In Type-Driven Development with Idris ch. 4, they say
The Prelude also defines functions and notation to allow Nat to be used like any other numeric type, so rather than writing S (S (S (S Z))), you can simply write 4.
and similarly for Fin. How does it achieve that? I've looked at the source but I can't figure it out.
from where you linked notice fromIntegerNat:
||| Convert an Integer to a Nat, mapping negative numbers to 0
fromIntegerNat : Integer -> Nat
fromIntegerNat 0 = Z
fromIntegerNat n =
if (n > 0) then
S (fromIntegerNat (assert_smaller n (n - 1)))
else
Z
and fromInteger in the Num implementation of Nat:
Num Nat where
(+) = plus
(*) = mult
fromInteger = fromIntegerNat
and Cast Integer Nat
||| Casts negative `Integers` to 0.
Cast Integer Nat where
cast = fromInteger
In the case of Idris1 it will attempt to cast from a literal (such as Char, String or Integer) into whatever type is required via those "fromFunctions" (as noted in a comment in one of the above sources: [...] '-5' desugars to 'negate (fromInteger 5)') and in general Idris1 supports implicit casting for any two types. ( http://docs.idris-lang.org/en/latest/tutorial/miscellany.html#implicit-conversions )
In the case of Idris2, there are some pragmas (%charLit fromChar, %stringLit fromString, %integerLit fromInteger) to hint the compiler to use some cast function from a literal into any other type.
Related
I'm learning OCaml, and the docs and books I'm reading aren't very clear on some things.
In simple words, what are the differences between
-10
and
~-10
To me they seem the same. I've encountered other resources trying to explain the differences, but they seem to explain in terms that I'm not yet familiar with, as the only thing I know so far are variables.
In fact, - is a binary operator, so some expression can be ambigous : f 10 -20 is treated as (f 10) - 20. For example, let's imagine this dummy function:
let f x y = (x, y)
If I want produce the tuple (10, -20) I naïvely would write something like that f 10 -20 which leads me to the following error:
# f 10 -20;;
Error: This expression has type 'a -> int * 'a
but an expression was expected of type int
because the expression is evaluated as (f 10) - 20 (so a substract over a function!!) so you can write the expression like this: f 10 (-20), which is valid or f 10 ~-20 since ~- (and ~+ and ~-. ~+. for floats) are unary operators, the predecense is properly respected.
It is easier to start by looking at how user-defined unary (~-) operators work.
type quaternion = { r:float; i:float; j:float; k:float }
let zero = { r = 0.; i = 0.; j = 0.; k = 0. }
let i = { zero with i = 1. }
let (~-) q = { r = -.q.r; i = -.q.i; j = -. q.j; k = -. q.k }
In this situation, the unary operator - (and +) is a shorthand for ~- (and ~+) when the parsing is unambiguous. For example, defining -i with
let mi = -i
works because this - could not be the binary operator -.
Nevertheless, the binary operator has a higher priority than the unary - thus
let wrong = Fun.id -i
is read as
let wrong = (Fun.id) - (i)
In this context, I can either use the full form ~-
let ok' = Fun.id ~-i
or add some parenthesis
let ok'' = Fun.id (-i)
Going back to type with literals (e.g integers, or floats), for those types, the unary + and - symbol can be part of the literal itself (e.g -10) and not an operator. For instance redefining ~- and ~+ does not change the meaning of the integer literals in
let (~+) = ()
let (~-) = ()
let really = -10
let positively_real = +10
This can be "used" to create some quite obfuscated expression:
let (~+) r = { zero with r }
let (+) x y = { r = x.r +. y.r; i = x.i +. y.i; k = x.k +. x.k; j =x.k +. y.j }
let ( *. ) s q = { r = s *. q.r; i = s *. q.i; j = s *. q.j; k = s *. q.k }
let this_is_valid x = +x + +10. *. i
OCaml has two kinds of operators - prefix and infix. The prefix operators precede expressions and infix occur in between the two expressions, e.g., in !foo we have the prefix operator ! coming before the expression foo and in 2 + 3 we have the infix operator + between expressions 2 and 3.
Operators are like normal functions except that they have a different syntax for application (aka calling), whilst functions are applied to an arbitrary number of arguments using a simple syntax of juxtaposition, e.g., f x1 x2 x3 x41, operators can have only one (prefix) or two (infix) arguments. Prefix operators are very close to normal functions, cf., f x and !x, but they have higher precedence (bind tighter) than normal function application. Contrary, the infix operators, since they are put between two expressions, enable a more natural syntax, e.g., x + y vs. (+) x y, but have lower precedence (bind less tight) than normal function application. Moreover, they enable chaining several operators in a row, e.g., x + y + z is interpreted as (x + y) + z, which is much more readable than add (add (x y) z).
Operators in OCaml distinguished purely syntactically. It means that the kind of an operator is fully defined by the first character of that operator, not by a special compiler directive, like in some other languages (i.e., there is not infix + directive in OCaml). If an operator starts with the prefix-symbol sequence, e.g., !, ?#, ~%, it is considered as prefix and if it starts with an infix-symbol then it is, correspondingly, an infix operator.
