### Re: [Epigram] Definitional equality in observational type theory

```
Hi Bas,

thank you - indeed this is a very interesting observation. This was
the proof I had in mind when I first claimed that the axiom of choice
is provable in OTT. However, I was deluded in believing that OTT/
predicative topoi exactly characterize the theory of the setoid
model. As you clearly point out this is not the case. To me this is
an incompleteness of the formal system (OTT/predicative topoi) wrt
the intended interpretation. This should be fixable, leading to OTT+X
or predicative topoi + X which should be complete wrt the setoid
model. More categorically we could say that X characterizes the
predicative topoi obtained by an exact completion of an LCCC - I think.

Indeed, in the setoid model we can construct a function

f : A - [B]
--
lift f : [A - B]

if the setoid A is trivial, i.e. has the identity as its
equivalence relation. The construction is exactly the one you point
out  it actually corresponds to my previous informal explanation why
this is not unreasonable:

Note that this is not completely unreasonable: we observe the
hidden choice made by f, but we compensate by this by hiding our
knowledge.

Which setoids have a trivial equality? Certainly all first order
types. However, if we start with an extensional theory (which can be
justified with the setoid model) than also higher types N - N have a
trivial equality (indeed the extensional equality here is the
finest equality). Hence we certainly get considerable more than
countable choice.

Actually, instead of using only [..] we can formulate a more general
operator for quotient types, given an equivalence relation ~ : B - B
- Prop, we define ~' : (A - B) - (A-B)-Prop as f ~' g = forall
a:A.f a ~ g a. We obtain the following generalisation:

f : A - B/~
-
lift f : (A - B)/~'

But hang on - what stops us from doing a Diaconescu? Excluded middle
doesn't hold in the setoid model, even though we only get P \/ not P
where A \/ B = [A + B].  But this would still require that we have P
+ not P in the underlying set. So what goes wrong?

We define the type of non-empty subsets of Bool as

NE = Sigma Q : Bool- Prop.Sigma x:Bool.Q x

We define the equivalence relation of extensional equality of subsets

~ : NE - NE - Prop

(Q,_) ~ (R,_) = forall b:Bool.Q b - R b

Now we derive:

h : NE/~ - [ NE ]

which is just the indentity on the underlying elements. The point is
that obviously ~ implies the trivial equality of [..].

Now, if we were able to lift h we get

lift h : [ NE/~ - NE]

Obviously, NE/~ isn't a setoid with a trivial equality!

The Diaconescu argument shows that we can prove for any P:Prop

H : NE/~ - NE
--
Dia H : P \/ not P = [P + not P]

and combining the two using bind we get

(lift h) = Dia : P \/ not P

see below for a Epigram 2+n style proof of Dia.

We can also see what goes wrong in general: The principle

f : A - B/~
-
lift f : (A - B)/~'

fails because given the setoid A = (A0,~A) the premise gives us an
underlying function f0:A0 - B (for simplicity we assume that B is
trivial) s.t. a ~A b implies f0 a ~B f0 b, where to use the same
function to construct an element of  A - B we need to show that  a
~A b implies f0 a = f0 b, and there is no reason to believe this -
unless ~A is the equality.

The main question is how to characterize abstractly the types A for
which lift is valid. Syntactically we could say All types not
containing quotients but this is not very nice. I'd like a semantic
condition for the type.

Cheers,
Thorsten

P.S.
For completeness: Dia itself can be constructed as in the COQ script:
we use

T,F : Bool - Prop

T b = (b=true) \/ P
F b = (b=false) \/ P

by applying H to the equivalence classe T,F and projecting out
the components we get

t,f : Bool
t = fst (H T)
f = fst (H F)
snd (H T) : t=true \/ P
snd (H F) : f=false \/ P

By analyzing the cases of the propositional components we get two
cases in which we can prove P (hence we are done with P \/ not P) and
one where we have

t=true
f=false

In this case we can prove not P: we assume p:P and using this we can
prove T b, F b for any b and hence T b - F b and therefore
T=F but then t=f and true=false and we have derived a contradiction.

On 1 Feb 2007, at 03:10, Bas Spitters wrote:

Hi Thorsten and others,

On Tuesday 30 January 2007 21:22:53 [EMAIL PROTECTED] wrote:
I do indeed think that Observational Type Theory with quotient
types should
be the language of a Predicative Topos. I don't see in the moment
that the
setoid model would introduce anything which isn't provable in the
Type
Theory and at least in the moment I don't see how to prove the
countable

axiom of choice in OTT. Bas, could you ```

### Re: [Epigram] Definitional equality in observational type theory

