The Haskell Sequence - Type System
http://sequence.complete.org/taxonomy/term/1/0
enRefuting Myths about Polymorphism
http://sequence.complete.org/node/381
<p> I am addressing an article written by Mark Dominus named <a href="http://blog.plover.com/oops/subtypes.html">Subtypes and
polymorphism</a>:
<h1>References and Type Safety</h1>
<p> In the second part of the article, Mark points out an example
which he says demonstrates where Standard ML breaks type safety. In
fact, he is passing along a common myth about the "value restriction"
in Standard ML and other Hindley-Milner based languages. To state it
quite simply up-front: the "value restriction" is not necessary to
protect type safety. It is simply there to preserve an intuition
about how code evaluates.
<p> This is the example he presents. If you give this to a Standard
ML compiler, it will fail on the line beginning with <code>val
a</code> because of the "value restriction."
<p>
<code>
fun id x = x (* id : α → α *)
val a = ref id; (* a : ref (α → α) *)
fun not true = false
| not false = true ; (* not: bool → bool *)
a := not;
(!a) 13
</code>
<p> If the value restriction were to be lifted, the code would still
be type-safe. Mark's interpretation of the code is incorrect. But
the thinking process which led him to his incorrect interpretation is
itself the reason for the value restriction.
<h3>The Value Restriction</h3>
<p>
The restriction itself is simple to describe: if you have code that looks like this:
<p>
<code>
val x = ...
</code>
<p>
and the <code>...</code> has a type involving polymorphism, then it must be a value syntactically.
<p><code>ref id</code> of type <code>(α → α) ref</code> is not a value -- it will
evaluate to a location-value. Also, it is polymorphic. Therefore the
value restriction applies, and the code is not permitted.
<p> The value restriction is seemingly helping us preserve type
safety. But that is a misconception. If you removed the value
restriction, you could compile the above code, and it would still be
type-safe.
<p> The short answer to what would happen is this: the <code>a</code>
on the line <code>a:=not</code> and other one on the line <code>(!a)
13</code> are not evaluating to same location-value! Therefore, you
are not, in fact, applying <code>not</code> to an integer. You are
applying <code>id</code> to an integer, which is perfectly fine.
<h3>Under the hood</h3>
<p> To explain why this is the case, I need to show you that the code
above is omitting important details about polymorphism. Those details
are automatically reconstructed and inserted by the typechecker. Here
is a version of the code that is decorated with the details after
typechecking:
<p>
<code>
val id : ∀α. α → α = Λα. fn (x:α) => x
val a : ∀α. (α → α) ref = Λα. (ref{α → α}) (id{α})
val not : bool → bool
a{bool} := not;
((!{int → int}) (a{int})) 13
</code>
<p> Whew! Now you can see why people prefer to let the typechecker
deal with this stuff. But now the Hindley-Milner style types have
been translated into a System F (aka polymorphic lambda-calculus)
style language.
<p> When you see the type α → α in H-M types, it's omitting the
universal quantifier. Haskell folks are already familiar with it,
because a popular extension allows the explicit use of ∀ (aka
"forall"). All the type variables must be quantified in a ∀ placed at
the beginning of the type.
<p> The term which has a type beginning with ∀ is called Λ (aka
"type-lambda"). Therefore, this term must be introduced everywhere a ∀
appears in the type. The typechecker rewrites your code to insert the
necessary type-lambdas.
<p> In order to use a polymorphic value (a term wrapped in a
type-lambda), you must apply a type to the type-lambda. This is
denoted <code>t{T}</code> in the above example. For example, to use
the polymorphic function <code>id : ∀a. α → α</code>, you must first
apply its type-lambda to the type that you want: <code>(id{int}) 13</code> steps to 13.
<p> Now, I can explain what is happening behind the scenes when you
use <code>a</code>. Since <code>a</code> is polymorphic, that means
it is wrapped in a type-lambda. In order to be used as a reference,
it must be applied to a type. For example:
<p>
<code>a{bool}</code> steps to <code>(ref{bool → bool}) (id{bool})</code> which then steps to <code>l1</code>
<p>
where <code>l1</code> of type <code>(bool → bool) ref</code>
is a location-value –- the result of evaluating a call to
<code>ref</code>.
