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Hi Roberto,

BCCCs seem to come up a lot at the intersection between database theory and 
programming language theory.  For example, if you say that a database schema is 
a finitely presented category, then the category of schemas and mappings 
between them is a BCCC.  For this particular BCCC, I believe that equality is 
semi-decidable, so I would like to say that its semi-decidability comes from 
its particular nature, and is not inherited from the free BCCC.  BCCCs also 
come up when you try to connect higher-order logic to the nested relational 
calculus via their categorical (topos) semantics.  I'd be happy to talk more 
about it offline.

Thanks,
Ryan

On Sep 21, 2013, at 9:34 AM, Roberto Di Cosmo <[email protected]> wrote:

> Dear Ryan,
>     beta-eta equality in the presence of strong sums is a tricky subject that 
> has been object of some attention for more than 20 years, both "per se" in 
> the rewriting community, that looked for an elegant confluent rewriting 
> system, and as a "tool" for researchers interested in type theory and 
> semantics, that wanted to know for example whether one can characterise 
> object isomorphism in a BiCCC, problem that one can try to attack by studying 
> invertible terms in the corresponding lambda calculus with strong types.
> 
> If one stays in the 1,*,-> fragment, all is wonderful: you get a nice 
> confluent and normalising rewriting system for the lambda calculs based on 
> eta expansion (see a blazingly short proof in 
> http://www.dicosmo.org/Articles/POD.ps), the type isomorphisms are finitely 
> axiomatizable
> and exactly correspond to the equations one knows from high schools for 
> natural numbers equipped with product, unit and exponentiation, related to 
> Tarski's High School algebra problem
> (a nice proof through Finite Sets is due to Sergei Soloviev in 1988, I gave 
> one based on invertible terms in my PhD thesis back in 1992).
> 
> You may add weak sums to the calculus, and still get a nice confluent and 
> normalising rewriting system, that can also accomodate bounded recursion 
> (http://www.pps.univ-paris-diderot.fr/~kesner/papers/icalp93.ps.gz).
> 
> As soon as you add strong sums, though, things get very hairy: Neil Ghani 
> proposed in his PhD thesis back in 1995 a decision procedure for lambda terms 
> with strong 1,+,*,-> (no zero, see 
> https://personal.cis.strath.ac.uk/neil.ghani/papers/yellowthesis.ps.gz): a 
> rewriting system is used to bring terms in this calculus to a normal form 
> which is not unique, and then it is shown that all such normal forms are 
> equivalent modulo a sort of commuting conversions, so to decide equality of 
> two terms, you first normalise them, and then check whether they are 
> equivalent w.r.t. conversion (the equivalence classes being finite, this is 
> decidable). The NBE appoach of Thorsten Altenkirch (et al.) of your LICS01 
> reference seems to provide a more straightforward way to decide equality, 
> even if somebody coming from the rewriting world might like a more intuitive 
> description of what the normal form actually look like.
> 
> If you add the zero, you are in for more surpises: looking at the special 
> BiCCC formed by the natural numbers equipped with 0,1,+,* and exponentiation, 
> you discover that equality is no longer finitely axiomatisable, and yet a 
> decision procedure does exist (see 
> http://www.dicosmo.org/Papers/zeroisnfa.pdf),
> which solves in the negative, but with a positive note, Tarski's problem for 
> this system.
> 
> But then, in BiCCCs, the connection between numerical equalities and type 
> isomorphisms breaks down and the nice result above says nothing about what 
> happens for the general case. In http://www.dicosmo.org/Papers/lics02.pdf we 
> proved that isos are non finitely axiomatisable in BiCCC, but we know nothing 
> on decidability.
> 
> As for deciding beta-eta equality, the work of Marcelo Fiore and Vincent 
> Balat http://www.cl.cam.ac.uk/~mpf23/papers/Types/tdpe.pdf shows how to 
> compute normal forms in full BiCCCs, and I kind of remember that we were 
> convinced at some point that these normal forms must be unique up to some 
> sort of commuting conversions (you can see this in the examples of the 
> paper), but we did not prove it at that moment, and it is true that one would 
> like to see this last step done (or a proof that it cannot be done).
> 
> I wonder whether Marcelo, Neil and Thorsten might shed some light on this 
> (Vincent and I started working on quite different areas a while ago)
> 
> By the way, out of curiosity, would you mind telling us how this result would 
> be useful your research?
> 
> --
> Roberto
> 
> P.S.: for those interested in type isomorphism, 
> http://www.dicosmo.org/Paper/mscs-survey.pdf provides a concise overview, 
> even if no longer up to date.
> 
> 
> 
> 2013/9/20 Ryan Wisnesky <[email protected]>
> [ The Types Forum, http://lists.seas.upenn.edu/mailman/listinfo/types-list ]
> 
> Hello,
> 
> I am trying to find a reference that states whether or not the free 
> bi-cartesian closed category is decidable.  That is, I am wondering if 
> beta-eta equality of the simply typed lambda calculus extended with strong 
> 0,1,+,* types is decidable.  (In particular, I am interested in the full 
> calculus, including the eliminator for 0, introduction for 1, and eta for all 
> types).
> 
> So far, my search through the literature says that the answer to this 
> question is "almost" or "maybe" :
> 
> - the STLC with strong 1,+,* types is decidable (omits 0, the empty type): 
> http://www.cs.nott.ac.uk/~txa/publ/lics01.pdf
> 
> - the STLC with strong 0,1,+,* types has normal forms, but equivalent terms 
> may have different such normal 
> forms:http://www.cl.cam.ac.uk/~mpf23/papers/Types/tdpe.pdf
> 
> - work on mechanized proofs of correctness for deciding co-product equality 
> may or may not include the "absurd" eliminator for the empty type: 
> http://www.cis.upenn.edu/~sweirich/wmm/wmm07/programme/licata.pdf
> 
> I feel like the answer to this question is probably well-known, but I can't 
> seem to find it in the literature.  Any help would be appreciated.
> 
> Thanks,
> Ryan
> 
> 
> 
> -- 
> Roberto Di Cosmo
>  
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