```Hi Thorsten,

On Thursday 01 February 2007 08:55:30 Thorsten Altenkirch wrote:
thank you - indeed this is a very interesting observation. This was
the proof I had in mind when I first claimed that the axiom of choice
is provable in OTT. However, I was deluded in believing that OTT/
predicative topoi exactly characterize the theory of the setoid
model. As you clearly point out this is not the case. To me this is
an incompleteness of the formal system (OTT/predicative topoi) wrt
the intended interpretation. This should be fixable, leading to OTT+X
or predicative topoi + X which should be complete wrt the setoid
model. More categorically we could say that X characterizes the
predicative topoi obtained by an exact completion of an LCCC - I think.

First of all, I am not sure whether a fix is needed. My recent experience in
constructive mathematics is that it is actually quite pleasant to work
without countable choice. Initial experience shows that it may lead to better
algorithms implicit in the proofs. (Most of this is in my work with Thierry
Coquand.) As you know, choice breaks abstract datatypes. See for instance the
encoding of ADT as existential types by Mitchell and Plotkin. This is not
possible in dependent type theory precisely because we have choice (this was
pointed out to me by Jesper Carlstrom some time ago). I understand one
motivation for your work on OTT as an attempt to bring back (some?) ADTs in
type theory. It would be a pity to throw them out again without a good
reason.

Having said that here are some quick remarks on your proposal below.
Having choice for N-N as somewhat uncommon. It is not present in Bishop's
constructive mathematics, i.e. more or less the setoid model of ML type
theory. However, it is present in Brouwerian intuitionism and in some
realizability models. Brouwer does not have choice for (N-N)-N.

Martin Hofmann has some discussion on choice and extensionality in his thesis,
but uses a syntactic criterion somewhat like the one that you outline.

One semantic candidate for X could be `all types have a projective cover'.
This is what is used in realizability theory and what Erik Palmgren
translated to type theory. A type P is projective if for all f:A-B and every
function g:P-B there is a function h:P-A with f o h = g. Projective
types have choice. The idea is that the projective types are Bishop's
pre-sets or the types in the setoid construction.

Best,

Bas

---
Research Group Foundations/ Institute for Computing and Information Sciences

```

### Re: [Epigram] Definitional equality in observational type theory

```
Hi Thorsten

I'm not sure I understand what's going on here.

Thorsten Altenkirch wrote:

Indeed, in the setoid model we can construct a function

f : A - [B]
--
lift f : [A - B]

if the setoid A is trivial, i.e. has the identity as its equivalence
relation.

I can do it if A is decided. If you give me a : A, then I pick

lift f = iI const (f a) Ii

or, in less idiomatic longhand

lift f = do
c - return const
b - f a
return (c b)

If you give me A - 0, it's easy.

f : A - B/~
-
lift f : (A - B)/~'

Again, if you give me some a : A, I say

lift f = const (f a)  -- const respects the equivalence because  const
b a ~' const b' a if b ~ b'

But hang on - what stops us from doing a Diaconescu? Excluded middle
doesn't hold in the setoid model, even though we only get P \/ not P
where A \/ B = [A + B].  But this would still require that we have P +
not P in the underlying set. So what goes wrong?

We define the type of non-empty subsets of Bool as

NE = Sigma Q : Bool- Prop.Sigma x:Bool.Q x

Note that NE is inhabited, eg by

moo = (const 1; true; ()).

We define the equivalence relation of extensional equality of subsets

~ : NE - NE - Prop

(Q,_) ~ (R,_) = forall b:Bool.Q b - R b

Now we derive:

h : NE/~ - [ NE ]

which is just the indentity on the underlying elements. The point is
that obviously ~ implies the trivial equality of [..].

Now, if we were able to lift h we get

lift h : [ NE/~ - NE]

return (const moo) : [ NE/~ - NE]

Obviously, NE/~ isn't a setoid with a trivial equality!

The Diaconescu argument shows that we can prove for any P:Prop

H : NE/~ - NE
--
Dia H : P \/ not P = [P + not P]

Dia (const moo) : P \/ not P

see below for a Epigram 2+n style proof of Dia.

Under the circumstances, I really hope you've cocked this up.