<p> On the other hand, later on we do:
<p>
<code>a{int}</code> steps to <code>(ref{int → int}) (id{int})</code> which then steps to <code>l2</code>
<p>
where <code>l2</code> is a different location-value. We've called
<code>ref</code> a second time, and it's given us a different
location.
<h3>The Unexpected</h3>
<p> When Mark wrote <code>val a = ref id</code> he clearly did not
expect <code>ref</code> to be called more than once. But I have
demonstrated that is exactly what would happen if his code were to be
compiled and run.
<p> This kind of behavior is type-safe; we never attempt to invoke
<code>not</code> on the integer 13 because the function
<code>not</code> is safely stored in an entirely different location.
It is possible to go ahead and prove the type safety of a language
like this with references. But clearly, the repeated computation of
<code>ref</code> is surprising, and might even be said to
be signs of a "leaky abstraction" in the Hindley-Milner type system.
It attempts to hide the details about type-lambdas, but they've
revealed themselves in the dynamic behavior of the code.
<p> The "value restriction" then, is simply a measure put in place to
prevent "surprising" behavior from occurring in the first place. By
allowing only values syntactically, you will not end up with any
evaluation after applying the type-lambda, and all is as expected.http://sequence.complete.org/node/381#commentsType SystemDiscussionSun, 02 Mar 2008 21:23:35 -0800mrd381 at http://sequence.complete.orgSubtyping and Polymorphism
http://sequence.complete.org/node/380
<p> I am addressing an article written by Mark Dominus named <a href="http://blog.plover.com/oops/subtypes.html">Subtypes and
polymorphism</a>:
<p>
There are two parts to this article, the first discusses the fact that
subtyping and polymorphism can lead to some unintuitive results. I
won't repeat his point here, but just note that the term for the
behavior of "List" here is called "invariant" and the incorrect
behavior which he had assumed is called "covariant."
<h1>Java's broken arrays</h1>
<p>
While Java generic Lists are "invariant," Java arrays are
"covariant." This is not due to some wonderful property of arrays,
but in fact is a gigantic flaw in the Java language which breaks
static type safety. As a result, arrays must carry run-time type
information in order to dynamically check the type of objects being
inserted. Consider this code:
<p>
<code>
String stringA[] = new String[1];
stringA[0] = "Hello";
Object objectA[] = stringA;
</code>
<p>
So far this seems okay -- String is a subtype of Object -- therefore
it seems safe to say that String arrays are a subtype of Object
arrays. This is called "covariant" behavior.
<p>
So if objectA is an array of Objects, and everything is a subtype of
Object, then it should be okay to put an Integer there instead, right?
<p>
<code>
objectA[0] = 1;
</code>
<p>
but wait, now <code>stringA[0]</code> has an Integer in it! A String
array cannot have an Integer in it. Before it seemed that arrays
might be "covariant," but mutation makes it seem that arrays must be
"contravariant." We can't be both at the same time, so it seems
mutable arrays are "invariant" after all.
<p>
If you try running the code given above, you get:
<p>
<code>
Exception in thread "main" java.lang.ArrayStoreException: java.lang.Integer
</code>
<p>
clearly, a run-time type-check is being performed.
<h3>About Covariance, Contravariance, and Invariance</h3>
<p> I've been using these terms, but haven't really defined them.
<p> A type-constructor <code>C</code> is covariant iff:
for all types <code>S,T</code>, if <code>S</code> is a subtype of
<code>T</code> then <code>C<S></code> is a subtype of
<code>C<T></code>.
<p> Similarly, for contravariance, if <code>S</code> is a subtype of
<code>T</code> then <code>C<T></code> is a subtype of
<code>C<S></code>. Note that <code>S</code> and <code>T</code>
have swapped.
<p> Invariance (the name is confusing, I know) is the lack of
covariance or contravariance.
<p> It may be of interest to know that functions are contravariant in
the input, but covariant in the output. Perhaps that is why the Java
folks are not so eager to have function types!
<h2>Broken arrays and generics</h2>
<p>
The brokenness of Java arrays has impacted on the design of Java
generics as well. Consider the following code that might be written
if you were designing your own generic container of some sort:
<p>
<code>
public class Container<T> {
private T[] a;
</code>
<p>
So far so good, right? An array of some parameterized type seems
reasonable.
