# Re: [isabelle] Quotients and code generation for higher-order functions

```Dear Andreas,

```
Many thanks for your patience and comprehensive explanations, I now do realize my previous mistakes. My current impression is that, given a term
```
Go (Îx. bind (c x) f)

```
"bind (c x) f" is of type foo. If "f" is of type "'a => 'b bar", we have to find a mapping of type "bar => foo" somehow and somewhere, which leads directly to the problem you were talking about. Therefore, it seems that the problem is really about the function "f" of type "'a => 'b bar".
```
A workaround may be to let "f" be of type "'a => 'a foo". That is,

```
lift_definition bind_bar :: "'a bar â ('a â 'b foo) â 'b bar" is bind sorry
```
should have it code equations directly.

```
Let me know if there is any other solutions. I am really interested in this topic :-)
```
Best,
Wenda

On 2015-07-24 13:16, Andreas Lochbihler wrote:
```
```Dear Wenda,

code_abstype and code abstract are normally used for types carved out
as a subset of some other type. For rat, e.g., the subset consists of
all the relatively prime pairs of integers where the second component
is positive. Although rat is constructed as a quotient from pairs of
integers, the code generator setup considers rat as isomorphic to the
subset of pairs described in the previous sentence, i.e., we have
carved out a subset. All code equations ensure that only such
normalised pairs occur at run-time, i.e., we have canonical
representatives. Conversely, the code equations can exploit that
rational numbers given as arguments are always normalised.

With code_datatype, you do not need canonical representatives. In
fact, you cannot even exploit them. As the code equations
pattern-match on the pseudo-constructors, all elements in the domain
of the pseudo-constructor are considered valid representations. Thus,
you cannot rely in the code equation on any invariant. For
pseudo-constructors, the challenge lies in defining the functions on
the abstract type such that the desired code equation is provable.

```
lemma [code]:"bind_bar (abs_bar x) f = abs_bar (bind x (Îu. rep_bar (f u)))" sorry
```
If we declare code_datatype abs_bar, then we can use this lemma as a
code equation. However, we also must provide a code equation for
rep_bar. Obviously, we would like to have the following equation

rep_bar (abs_bar x) = x

However, this equation expresses that abs_bar does not discard any
information in x, i.e., abs_bar is injective. As bar is a
(non-trivial) quotient of foo, abs_bar clearly is not injective, as it
maps every x to the equivalence class of x. In fact, we cannot define
any function f such that f (abs_bar x) = x. Thus, we cannot find any
such unpacking function rep_bar at all. The only reason giving me some
hope in the case above is that ultimately, we are applying abs_bar on
the right-hand side again, i.e., we are ultimately throwing away all
the extra information contained in the raw type.

Best,
Andreas

PS: As bar is a quotient of foo, we know that "abs_bar (rep_bar x) =
x", but this kind of equation is only suitable for code_abstract,
which requires canonical representatives.

On 24/07/15 13:38, Wenda Li wrote:
```
```Dear Andreas,

```
Thanks for reminding me of the non-canonical representation, I will have a similar issue in my formalization :-) However, in my understanding, a canonical representation is important when using "code abstype"/"code abstract" to restore executability (e.g. Rat.thy Polynomial.thy), while with "code_datatype" it seems that we can deal with executability
```in a more flexible way (e.g. Real.thy).

In this case, if we can prove

```
lemma [code]:"bind_bar (abs_bar x) f = abs_bar (bind x (Îu. rep_bar (f u)))" sorry
```
we should be able to evaluate

value "bind_bar (abs_bar (Stop (1::nat))) (Îu. abs_bar(Stop (u+2)))"

```
value "bind_bar (abs_bar (Go (Îx::nat. Stop (x+2)))) (Îu. abs_bar(Stop (u+2)))"
```
Please correct me if I am wrong at any point.

Wenda

On 2015-07-24 12:14, Andreas Lochbihler wrote:
```
```Hi Wenda,

On 24/07/15 13:05, Wenda Li wrote:
```
lemma [code]:"bind_bar (abs_bar x) f = abs_bar (bind x (Îu. rep_bar (f u)))" sorry
```
Can you prove this in your theory?
```
```Of course, this type-checks and is provable. However, I'd need a code
equation for rep_bar in the form "rep_bar (Abs_bar x) = ...". And for
this, I'd need a notion of canonical representative in the raw type,
which I don't have at the moment. It would require a lot of work (in
Isabelle) to define such a notion. Moreover, the generated code would
have to transform everything into the normal form, which can be
computationally prohibitive.

Best,
Andreas

```
```On 2015-07-24 07:51, Andreas Lochbihler wrote:
```
```Dear all,

I am currently stuck at setting up code generation for a quotient
```
type. To that end, I have declared an embedding from the raw type to the quotient type as pseudo-constructor with code_datatype and am now
```trying to prove equations that pattern-match on the
```
pseudo-constructor. There are no canonical representatives in my raw type, so I cannot define an executable function from the quotient type
```to the raw type.

```
I am stuck at lifting functions in which the quotient type occurs as
```the result of higher-order functions. The problem is that I do not
```
know how to pattern-match on a pseudo-constructor in the result of a
```function. Here is an example:

datatype 'a foo = Stop 'a | Go "nat â 'a foo"

primrec bind :: "'a foo â ('a â 'b foo) â 'b foo" where
"bind (Stop x) f = f x"
| "bind (Go c) f = Go (Îx. bind (c x) f)"

```
axiomatization rel :: "'a foo â 'a foo â bool" where rel_eq: "equivp rel"
```  quotient_type 'a bar = "'a foo" / rel by(rule rel_eq)
code_datatype abs_bar

```
lift_definition bind_bar :: "'a bar â ('a â 'b bar) â 'b bar" is bind sorry
```
```
My problem is now to state and prove code equations for bind_bar. Obviously,
```
lemma "bind_bar (abs_bar x) f = abs_bar (bind x f)"

```
does not work, as bind expects f to return a foo, but f returns a bar. My next attempt is to inline the recursion of bind. The case for Stop
```is easy, but I am out of ideas for Go.

lemma "bind_bar (abs_bar (Stop x)) f = f x"
"bind_bar (abs_bar (Go c)) f = ???"

```
Is there a solution to my problem? Or am I completely on the wrong track.
```
Thanks for any ideas,
Andreas
```
```
```
```
```
```
--
Wenda Li
PhD Candidate
Computer Laboratory
University of Cambridge

```

This archive was generated by a fusion of Pipermail (Mailman edition) and MHonArc.