We can also see what goes wrong in general: The principle

f : A - B/~
-
lift f : (A - B)/~'

fails because given the setoid A = (A0,~A) the premise gives us an
underlying function f0:A0 - B (for simplicity we assume that B is
trivial) s.t. a ~A b implies f0 a ~B f0 b, where to use the same
function to construct an element of  A - B we need to show that  a ~A
b implies f0 a = f0 b, and there is no reason to believe this - unless
~A is the equality.

I just about swallow that. But if A0 is decided, you need to use the
spec part (which you're eliding for simplicity) to force me to use f0 in
the way you're suggesting.

P.S.
For completeness: Dia itself can be constructed as in the COQ script:
we use

T,F : Bool - Prop

T b = (b=true) \/ P
F b = (b=false) \/ P

by applying H to the equivalence classe T,F and projecting out the
components we get

t,f : Bool
t = fst (H T)
f = fst (H F)
snd (H T) : t=true \/ P
snd (H F) : f=false \/ P

Are you sure? I thought

H had return type NE, giving
(P; a; p) = fst (H T), (Q, b, q) := fst(H F) : Bool - Prop
where a,b : Bool, p : P a, q : Q b

but no connection necessary between T and P, F and Q.

Now, I know that my pathological functions are not the functions you're
thinking of, but it does rather show that the real action, whatever it
is, lies in the spec parts which you're throwing away.

More later

Conor

PS I blogged a bit about implementing OTT...

```

### Re: [Epigram] Definitional equality in observational type theory

```
Hi Conor,

I'm not sure I understand what's going on here.

This happens if you don't type check your definitions. I got carried
away with my non-dependent simplification of the story.

Thank you for actually reading it.

Indeed, in the setoid model we can construct a function

f : A - [B]
--
lift f : [A - B]
if the setoid A is trivial, i.e. has the identity as its
equivalence relation.

I can do it if A is decided. If you give me a : A, then I pick

lift f = iI const (f a) Ii

Indeed, however your trick won't work for the dependent version:

f : Pi a:A.[B a]
-
lift f: [Pi a:A.B a]

f : A - B/~
-
lift f : (A - B)/~'

Again, if you give me some a : A, I say

lift f = const (f a)  -- const respects the equivalence because
const b a ~' const b' a if b ~ b'

Dito. We need

f : Pi a:A.(B a)/(~ a)
-
lift f: (Pi a:A.B a)/~'

where ~' is defined as before: f ~' g = Pi a:A.f a ~ g a but for
f,g : Pi a:A.B a and ~ is actually a family of equivalence relations
~' : Pi a:A.(B a) - (B a) - Prop.

We define the type of non-empty subsets of Bool as

NE = Sigma Q : Bool- Prop.Sigma x:Bool.Q x

Note that NE is inhabited, eg by

moo = (const 1; true; ()).

We define the equivalence relation of extensional equality of subsets

~ : NE - NE - Prop

(Q,_) ~ (R,_) = forall b:Bool.Q b - R b

Now we derive:

h : NE/~ - [ NE ]

We define

h0 : Pi (P,_):NE . [ Sigma b:2.P b ]
h0 (P,p) = return p

and observe that this trivially preserves ~ and hence we obtain

h : Pi (P,_):NE/~ . [ Sigma b:2.P b ]

Now, if we were able to lift h we get

lift h : [ NE/~ - NE]

return (const moo) : [ NE/~ - NE]

lift h : [ Pi (P,_):NE/~ . Sigma b:2.P b ]

and your version ceases to work.

Obviously, NE/~ isn't a setoid with a trivial equality!

The Diaconescu argument shows that we can prove for any P:Prop

H : NE/~ - NE
--
Dia H : P \/ not P = [P + not P]

Corrected:

H : Pi (P,_):NE/~ . Sigma b:2.P b

Dia H : P \/ not P = [P + not P]

Dia (const moo) : P \/ not P

see below for a Epigram 2+n style proof of Dia.

Under the circumstances, I really hope you've cocked this up.

I have.

We can also see what goes wrong in general: The principle

f : A - B/~
-
lift f : (A - B)/~'

fails because given the setoid A = (A0,~A) the premise gives us an
underlying function f0:A0 - B (for simplicity we assume that B is
trivial) s.t. a ~A b implies f0 a ~B f0 b, where to use the same
function to construct an element of  A - B we need to show that
a ~A b implies f0 a = f0 b, and there is no reason to believe this
- unless ~A is the equality.

I just about swallow that. But if A0 is decided, you need to use
the spec part (which you're eliding for simplicity) to force me to
use f0 in the way you're suggesting.