The - and -. operators are treated specially and can appear both as prefix and infix. E.g., in 1 - -2 we have - occurring both in the infix and prefix positions. However, it is only possible to disambiguate between the infix and the prefix versions of the - (and -.) operators when they occur together with other operators (infix or prefix), but when we have a general expression, the - operator is treated as infix. E.g., max 0 -1 is interpreted as (max 0) - 1 (remember that operator has lower precedence than function application, therefore when they two appear with no parentheses then functions are applied first and operators after that). Another example, Some -1, which is interpreted as Some - 1, not as Some (-1). To disambiguate such code, you can use either the parentheses, e.g., max 0 (-1), or the prefix-only versions, e.g, Some ~-1 or max 0 ~-1.
As a matter of personal style, I actually prefer parentheses as it is usually hard to keep these rules in mind when you read the code.
1) Purists will say that functions in OCaml can have only one argument and f x1 x2 x3 x4 is just ((f x1) x2) x3) x4, which would be a totally correct statement, but a little bit irrelevant to the current discussion.
I'm trying to check the equality between two integers in Coq, but I get this error: "The term "first = second" has type "Prop" which is not a (co-)inductive type.". Is there any library in Coq that provides equality checking? Here is my code:
Definition verify_eq (first : Z) (second : Z) : Z :=
if first = second then 0 else 1.
You're in luck! In the same module where Z is defined (I'm assuming ZArith in the standard library), there's a term Z.eqb : Z -> Z -> bool that gives a a boolean test for integer equality (technically it's in the submodule Z — that's why there's a Z in the name).
Require Import ZArith. (* I assume you already imported this, since you're using Z *)
Definition verify_eq (first : Z) (second : Z) : Z :=
if Z.eqb first second then 0 else 1.
It is possible to represent to some how make unFix total? Possibly by restricting what f is?
record Fix (f : Type -> Type) where
constructor MkFix
unFix : f (Fix f)
> :total unFix
Fix.unFix is possibly not total due to:
MkFix, which is not strictly positive
The problem here is that Idris has no way of knowing that the base functor you are using for your datatype is strictly positive. If it were to accept your definition, you could then use it with a concrete, negative functor and prove Void from it.
There are two ways to represent strictly positive functors: by defining a universe or by using containers. I have put everything in two self-contained gists (but there are no comments there).
A Universe of Strictly Positive Functors
You can start with a basic representation: we can decompose a functor into either a sigma type (Sig), a (strictly-positive) position for a recursive substructure (Rec) or nothing at all (End). This gives us this description and its decoding as a Type -> Type function:
-- A universe of positive functor
data Desc : Type where
Sig : (a : Type) -> (a -> Desc) -> Desc
Rec : Desc -> Desc
End : Desc
-- The decoding function
desc : Desc -> Type -> Type
desc (Sig a d) x = (v : a ** desc (d v) x)
desc (Rec d) x = (x, desc d x)
desc End x = ()
Once you have this universe of functors which are guaranteed to be strictly positive, you can take their least fixpoint:
-- The least fixpoint of such a positive functor
data Mu : Desc -> Type where
In : desc d (Mu d) -> Mu d
You can now define your favourite datatype.
Example: Nat
We start with a sum type of tags for each one of the constructors.
data NatC = ZERO | SUCC
We then define the strictly positive base functor by storing the constructor tag in a sigma and computing the rest of the description based on the tag value. The ZERO tag is associated to End (there is nothing else to store in a zero constructor) whilst the SUCC one demands a Rec End, that is to say one recursive substructure corresponding to the Nat's predecessor.
natD : Desc
natD = Sig NatC $ \ c => case c of
ZERO => End
SUCC => Rec End
Our inductive type is then obtained by taking the fixpoint of the description:
nat : Type
nat = Mu natD
We can naturally recover the constructors we expect:
zero : nat
zero = In (ZERO ** ())
succ : nat -> nat
succ n = In (SUCC ** (n, ()))
References
This specific universe is taken from 'Ornamental Algebras, Algebraic Ornaments' by McBride but you can find more details in 'The Gentle Art of Levitation' by Chapman, Dagand, McBride, and Morris.
Strictly Positive Functors as the Extension of Containers
The second representation is based on another decomposition: each inductive type is seen as a general shape (i.e. its constructors and the data they store) plus a number of recursive positions (which can depend on the specific value of the shape).
record Container where
constructor MkContainer
shape : Type
position : shape -> Type
Once more we can give it an interpretation as a Type -> Type function:
container : Container -> Type -> Type
container (MkContainer s p) x = (v : s ** p v -> x)
And take the fixpoint of the strictly positive functor thus defined:
data W : Container -> Type where
In : container c (W c) -> W c
You can once more recover your favourite datatypes by defining Containers of interest.
