid_op: M_scope.
Open Scope M_scope.
We can now give a new definition of Monoid, using the type monoid_binop A instead of A → A → A, and the infix notation x * y instead of monoid_op x y :
Class Monoid (A:Type) (dot : monoid_binop A) (one : A) : Type := {
dot_assoc : forall x y z:A, x*(y*z) = x*y*z;
one_left : forall x, one * x = x;
one_right : forall x, x * one = x
}.
There's probably a good reason why Pierre Castéran and Matthiu Sozeau deal with it that way.
But wouldn't
Definition group_op {S op} {G : #Group S op} := op.
Infix "*" := group_op.
also work here? (I only tried on two very basic test cases.)
This would spare you changing definition of Group.
I am reading the book Introduction to Mathematics Philosophy by B.Russell and trying to formalize the definitions. Whereas I got stuck on proving the equivalence between the two definitions of similarity posed in the book.
Here are the text quoted from the book. (context)
1) Defining similarity directly:
We may define two relations P and Q as “similar,” or as having
“likeness,” when there is a one-one relation S whose domain is the
field of P and whose converse domain is the field of Q, and which is
such that, if one term has the relation P to another, the correlate of
the one has the relation Q to the correlate of the other, and vice
versa.
Here's my comprehension of the above text:
Inductive similar {X} (P : relation X) (Q : relation X) : Prop :=
| similar_intro : forall (S : relation X),
one_one S ->
(forall x, field P x <-> domain S x) ->
(forall x y z w, P x y -> S x z -> S y w -> Q z w) ->
(forall x y z w, Q z w -> S x z -> S y w -> P x y) ->
similar P Q.
2) Defining similarity through the concept of 'correlator':
A relation S is said to be a “correlator” or an “ordinal correlator”
of two relations P and Q if S is one-one, has the field of Q for its
converse domain, and is such that P is the relative product of S and Q
and the converse of S.
Two relations P and Q are said to be “similar,” or to have “likeness,”
when there is at least one correlator of P and Q.
My definition to this is:
Inductive correlator {X} (P Q : relation X) : relation X -> Prop :=
| correlator_intro : forall (S : relation X),
one_one S ->
(forall x, field P x <-> domain S x) ->
(forall x y, relative_product (relative_product S Q) (converse S) x y <-> P x y) ->
correlator P Q S.
Inductive similar' {X} (P Q : relation X) : Prop :=
| similar'_intro : forall S, correlator P Q S -> similar' P Q.
But I couldn't prove that similar is equivalent to similar', where did I make the mistake? Thanks a lot.
I want to proof this lemma in Coq:
a : Type
b : Type
f : a -> b
g : a -> b
h : a -> b
______________________________________(1/1)
(forall x : a, f x = g x) ->
(forall x : a, g x = h x) -> forall x : a, f x = h x
I know that Coq.Relations.Relation_Definitions defines transitivity for relations:
Definition transitive : Prop := forall x y z:A, R x y -> R y z -> R x z.
Simply using the proof tactic apply transitivity obviously fails. How can I apply the transitivity lemma to the goal above?
The transitivity tactic requires an argument, which is the intermediate term that you want to introduce into the equality. First call intros (that's almost always the first thing to do in a proof) to have the hypotheses nicely in the environment. Then you can say transitivity (g x) and you're left with two immediate applications of an assumption.
intros.
transitivity (g x); auto.
You can also make Coq guess which intermediate term to use. This doesn't always work, because sometimes Coq finds a candidate that doesn't work out in the end, but this case is simple enough and works immediately. The lemma that transitivity applies is eq_trans; use eapply eq_trans to leave a subterm open (?). The first eauto chooses a subterm that works for the first branch of the proof, and here it also works in the second branch of the proof.
intros.
eapply eq_trans.
eauto.
eauto.
This can be abbreviated as intros; eapply eq_trans; eauto. It can even be abbreviated further to
eauto using eq_trans.
eq_trans isn't in the default hint database because it often leads down an unsuccessful branch.
Ok, I was on the wrong track. Here is the proof of the lemma:
Lemma fun_trans : forall (a b:Type) (f g h:a->b),
(forall (x:a), f x = g x) ->
(forall (x:a), g x = h x) ->
(forall (x:a), f x = h x).
Proof.
intros a b f g h f_g g_h x.
rewrite f_g.
rewrite g_h.
trivial.
Qed.
Is there an extensible, efficient way to write existential statements in Haskell without implementing an embedded logic programming language? Oftentimes when I'm implementing algorithms, I want to express existentially quantified first-order statements like
∃x.∃y.x,y ∈ xs ∧ x ≠ y ∧ p x y
where ∈ is overloaded on lists. If I'm in a hurry, I might write perspicuous code that looks like
find p [] = False
find p (x:xs) = any (\y -> x /= y && (p x y || p y x)) xs || find p xs
or
find p xs = or [ x /= y && (p x y || p y x) | x <- xs, y <- xs]
But this approach doesn't generalize well to queries returning values or predicates or functions of multiple arities. For instance, even a simple statement like
∃x.∃y.x,y,z ∈ xs ∧ x ≠ y ≠ z ∧ f x y z = g x y z
requires writing another search procedure. And this means a considerable amount of boilerplate code. Of course, languages like Curry or Prolog that implement narrowing or a resolution engine allow the programmer to write statements like:
find(p,xs,z) = x ∈ xs & y ∈ xs & x =/= y & f x y =:= g x y =:= z
to abuse the notation considerably, which performs both a search and returns a value. This problem arises often when implementing formally specified algorithms, and is often solved by combinations of functions like fmap, foldr, and mapAccum, but mostly explicit recursion. Is there a more general and efficient, or just general and expressive, way to write code like this in Haskell?
There's a standard transformation that allows you to convert
∃x ∈ xs : P
to
exists (\x -> P) xs
If you need to produce a witness you can use find instead of exists.
The real nuisance of doing this kind of abstraction in Haskell as opposed to a logic language is that you really must pass the "universe" set xs as a parameter. I believe this is what brings in the "fuss" to which you refer in your title.
Of course you can, if you prefer, stuff the universal set (through which you are searching) into a monad. Then you can define your own versions of exists or find to work with the monadic state. To make it efficient, you can try Control.Monad.Logic, but it may involve breaking your head against Oleg's papers.
Anyway, the classic encoding is to replace all binding constructs, including existential and universal quantifiers, with lambdas, and proceed with appropriate function calls. My experience is that this encoding works even for complex nested queries with a lot of structure, but that it always feels clunky.
Maybe I don't understand something, but what's wrong with list comprehensions? Your second example becomes:
[(x,y,z) | x <- xs, y <- xs, z <- xs
, x /= y && y /= z && x /= z
, (p1 x y z) == (p2 x y z)]
This allows you to return values; to check if the formula is satisfied, just use null (it won't evaluate more than needed because of laziness).