<p>
<code>
public Container() {
a = (T[])new T[10];
}
}
</code>
<p>
Uh oh!
<p>
<code>
Container.java:5: generic array creation
a = (T[])new T[10];
^
Note: Container.java uses unchecked or unsafe operations.
Note: Recompile with -Xlint:unchecked for details.
1 error
</code>
<p>
No can do. Java is jumping down my throat over this seemingly
reasonable code. Why? Generic arrays are not permitted like this,
because -- as described above -- Java arrays must carry run-time type
information. But Generics promise total type erasure -- meaning NO
run-time type information is kept. Sounds like they designed
themselves into a corner when creating Java, and later, Generics.
<h1>Polymorphism is not Subtyping</h1>
<p>
Returning to the second part of the article, I want to dispel a few
common misperceptions about polymorphism.
<p>
Mark Dominus:
<cite> Then we define a logical negation function, not, which
has type bool → bool, that is, it takes a boolean argument and
returns a boolean result. Since this is a subtype of α → α, we can
store this function in the cell referenced by a.
</cite>
<p> I want to make it absolutely clear that bool → bool is not a
subtype of α → α. There is no notion of subtyping in basic
Hindley-Milner, and it is still not the case even in a "loose" sense.
<p> The easiest way to verify this for yourself is to find your
nearest Standard ML (or similar) prompt and type this:
<p>
<code>
not : 'a -> 'a
</code>
<p> and get a type error. <code>not : bool → bool</code> and it is
not of type α → α.
<p> Now, his example does not typecheck in Standard ML because of the
"value restriction." The value restriction does not exist to preserve
type safety at all, but only to prevent programmer confusion.
<p> This post has gone on too long about Java, so I am afraid it will
wash out the more important points about polymorphism and the value
restriction. I will leave it then, to the next post.http://sequence.complete.org/node/380#commentsType SystemDiscussionSun, 02 Mar 2008 19:54:23 -0800mrd380 at http://sequence.complete.orgStrongly Specified Functions
http://sequence.complete.org/node/363
<h3>How to make writing easy functions hard.</h3>
<p>
Kidding aside, a strongly specified function in Coq not only computes but simultaneously provides a proof of some property.
The construction of the proof is integrated with the body of the function. This differentiates it from "weakly" specified functions, which are written separately from the proof.
</p>
<p>
Proving properties about functions is one of the main practical uses of Coq. Unfortunately, it's more difficult than simply writing code. But it is more interesting (and useful) than, say, simple proofs about the monad laws. One thing I learned, the hard way, is that it's better to start small.
</p>
<p>
The first example to talk about is
<a href="http://www.decidable.org/coq/strongspecs/reverse.html">reverse.</a>
It has a simple inductive definition, so that isn't a big deal, like it can be for some things. It uses basic structural induction on lists, so I don't need to define my own well founded induction principle. Hopefully, it illustrates the meaning of "integrated computation and proof."
</p>
<p>
I've split the property up into two inductive definitions. The first one is the actual "reversed" relation between two lists. Notice that it is not necessarily in <code>Set</code>. The second definition is what actually gets constructed by the function. It says: in order to construct a reversed list of <code>xs</code>, you need to supply a list <code>xs'</code> and a proof of <code>reversed xs xs'</code>.
</p>
<p>
The definition is slightly different than other functions you may have seen in Coq (or not). It actually looks more like a theorem proof than a function. This is a good thing.
</p>
<p>
I use the <code>refine</code> tactic now extensively. This tactic allows you to specify an exact proof term, like <code>exact</code>, but also to leave "jokers" to be filled in by further elaboration. The two spots I've left correspond to the base case and the inductive case, which you should be able to figure out from the <code>match</code>. Side note: the <code>return</code> type-annotation is necessary, otherwise type reconstruction won't figure it out.
</p>
<p>
At this point, I've told Coq that there is a <code>nil</code> case and a "cons" case. The <code>nil</code> case is easy, so I just supplied an exact proof term. You can read that as saying: "The reverse of <code>nil</code> is <code>nil</code> and a proof of this is <code>reversed_nil</code>."
