### 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 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

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, 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.