as Manuel indicated, you'll likely have to define your function in
One way would be to use mutual recursion:
interesting :: "nat ⇒ t ⇒ t" and
boring :: "nat ⇒ t ⇒ t"
"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.
shows "interesting_boring_dom (Inl (k, t)) ⟹ size (interesting k t) =
and "interesting_boring_dom (Inr (k, t)) ⟹ 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 ⇒ t) ⇒ t ⇒ 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:
shows "interesting k s =
(case s of
N n ⇒ N (n + k)
| A t ⇒ A (interesting (k + 1) t)
| t ⇒ boring' (interesting k) t)"
and "boring k s = boring' (interesting k) s"
by (induct k s and k s rule: interesting_boring.induct) auto
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? ;)
Post by Manuel Eberl
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" – that means
that you are relying on the fact that the "interesting" function does
not increase the size of its argument.
assumes "⋀y. size (f y) = size y"
shows "size (boring f x) = size x"
using assms by (induction x) simp_all
assumes "interesting_dom (k, t)"
shows "size (interesting k t) = size t"
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 – as you discovered yourself – 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 – 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.