</p>
<p>
For the inductive case, I will use <code>refine</code> again to drill deeper into the function and the proof. Notice that the recursive call is made with <code>match</code>. This allows me to separate the recursively computed value <code>xs'</code> from the usage of the induction hypothesis <code>Hxs'</code>. There is a little trap here -- if you use automation prior to making the recursive call you may get a "Proof completed" message. But you won't be able to save the proof, because the "termination checker" will yell at you. The same thing happens if you incorrectly invoke the recursion.
</p>
<p>
The "proof completed" message isn't bogus -- it's just that the function the <code>auto</code> tactic had proven was not the function for which we were looking.
</p>
<p>
The <code>refine</code> statement provides the necessary hint to the theorem prover. I construct the returned data in a separate <code>refine</code> just to illustrate how you can drill down with the tool. This just leaves the simple proof that <code>xs'++(x::nil)</code> is in fact, the reverse.
</p>
<p>
<code>Recursive Extraction</code> pulls out all the computation-related (stuff in <code>Set</code>) bits while leaving behind the proof-related bits. If you want it in Haskell, as always, change the language with <code>Extraction Language Haskell.</code>
</p>
<p>
(All examples were written using Coq 8.1pl1)
</p>http://sequence.complete.org/node/363#commentsType SystemDiscussionSun, 23 Sep 2007 20:37:24 -0700mrd363 at http://sequence.complete.orgCoq and The Monad Laws: The Third
http://sequence.complete.org/node/360
<h2>The Third Monad Law</h2>
<p>
The previous two articles introduced Coq and the first two Monad Laws.
I am discussing the third one separately because it will take longer
to prove.
</p>
<p>
The proof for the third law will proceed at first like the others,
induction and some simplification.
</p>
<p><code>
Theorem third_law :
forall (A B C : Set) (m : list A)
(f : A -> list B) (g : B -> list C),
bind B C (bind A B m f) g =
bind A C m (fun x => bind B C (f x) g).
Proof.
induction m.
(* base case *)
simpl.
trivial.
(* inductive case *)
intros f g.
simpl.
unfold bind.
unfold bind in IHm.
</code></p>
<p>Which brings us to this state in the interactive theorem prover.</p>
<p><code>
1 subgoal
A : Set
B : Set
C : Set
a : A
m : list A
IHm : forall (f : A -> list B) (g : B -> list C),
flat_map g (flat_map f m) =
flat_map (fun x : A => flat_map g (f x)) m
f : A -> list B
g : B -> list C
============================
flat_map g (f a ++ flat_map f m) =
flat_map g (f a) ++ flat_map (fun x : A => flat_map g (f x)) m
</code></p>
<p>
At this point, if we could rewrite
<code>flat_map g (f a ++ flat_map f m)</code>
into
<code>flat_map g (f a) ++ flat_map g (flat_map f m))</code>
then we would be able to apply the Inductive Hypothesis and
be home free.
</p>
<p>
The "cut" tactic allows you to make an assumption, and then later
come back and prove your assumption correct. Using "cut",
</p>
<p><code>
cut (flat_map g (f a ++ flat_map f m) =
flat_map g (f a) ++ flat_map g (flat_map f m)).
intro Distrib.
rewrite Distrib.
rewrite IHm.
reflexivity.
</code></p>
<p>
the original goal is easily solved. But Coq has generated an
additional subgoal: we must now prove that this cut is correct.
</p>
<p><code>
1 subgoal
A : Set
B : Set
C : Set
a : A
m : list A
IHm : forall (f : A -> list B) (g : B -> list C),
flat_map g (flat_map f m) =
flat_map (fun x : A => flat_map g (f x)) m
f : A -> list B
g : B -> list C
============================
flat_map g (f a ++ flat_map f m) =
flat_map g (f a) ++ flat_map g (flat_map f m)
</code></p>
<p>
We'll proceed by induction on <code>f a</code> which
has inductive type <code>list B</code>.
</p>
<p><code>
induction (f a).
(* base case *)
simpl.
reflexivity.
(* inductive case *)
simpl.
rewrite IHl.
rewrite app_ass.
reflexivity.
Qed.
End ListMonad.
</code></p>
<p>
All done. We only needed the association property of list append,
which I found by querying <code>SearchAbout app.</code>
</p>
<h2>Summary</h2>
<p>
Here is a much shorter proof which takes advantage of some of Coq's
automated tactics.