You are right. The reasoning only works for the dependent version.

P.S.
For completeness: Dia itself can be constructed as in the COQ
script: we use

T,F : Bool - Prop

T b = (b=true) \/ P
F b = (b=false) \/ P

by applying H to the equivalence classe T,F and projecting out
the components we get

t,f : Bool
t = fst (H T)
f = fst (H F)
snd (H T) : t=true \/ P
snd (H F) : f=false \/ P

Are you sure? I thought

H had return type NE, giving
(P; a; p) = fst (H T), (Q, b, q) := fst(H F) : Bool - Prop
where a,b : Bool, p : P a, q : Q b

but no connection necessary between T and P, F and Q.

This should now work with the correct type of H.

Cheers,
Thorsten

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may still contain software viruses, which could damage your computer system:
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University of Nottingham may be monitored as permitted by UK legislation.

```

### Re: [Epigram] Definitional equality in observational type theory

```
Hi Bas,

thank you - indeed this is a very interesting observation. This was
the proof I had in mind when I first claimed that the axiom of choice
is provable in OTT. However, I was deluded in believing that OTT/
predicative topoi exactly characterize the theory of the setoid
model. As you clearly point out this is not the case. To me this is
an incompleteness of the formal system (OTT/predicative topoi) wrt
the intended interpretation. This should be fixable, leading to OTT+X
or predicative topoi + X which should be complete wrt the setoid
model. More categorically we could say that X characterizes the
predicative topoi obtained by an exact completion of an LCCC - I
think.

First of all, I am not sure whether a fix is needed. My recent
experience in

constructive mathematics is that it is actually quite pleasant to work
without countable choice. Initial experience shows that it may lead
to better
algorithms implicit in the proofs. (Most of this is in my work with
Thierry
Coquand.) As you know, choice breaks abstract datatypes. See for
instance the
encoding of ADT as existential types by Mitchell and Plotkin. This
is not
possible in dependent type theory precisely because we have choice
(this was

pointed out to me by Jesper Carlstrom some time ago). I understand one
motivation for your work on OTT as an attempt to bring back (some?)

type theory. It would be a pity to throw them out again without a good
reason.

I'd like to understand this better - in the moment I am unconvinced.

f : Pi a:A.(B a)/(~ a)
-
lift f: (Pi a:A.B a)/~'

for discrete types A, which means that I work in the internal
language of the setoid model. Quotient types, even with lift, seem to
be a good way to capture ADTs.

Also if you want to reject lift, it would be good to have a model
construction (maybe like the setoid model) which refutes it. Other
things which you want may become true in this model.

Having said that here are some quick remarks on your proposal below.
Having choice for N-N as somewhat uncommon. It is not present in
Bishop's
constructive mathematics, i.e. more or less the setoid model of ML
type

theory. However, it is present in Brouwerian intuitionism and in some
realizability models. Brouwer does not have choice for (N-N)-N.

Martin Hofmann has some discussion on choice and extensionality in
his thesis,

but uses a syntactic criterion somewhat like the one that you outline.

I alreay realized that I should reread Martin's thesis.

One semantic candidate for X could be `all types have a projective
cover'.

This is what is used in realizability theory and what Erik Palmgren
translated to type theory. A type P is projective if for all f:A-B
and every

function g:P-B there is a function h:P-A with f o h = g. Projective
types have choice. The idea is that the projective types are
Bishop's

pre-sets or the types in the setoid construction.

Cheers,
Thorsten

This message has been checked for viruses but the contents of an attachment
may still contain software viruses, which could damage your computer system:
you are advised to perform your own checks. Email communications with the
University of Nottingham may be monitored as permitted by UK legislation.

```

### Re: [Epigram] Definitional equality in observational type theory

```Hi Thorsten and others,

On Tuesday 30 January 2007 21:22:53 [EMAIL PROTECTED] wrote:
I do indeed think that Observational Type Theory with quotient types should
be the language of a Predicative Topos. I don't see in the moment that the
setoid model would introduce anything which isn't provable in the Type
Theory and at least in the moment I don't see how to prove the countable
axiom of choice in OTT. Bas, could you explain, please?

Assuming that OTT is the internal type theory of a predicative topos, say a
PiW-pretopos (which it seems to be the case), it is not possible to prove
countable choice in OTT. The simple reason is that every topos is a
PiW-pretopos and that countable choice(CAC) does not hold in the topos of
sheaves over the reals. Therefore CAC can not hold in the internal type
theory (i.e. OTT).

