*To*: cl-isabelle-users at lists.cam.ac.uk*Subject*: Re: [isabelle] Missing transfer rules in quotient type*From*: Ondřej Kunčar <kuncar at in.tum.de>*Date*: Tue, 17 Dec 2013 17:31:15 +0100*In-reply-to*: <CAFzz2xR8XcBZ1r59fJS_74TxsLrBLMqyZbo0XdLfQn1j6a1zCA@mail.gmail.com>*References*: <CAFzz2xR8XcBZ1r59fJS_74TxsLrBLMqyZbo0XdLfQn1j6a1zCA@mail.gmail.com>*User-agent*: Mozilla/5.0 (X11; Linux x86_64; rv:24.0) Gecko/20100101 Thunderbird/24.1.0

Hi Daniel, sorry for the inconvenience.

context begin interpretation lifting_syntax . ... end Hope this helps. Ondrej On 12/17/2013 04:17 PM, Daniel Raggi wrote:

Dear transfer/lifting experts, I've been experimenting with transfer and lifting and there is something that I just haven't managed to understand. I didn't have this problem before I upgraded from Isabelle2013 to Isabelle2013-2 (and I suspect I would have also had it with Isabelle2013-1, but I never used it) Things similar to what I'm about to describe happen every time I define a new type using quotient_type or typedef. I'll give a concrete example with quotient_type (a type with two elements, as a quotient from integers): quotient_type bin = int / "(λx y. [x = y] (mod 2))" <proof> Then I instantiate bin to the numeral class successfully and prove the transfer rule that lets one transfer every int numeral to the 'same' bin numeral: lemma cr_bin_numeral [transfer_rule]: "cr_bin (numeral x) (numeral x)". As far as I understand, this should be enough to do: lemma "2 = (4::bin)" apply transfer to get subgoal "[2 = (4::int)] (mod 2)". However, I get: 1. ?a5 (2::int) (4::int) 2. Transfer.Rel (cr_bin ===> cr_bin ===> op =) ?a5 op = Here, the obvious value for ?a5 is the relation "λx y. [x = y] (mod 2)", and actually this is by default with every quotient type definition, so I don't understand why the transfer rule was not generated automatically in the first place when I defined the new type. I thought it was already automatic in Isabelle2013, and my theories actually broke down once I tried running them on Isabelle2013-2 because of this issue. The issue doesn't stop here, because even if I define the transfer rule (cr_bin ===> cr_bin ===> op =) (λx y. [x = y] (mod 2)) op = by hand I still get a problem when I try: lemma "1 = (3::bin)" apply transfer What I get is 1. ?a5 1 3 2. Transfer.Rel (*pcr_bin *===> cr_bin ===> op =) ?a5 op = This is because 1 was actually lifted so that I could instantiate bin in the numeral class, so it's transfer rule is generated differently. I'm not familiar with the pcr_ thing. Can anyone explain this? So, If I wanted to write the transfer rules for equality manually I would have to make four rules: (cr_bin ===> cr_bin ===> op =) (λx y. [x = y] (mod 2)) op = (pcr_bin ===> cr_bin ===> op =) (λx y. [x = y] (mod 2)) op = (cr_bin ===> pcr_bin ===> op =) (λx y. [x = y] (mod 2)) op = (pcr_bin ===> pcr_bin ===> op =) (λx y. [x = y] (mod 2)) op = I keep thinking that there's something essential I'm not understanding, but maybe it's not me. I would greatly benefit from your feedback. I'm attaching a test file with this experiment, if anyone wants to have a look. Best, Daniel

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