</p>
<p><code>
Theorem third_law' :
forall (A B C : Set) (m : list A)
(f : A -> list B) (g : B -> list C),
bind B C (bind A B m f) g =
bind A C m (fun x => bind B C (f x) g).
Proof.
induction m; simpl; intuition.
replace (bind B C (f a ++ bind A B m f) g)
with (bind B C (f a) g ++ bind B C (bind A B m f) g);
[ rewrite IHm
| induction (f a); simpl; auto;
rewrite app_ass; rewrite IHl
]; auto.
Qed.
</code></p>
<p>
On a final note, Coq has the ability to extract code into
several different languages.
</p>
<p><code>
Extraction Language Haskell.
Recursive Extraction bind ret.
</code></p>
<p>results in</p>
<p><code>
module Main where
import qualified Prelude
data List a = Nil | Cons a (List a)
app l m = case l of
Nil -> m
Cons a l1 -> Cons a (app l1 m)
flat_map f l = case l of
Nil -> Nil
Cons x t -> app (f x) (flat_map f t)
ret :: a1 -> List a1
ret a = Cons a Nil
bind :: (List a1) -> (a1 -> List a2) -> List a2
bind m f = flat_map f m
</code></p>http://sequence.complete.org/node/360#commentsType SystemDiscussionThu, 16 Aug 2007 13:18:03 -0700mrd360 at http://sequence.complete.orgCoq and The Monad Laws: The First and Second
http://sequence.complete.org/node/359
<h2>The First Monad Law</h2>
<p>
In the previous article, I gave a brief introduction to Coq and the List Monad.
I listed three Theorems but did not prove them. Now I will show how to prove the
first two Theorems.
</p>
<p>
Here is the context of our code:
</p>
<p>
<code>
Section ListMonad.
Require Import List.
Definition ret (A : Set) (a : A) := a :: nil.
Definition bind (A B : Set) (m : list A) (f : A -> list B) :=
flat_map f m.
</code>
</p>
<p>The first Theorem, restated:</p>
<p>
<code>
Theorem first_law : forall (A B : Set) (a : A) (f : A -> list B),
bind A B (ret A a) f = f a.
</code>
</p>
<p>At this point, Coq will enter an interactive proof mode. In
<code>coqtop</code>, the prompt will change accordingly. In
ProofGeneral Emacs, a new window will pop up. In either case,
you will be presented with a view of the current assumptions and goals
that looks like this:
</p>
<p>
<code>
1 subgoal
============================
forall (A B : Set) (a : A) (f : A -> list B), bind A B (ret A a) f = f a
</code>
</p>
<p>There are no assumptions, and one goal to prove.</p>
<p>
<code>
Proof.
intros A B a f.
</code>
</p>
<p>Coq proofs can proceed with "tactics" which instruct the
proof engine to perform a certain operation, or series of operations.
In this case, we are introducing assumptions named <code>A,B,a,f</code>
drawn from the antecedents of the goal. Now the proof view looks like this:</p>
<p><code>
1 subgoal
A : Set
B : Set
a : A
f : A -> list B
============================
bind A B (ret A a) f = f a
</code></p>
<p>The "unfold" tactic takes the definition of the given
argument and tries to substitute it directly into the goal.</p>
<p><code>
unfold bind.
unfold ret.
unfold flat_map.
</code></p>
<p>Now we have a vastly simplified goal.</p>
<p><code>
1 subgoal
A : Set
B : Set
a : A
f : A -> list B
============================
f a ++ nil = f a
</code></p>
<p>Everyone knows that appending <code>nil</code> to a list returns the same list.
But how do we explain this to Coq? Search and find out if Coq already
knows about this fact.</p>
<p><code>
SearchRewrite (_ ++ nil).
</code></p>
<p><code>SearchRewrite</code> is used to search for theorems of the form
<code>a = b</code> which can be used to rewrite portions of your goal.