Example: Nat
Natural numbers have two constructors, each storing nothing at all. So the shape will be a Bool. If we are in the zero case then there are no recursive positions (Void) and in the succ one there is exactly one (()).
natC : Container
natC = MkContainer Bool (\ b => if b then Void else ())
Our type is obtained by taking the fixpoint of the container:
nat : Type
nat = W natC
And we can recover the usual constructors:
zero : nat
zero = In (True ** \ v => absurd v)
succ : nat -> nat
succ n = In (False ** \ _ => n)
References
This is based on 'Containers: Constructing Strictly Positive Types' by Abbott, Altenkirch, and Ghani.
This is a follow up to this question. Thanks to Kwartz I now have a state of the proposition if b divides a then b divides a * c for any integer c, namely:
alsoDividesMultiples : (a, b, c : Integer) ->
DivisibleBy a b ->
DivisibleBy (a * c) b
Now, the goal has been to prove that statement. I realized that I do not understand how to operate on dependent pairs. I tried a simpler problem, which was show that every number is divisible by 1. After a shameful amount of thought on it, I thought I had come up with a solution:
-- All numbers are divisible by 1.
DivisibleBy a 1 = let n = a in
(n : Integer ** a = 1 * n)
This compiles, but I was had doubts it was valid. To verify that I was wrong, it changed it slightly to:
-- All numbers are divisible by 1.
DivisibleBy a 1 = let n = a in
(n : Integer ** a = 2 * n)
This also compiles, which means my "English" interpretation is certainly incorrect, for I would interpret this as "All numbers are divisible by one since every number is two times another integer". Thus, I am not entirely sure what I am demonstrating with that statement. So, I went back and tried a more conventional way of stating the problem:
oneDividesAll : (a : Integer) ->
(DivisibleBy a 1)
oneDividesAll a = ?sorry
For the implementation of oneDividesAll I am not really sure how to "inject" the fact that (n = a). For example, I would write (in English) this proof as:
We wish to show that 1 | a. If so, it follows that a = 1 * n for some n. Let n = a, then a = a * 1, which is true by identity.
I am not sure how to really say: "Consider when n = a". From my understanding, the rewrite tactic requires a proof that n = a.
I tried adapting my fallacious proof:
oneDividesAll : (a : Integer) ->
(DivisibleBy a 1)
oneDividesAll a = let n = a in (n : Integer ** a = b * n)
But this gives:
|
12 | oneDividesAll a = let n = a in (n : Integer ** a = b * n)
| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
When checking right hand side of oneDividesAll with expected type
DivisibleBy a 1
Type mismatch between
Type (Type of DPair a P)
and
(n : Integer ** a = prim__mulBigInt 1 n) (Expected type)
Any help/hints would be appreciated.
First off, if you want to prove properties on number, you should use Nat (or other inductive types). Integer uses primitives that the argument can't argue further than prim__mulBigInt : Integer -> Integer -> Integer; that you pass two Integer to get one. The compiler doesn't know anything how the resulting Integer looks like, so it cannot prove stuff about it.
So I'll go along with Nat:
DivisibleBy : Nat -> Nat -> Type
DivisibleBy a b = (n : Nat ** a = b * n)
Again, this is a proposition, not a proof. DivisibleBy 6 0 is a valid type, but you won't find a proof : Divisible 6 0. So you were right with
oneDividesAll : (a : Nat) ->
(DivisibleBy a 1)
oneDividesAll a = ?sorry
With that, you could generate proofs of the form oneDividesAll a : DivisibleBy a 1. So, what comes into the hole ?sorry? :t sorry gives us sorry : (n : Nat ** a = plus n 0) (which is just DivisibleBy a 1 resolved as far as Idris can). You got confused on the right part of the pair: x = y is a type, but now we need a value – that's what's your last error cryptic error message hints at). = has only one constructor, Refl : x = x. So we need to get both sides of the equality to the same value, so the result looks something like (n ** Refl).
As you thought, we need to set n to a:
oneDividesAll a = (a ** ?hole)
For the needed rewrite tactic we check out :search plus a 0 = a, and see plusZeroRightNeutral has the right type.
oneDividesAll a = (a ** rewrite plusZeroRightNeutral a in ?hole)
Now :t hole gives us hole : a = a so we can just auto-complete to Refl:
oneDividesAll a = (a ** rewrite plusZeroRightNeutral a in Refl)
A good tutorial on theorem proving (where it's also explained why plus a Z does not reduce) is in the Idris Doc.
Is there a way to tell Idris to interpret decimal strings such as 2, 10, etc as Nat? The default behavior in the repl is to interpret them as Integer. In Coq, for example, it's possible to specify an interpretation scope with % to disambiguate notations, so I guess I'm hoping that something like 10%Nat exists. Is there something like that in Idris?
The standard prelude contains
the : (a : Type) -> (value: a) -> a
the _ = id
which can be used to give explicit types:
the Integer 10
the Nat 6
the (Vect 3 Int) [1,2,3]
There's also with [namespace] [expr], which privileges namespace inside expr. This seems closer to %, but the seems more often used.
with Vect [['a', 'b']] -- Vect 1 (Vect 2 Char)
with List [['a', 'b']] -- List (List Char)
You can make a syntax extension for the:
syntax [expr] "%" [type] = the type expr
5%Nat
10%Int
But not for with.