Now consider the setoid model. The crucial observation is that the setoid
Nat=(N,=_Nat) carries the finest possible equality.
Let X=(X',=_X) be a setoid.
If we know:
forall n in Nat there exists x in X such that (R n x),
then, by the construction of the setoid model, there is a function on the
underlying types f:N-X' such that
forall n, (R n (f n))
because Nat carries the finest equality f lifts to a function on the setoid
level since:
If n=_Nat m, then (f n)=_X (f m).
Write f':Nat = X for this lifted function. Then
forall n in Nat, (R n (f' n)).
Thus proving countable choice in the setoid model.

This is a well-known argument in Bishop's constructive mathematics. Bishop's
treatment of sets lead to the development of setoids by Hofmann. The lecture
notes of the TYPES summer school in Chalmers by Erik Palmgren give a nice
presentation of Bishop set theory in type theory.

Bas

```

### Re: [Epigram] Definitional equality in observational type theory

```
Hi Robin,

The paper says Observational reasoning can be simulated in
Intensional

Type Theory by the use of setoids, i.e. types with an explicit
equivalence relation.

I'm not clear on what the differences are between OTT and simulating
observational reasoning with setoids in ITT. Clearly OTT has an
explicit equivalence relation (=) which is polymorphic over
(certain)

types. So does the paper mean explicit in some other sense?

I am not sure I understand your question. Let my try to answer anyway.

In Intensional Type Theory one uses setoids to simulate extensional
reasoning, a setoid is a type together with an equivalence relation
(i.e. you can form the large type SETOID):

A : *
~_AA : A - A - *
eqR_A : EquivRel ~
-
(A,~_A,eqR_A) : SETOID

where EquivRel is formalizes that ~ is an equivalence relation. Given
setoids AA=(A,~A,eqR_A) and BB = (B,~B,eqR_B) we can form a new
setoid AA = BB  where the underlying set is given as

A=B = Sigma f : A-B . Pi a,b:A.(a ~A b) - f a ~ f b

and the equivalence relation

(f,_) ~A=B (g,_) = Pi a:A.f a ~B g a

and I leave it as an exercise to show that this is an equivalence
relation. Clearly, we can use A = B as if it were the function space
and ~ as if it were the propositional equality on the function type.
However, each time we want to substitute an extensionally equal
function by another one, we still have to prove that our predicate
preserves the setoid equality. This seems unnessessary because we
know that all predicates we can define preserve extensional equality.

In a different way this also applies to quotient types. While in
general not all function preserve the equality we quotient by, we can
introduce an elimination operator for quotients which requires us to
show that the operation preserves the equality. If this is the only
way we ever use quotient types we also know that any predicate will
preserve the equality.

E.g. if we want to formalize category theory we would like to define
a category simply as a structure:

Obj : *
Hom : Obj - Obj - *
id : Pi A:Obj.Hom A A
comp : Pi A,B,C:Hom B C - Hom A B - Hom A c
eqL : Pi A,B:*;f:Hom A B.comp f (id A) = f
...

However, we want be able to construct many categories this way,
because the equality we want to use may not be the propositional
equality of Intensional Type Theory. E.g. if we formalize the
category of groups a morphism is a function on the underlying sets
which commutes with the group structure. Hence for two morphisms to
be equal requires to show that they use the same proof that the
function commutes with the group structure!?

Hence we certainly should introduce a setoid of Homsets. Some people
have suggested that this is sufficent, however, in this setting it is
not clear how to perform certain constructions which turn homsets
into objects, such as the construction of the arrow category (objects
are morphisms, morphisms are commuting squares). It would be
consequent to use a setoid to model objects as well. Fine, however,
we now have to make precise that homsets are a family of setoids
indexed by a setoid. In particular we have to provide a family of
isomorphisms which assigns to equal objects an isomorphism of
homesets...

All this can be done, but it seems that we are reinventing Type
Theory on the level of setoids. An indeed this basically the idea of
OTT: to use a model construction to interpret all constructions of
Type Theory as taking place in the setoid interpretation. The
advantage is that you can use the setoid equality in the model in the
same way as you use propositional equality in ordinary Type Theory.
I.e. we use the relation ~A=B as defined above but we never have to
show that our predicates respect this relation because this is part
of our model construction (or more exactly part of our defintion what
it means to be a predicate over A=B). In the category theory example
we can just use the naive formalisation, but since we can use
arbitrary quotients we get all the categories we are interested in.