I use the wildcard <code>_</code> to indicate that I want to find any
theorem involving appending <code>nil</code> to anything. Coq finds:
</p>
<p><code>
app_nil_end: forall (A : Set) (l : list A), l = l ++ nil
</code></p>
<p>
We can use this with the "rewrite" tactic. We want to
rewrite our code which looks like <code>l ++ nil</code>
into <code>l</code>. That is a right-to-left rewrite,
which is specified like so:
</p>
<p><code>
rewrite <- app_nil_end.
</code></p>
<p>And finally we have a goal <code>f a = f a</code> which
Coq can trivially solve with</p>
<p><code>
reflexivity.
</code></p>
<p>Or any of the more automated tactics, like "trivial" or "auto".</p>
<p><code>
Qed.
</code></p>
<p>The first monad law is discharged and proven. You can run
<code>Check first_law.</code> to see the new theorem.</p>
<h2>The Second Monad Law</h2>
<p>This second proof will proceed by induction on the structure of the
list parameter <code>m</code>. The "induction" tactic causes Coq to
try a proof by induction. The proof will be split into two parts,
based on case breakdown of the list type.
<p><code>
Theorem second_law : forall (A : Set) (m : list A),
bind A A m (ret A) = m.
Proof.
induction m.
</code></p>
The first is the base case for <code>nil</code>, and
the second is the inductive case for <code>cons</code>.
<p><code>
2 subgoals
A : Set
============================
bind A A nil (ret A) = nil
subgoal 2 is:
bind A A (a :: m) (ret A) = a :: m
</code></p>
<p>The base case is easy, the "simpl" tactic
can take care of the unfolding and simplification.</p>
<p><code>
simpl.
reflexivity.
</code></p>
<p>The inductive case is also straightforward, because
after the "simpl" step, the Inductive Hypothesis is
already applicable. The appropriate <code>IHm</code>
assumption is provided by the "induction" tactic.</p>
<p><code>
simpl.
rewrite IHm.
reflexivity.
Qed.
</code></p>
<h2>Summary</h2>
<p>The full proofs, minus the commentary:</p>
<p><code>
Theorem first_law : forall (A B : Set) (a : A) (f : A -> list B),
bind A B (ret A a) f = f a.
Proof.
intros A B a f.
unfold bind.
unfold ret.
unfold flat_map.
rewrite <- app_nil_end.
reflexivity.
Qed.
Theorem second_law : forall (A : Set) (m : list A),
bind A A m (ret A) = m.
Proof.
induction m.
(* base case *)
simpl.
reflexivity.
(* inductive case *)
simpl.
rewrite IHm.
reflexivity.
Qed.
</code></p>
<p>Next time, the Third Monad Law, and a more complicated proof.</p>http://sequence.complete.org/node/359#commentsType SystemDiscussionWed, 15 Aug 2007 15:14:06 -0700mrd359 at http://sequence.complete.orgCoq and The Monad Laws: Introduction
http://sequence.complete.org/node/358
<p>
Over the past few months I've been spending some time with <a href="http://coq.inria.fr/">Coq</a>, the proof assistant.
I am still very much a beginner, so as an exercise I decided to try and prove the three <a href="http://haskell.org/haskellwiki/Monad_Laws">Monad laws</a> for the implementation of the List Monad.
</p>
<p>
In Haskell, the List Monad looks like this:
</p>
<p>
<code>
instance Monad [] where
return x = [x]
m >>= f = concatMap f m
</code>
</p>
<p>
Coq is a proof-assistant for a constructive logic, so it can also be used as a (dependently-typed) functional programming language.
You can open an interactive session and type the lines in using <code>coqtop</code>. However, I use the ProofGeneral Emacs-based interface and save my code in a file like "listmonad.v". You can then step through the lines using <code>C-c C-n</code>. There also exists a custom GUI called CoqIDE.
</p>
<p>
In Coq, I defined the Monad methods like so:
</p>
<p>
<code>
Section ListMonad.
Require Import List.
Definition ret (A : Set) (a : A) := a :: nil.
Definition bind (A B : Set) (m : list A) (f : A -> list B) :=
flat_map f m.
</code>
</p>
<p>
I opened up a new section (but did not close it yet). The definitions are fairly unremarkable, except that now the types A and B are explicit parameters. In Coq, a single colon is used for type annotations, and a double colon is used for list consing. <code>flat_map</code> is Coq's version of <code>concatMap</code>. I did not know that before, but found it by using <code>SearchAbout list.</code> at the interactive Coq prompt.
