*To*: <cl-isabelle-users at lists.cam.ac.uk>*Subject*: Re: [isabelle] Function definition*From*: <Thomas.Sewell at data61.csiro.au>*Date*: Wed, 28 Sep 2016 01:47:01 +0000*Accept-language*: en-US, en-AU*In-reply-to*: <062fbf84-f95d-2f12-f2e4-0ef225429fef@gmail.com>*References*: <CADBMspKbPobfFuis3n10UvLNkEuJJew1Qy+3o6BWnx56dsTHRw@mail.gmail.com> <d234e92f-9b22-0e15-f5fb-57eadfee4af7@in.tum.de> <062fbf84-f95d-2f12-f2e4-0ef225429fef@gmail.com>*Thread-index*: AQHSGJbxsrSXmOx2n0CkwUG5NQf1bqCMYDiAgAAwH4CAAOs/AA==*Thread-topic*: [isabelle] Function definition*User-agent*: Mozilla/5.0 (X11; Linux x86_64; rv:45.0) Gecko/20100101 Thunderbird/45.3.0

With some extra thought, you can make these definitions work by explicitly forcing "boring" to be size-preserving. datatype t = N nat | A t | B t t definition "size_constrain f v = (if size (f v) = size v then f v else v)" lemma size_size_constrain[simp]: "size (size_constrain f v) = size v" by (simp add: size_constrain_def) fun boring :: "(t => t) => t => t" where "boring f (N n) = N n" | "boring f (A t) = A (size_constrain f (boring f t))" | "boring f (B t1 t2) = B (size_constrain f (boring f t1)) (size_constrain f (boring f t2))" lemma boring_size[simp]: "size (boring f t) = size t" by (induct t, simp_all) lemma boring_fundef_cong[fundef_cong]: "â ât'. size t' < size t â f t' = f' t' â â boring f t = boring f' t" by (induct t, simp_all add: size_constrain_def) fun interesting :: "nat => t => t" where "interesting k (N n) = N (n + k)" | "interesting k (A t) = A (interesting (k + 1) t)" | "interesting k t = boring (interesting k) t" lemma interesting_size[simp]: "size (interesting k t) = size t" by (induct t arbitrary: k , simp_all) lemma interesting_size_constrain[simp]: "size_constrain (interesting k) = interesting k" by (simp add: size_constrain_def fun_eq_iff) That's a bit of a pain but it works with standard tools and gets you roughly what you want. It might also be possible to relax the size constraint slightly to allow size reduction, for instance. Cheers, Thomas. On 27/09/16 21:45, Christian Sternagel wrote: > Dear Ed, > > as Manuel indicated, you'll likely have to define your function in > another way. > > One way would be to use mutual recursion: > > function (sequential) > interesting :: "nat â t â t" and > boring :: "nat â t â t" > where > "interesting k (N n) = N (n + k)" > | "interesting k (A t) = A (interesting (k + 1) t)" > | "interesting k t = boring k t" > | "boring k (N n) = N n" > | "boring k (A t) = A (interesting k (boring k t))" > | "boring k (B t1 t2) = B (interesting k (boring k t1)) (interesting k > (boring k t2))" > by (pat_completeness) auto > > where your "boring" is replaced by a variant that only takes "k" as > parameter. Then we can prove that both functions are size-preserving, as > already suggested by Manuel. > > lemma [termination_simp]: > shows "interesting_boring_dom (Inl (k, t)) â size (interesting k t) = > size t" > and "interesting_boring_dom (Inr (k, t)) â size (boring k t) = size t" > by (induct k t and k t rule: interesting_boring.pinduct) > (simp_all add: interesting.psimps boring.psimps) > > Which incidentally suffices for termination: > > termination by (lexicographic_order) > > It remains to show that this actually corresponds to your original > function specification. Here, I use "boring'" for your "boring". > > fun boring' :: "(t â t) â t â t" where > "boring' f (N n) = N n" > | "boring' f (A t) = A (f (boring' f t))" > | "boring' f (B t1 t2) = B (f (boring' f t1)) (f (boring' f t2))" > > Your definition of "interesting" (modulo "case" on the right vs. > pattern-matching on the left) can be obtained by mutual induction: > > lemma > shows "interesting k s = > (case s of > N n â N (n + k) > | A t â A (interesting (k + 1) t) > | t â boring' (interesting k) t)" > and "boring k s = boring' (interesting k) s" > by (induct k s and k s rule: interesting_boring.induct) auto > > cheers > > chris > > PS: For those who care and know what I'm talking about: termination of > the TRS corresponding to "interesting" and "boring" is trivial for > modern termination tools. Maybe a reason to revive IsaFoR/TermFun? ;) > > On 09/27/2016 10:52 AM, Manuel Eberl wrote: >> Unfortunately, it's not that easy. >> >> You pass the "interesting" function to the "boring" function as a >> parameter, and the "boring" function applies that function that it is >> given to things. In order for this to work, you need to essentially show >> that the "interesting" function that you are defining is only called on >> values for which it terminates (i.e. that are smaller than the original >> argument that it got) >> >> Normally, this is done with a fundef_cong rule, but in this case, I >> don't see how that is possible, because the values that "interesting" is >> called on by "boring" are "boring f t1" and "boring f t2" â that means >> that you are relying on the fact that the "interesting" function does >> not increase the size of its argument. >> >> Showing that "interesting" is size-preserving is actually straightforward: >> >> lemma same_size_boring: >> assumes "ây. size (f y) = size y" >> shows "size (boring f x) = size x" >> using assms by (induction x) simp_all >> >> lemma size_interesting_aux: >> assumes "interesting_dom (k, t)" >> shows "size (interesting k t) = size t" >> using assms >> by (induction rule: interesting.pinduct) (simp_all add: >> interesting.psimps same_size_boring) >> >> However, I have no idea how you would then go on and prove termination. >> Termination proofs depend on the call graph that is computed for the >> recursive definition, and if you don't have a fundef_cong rule for >> boring, that call graph â as you discovered yourself â will be too >> coarse (i.e. you will not have the information that you need in your >> termination proof). However, as I see it, any cong rule for "boring" >> would have to be conditional, which is, as far as I am aware, not >> possible for fundef_cong rules. >> >> What you are trying to do may therefore very well be outside of what the >> function package can do. (although I'm not 100% sure about that â still, >> even if it is possible, I would bet it will be ugly) >> >> My advice would be: try to define your functions in a simpler way. >> Proofs involving nested recursion tend to get very ugly very quickly, >> because the termination of your function depends on the semantics of >> your function, and semantic properties are difficult to use without a >> full termination proof. >> >> Cheers, >> >> Manuel >>

**References**:**[isabelle] Function definition***From:*Edward Pierzchalski

**Re: [isabelle] Function definition***From:*Manuel Eberl

**Re: [isabelle] Function definition***From:*Christian Sternagel

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