So to summarize: OTT can be viewed as a convenient interface to use
setoids without having to realize all the underlying machinery. It's
goal is to have an extensional propositional equality but to retain
decidability and canonicity.

Cheers.
Thorsten

And doesn't OTT have the same problems (equality not automatically
substitutive, complications in formalising category theory, lots of
explicit coercions) that the paper says ITT has? I don't see how OTT
helps with any of those problems, but the way that OTT is contrasted
with ITT suggests that it should. I must be misunderstanding
something.

--
Robin

```

### Re: [Epigram] Definitional equality in observational type theory

```Thanks to all who replied.

I have another couple of questions, more fundamental this time.

The paper says Observational reasoning can be simulated in Intensional
Type Theory by the use of setoids, i.e. types with an explicit
equivalence relation.

I'm not clear on what the differences are between OTT and simulating
observational reasoning with setoids in ITT. Clearly OTT has an
explicit equivalence relation (=) which is polymorphic over (certain)
types. So does the paper mean explicit in some other sense?

And doesn't OTT have the same problems (equality not automatically
substitutive, complications in formalising category theory, lots of
explicit coercions) that the paper says ITT has? I don't see how OTT
helps with any of those problems, but the way that OTT is contrasted
with ITT suggests that it should. I must be misunderstanding something.

--
Robin

```

### Re: [Epigram] Definitional equality in observational type theory

```
Hi Robin,

I've been reading Towards Observational Type Theory. I'm new to type
theory, so I have not been able to understand big chunks of the paper;
hopefully as I read wider I'll understand more of it.

We should certainly strive to fix this...

However, one rule which jumped out at me is this, on page 9,
because it

seems silly (apologies if my email client mangles this even after I've
tried to fix it up - I've never tried pasting type rules from PDFs
into

it before):

Γ |- S≡T  Γ |- Q:S=T  Γ |- s:S
--
T
Γ |- s[QS ≡ s : T

It seems silly to include a definitional equality as a premise,
because
I _thought_ the whole point of definitional equality was that S≡T
means

that S and T are freely co-substitutable, and type-checking will
automatically substitute S for T or T for S _whenever_ required. (For
example, it is often noted that Coq will automatically reduce (i.e.
substitute) n+0 to n in a dependent type - but only if you write n+0
the right way round! I can't remember offhand which way round
works.)

You are right to say that this rule is unusual. In the implementation
it creates a dependency between defintional equaliy and reduction
which doesn't appear in conventional systems.

It can be justified by observing that in intensional Type Theory we
have tat

s [refl S ≡ s

where refl S : S=S. However, due to proof-irrelevance any term Q : S
= S will be definitionally equal to refl S, hence if p:S=S then

s [p ≡ s

So anyway, surely any derivation which needed that rule could also

Γ |- Q:S=S  Γ |- s:S
--
S
Γ |- s[QS ≡ s : S

(which is trivial), substituting S for T before applying it, and
substituting Ts for some of the Ss afterwards.

You are right - your rule is equivalent and more concise. However, a
type checker will have to implement the first rule. To check whether
s [Q reduces to s you have to verify whether the types in the type
of Q:S = T are definitionally equal, i.e. whether S ≡ T.

Where have I gone wrong?

Indeed nowhere.

There is some discussion of related issues in the Epigram developers
blog:

http://www.e-pig.org/epilogue/

is particular in the Epigram 2 design doc.

Cheers,
Thorsten
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### Re: [Epigram] Definitional equality in observational type theory

```
Hi Robin,

I just saw Thorsten answered your e-mail. I'd better finish what I
started though.

I'm not an expert on OTT, but here are a few thoughts.

The real rule for coercion is the one on page 3. It doesn't require
the types to be definitionally equal. Section 5 mainly speculates on

* Definitional proof irrelevance

* Definitionally defining all coherence proofs to be refl and
coercions within the same type to be the identity.

Like Thorsten says, this is a bit tricky and makes evaluation and
equality mutually recursive - which is a bit odd. If you're looking
it's a bit technical and the really important stuff is in the first
few sections.

I'm not sure if it helps, but here's a bit of Agda code
implementing OTT:

www.cs.nott.ac.uk/~wss/Misc/ott.agda

I certainly understood much more of the paper after writing the Agda.
I'm not sure if reading it will have quite the same effect, but it
might be worth a shot.

All the best,

Wouter

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