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 Radboud University Nijmegen www.cs.ru.nl/~spitters/
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 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 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?) ADTs in 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. Instead of choice, I'd like 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. I'll think about this. 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
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, thank you for your comments. Indeed we are currently revising this paper, hence your comments are especially welcome. 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 use this rule instead: 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 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 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 adding two extensions: * 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 to understand OTT, I wouldn't get too worked up about this rule - 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 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.