</p>
<p>
I can see what Coq thinks of my definitions:
</p>
<p>
<code>
Coq < Check ret.
ret
: forall A : Set, A -> list A
Coq < Check bind.
bind
: forall A B : Set, list A -> (A -> list B) -> list B
</code>
</p>
<p>
Now we are ready to state the three Monad Laws. First, in Haskell notation:
</p>
<p>
<code>
1. return a >>= f = f a
2. m >>= return = m
3. (m >>= f) >>= g = m >>= (\x -> f x >>= g)
</code>
</p>
<p>
The first two Monad laws, in essence, assert that return is an identity operation on the left and right sides of bind. The third says that bind is associative.
</p>
<p>
In Coq:
</p>
<p>
<code>
Theorem first_law :
forall (A B : Set) (a : A) (f : A -> list B),
bind A B (ret A a) f = f a.
Theorem second_law :
forall (A : Set) (m : list A),
bind A A m (ret A) = m.
Theorem third_law :
forall (A B C : Set) (m : list A)
(f : A -> list B) (g : B -> list C),
bind B C (bind A B m f) g =
bind A C m (fun x => bind B C (f x) g).
</code>
</p>
<p>
Entering any of these Theorems into Coq will engage the interactive proof mode. In the next article, I will examine the proofs of the first two laws.
</p>http://sequence.complete.org/node/358#commentsType SystemDiscussionWed, 15 Aug 2007 13:26:58 -0700mrd358 at http://sequence.complete.orgSearch by Type
http://sequence.complete.org/node/134
Here's a random thought. What if we had a way to search for Haskell functions by type signature. Take the <a href="http://www.haskell.org/ghc/docs/latest/html/libraries/index.html">Hierarchical Libraries</a>, slice/dice/index the type signatures so that you could search for "Num b => a -> b" and it would present you with a list of functions like <code>length :: [a] -> Int</code>. You'd also probably want to allow swizzling of the input arguments so that you could find "map" with a search like "[a] -> (a -> b) -> [b]". Then the next step would be to analyze your existing code to see if you've reinvented any wheels.http://sequence.complete.org/node/134#commentsType SystemMon, 05 Dec 2005 10:37:39 -0800Greg Buchholz134 at http://sequence.complete.orgComputer Assisted Programming
http://sequence.complete.org/node/124
<p>Thought I'd post a link to one of my <a href="http://kerneltrap.org/node/5591">static typing rants</a>, since there will probably be more "true believers" on this site.</p>
http://sequence.complete.org/node/124#commentsType SystemDiscussionThu, 20 Oct 2005 13:46:05 -0700Greg Buchholz124 at http://sequence.complete.orgHaskell puts the complexity of programming in the right place
http://sequence.complete.org/node/31
<p>Every useful programming language has an area of complexity that you must overcome to become comfortable with the language. For Perl, you need to relax your ideas about regularity and consistency and learn to read and write Perl instinctively and trust that all of the exceptions and special cases will make as much sense in Perl as in English.</p>
<p>Haskell is very much a "back-end" language. What I mean is that the language just sits back and waits until you have everything lined up in a clean chain of well-typed functions. It won't do anything but keep spitting back your code at your until you have your problem reduced to something expressible in expressions.</p>
<p>This means you spend a lot of time with the type checker. And possibly a lot of time with making sure that your IO can make it through the snake's tube of a Monad before getting into yor program. </p>
<p>So, Haskell can be a big turn-off to someone who needs a language which, by Haskell standards, oversteps it bounds. If you want to mix IO and your program, If you want to quickly setup a webshop and need easy CGI processing, or any of a number of things that are highly available in languages like Perl, Python, Ruby, Tcl, PHP, then Haskell seems like a huge stumbling block.</p>
<p>But the way that Haskell quarantines I/O. The hurdles that it puts you through are there for some very very good reasons. The more experienced you become with it's advanced features such as Monads and combinators, the more you see how to crisply and accurately separate wheat from chaff, cause from effect, and smoke from fire.</p>
http://sequence.complete.org/node/31#commentsType SystemDiscussionSun, 13 Mar 2005 06:39:39 -0800metaperl31 at http://sequence.complete.org