# Re: [isabelle] Finite_Set comp_fun_commute

Hi John.

Sorry for the confusion with the invariant rules. They hold (if the
property P admits only one result) but are probably not very useful to

To prove that (even a stronger one) you can exploit the fact that the
definition of fold does not depend on this strange locale with too
strong assumptions, but on an inductively defined predicate fold_graph,
that contains the results of all possible fold orderings.

Here is the proof of your fold_image_stronger, but you do not need the
first two assumptions at all!
Please excuse the aux-lemmas and the not very elaborated proofs:

theorem fold_image_even_stronger:
assumes I: "inj_on g A"
shows "Finite_Set.fold f x (gA) = Finite_Set.fold (f o g) x A"
oops (*Proof comes below, first need to show some aux-lemmas *)

lemma inj_img_insertE:
assumes I: "inj_on f s"
assumes N: "x\<notin>r"
assumes E: "insert x r = fs"
obtains x' s' where "s=insert x' s'" and "x'\<notin>s'"
and "x = f x'" and "r = fs'"
proof -
from E have "x\<in>fs" by auto
then obtain x' where "x'\<in>s" and "x = f x'" by auto
hence "s = insert x' (s - {x'})" by auto
have "r = f(s - {x'})" apply auto
apply (metis (no_types) Diff_insert_absorb E I N
s = insert x' (s - {x'}) x = f x' image_insert inj_on_insert
insertI1 insert_Diff1)
by (metis E I x = f x' x' \<in> s imageI inj_on_contraD insertE)
show ?thesis
apply rule
apply fact
apply simp
apply fact+
done
qed

lemma fold_graph_image:
"inj_on g s \<Longrightarrow> fold_graph f a (gs) r
\<longleftrightarrow> fold_graph (f o g) a s r"
proof
case goal1
from goal1(2,1) show ?case
proof (induct "gs" r arbitrary: s rule: fold_graph.induct)
case emptyI thus ?case by (auto intro: fold_graph.emptyI)
next
case (insertI x A r s)
with inj_img_insertE obtain x' A' where
"x'\<notin>A'" and [simp]: "s = insert x' A'" "x = g x'" "A=gA'"
by metis
with insertI.hyps(3)[of A'] insertI.prems
have FG: "fold_graph (f o g) a A' r"
by auto
from fold_graph.insertI[OF x'\<notin>A' FG]
show ?case by simp
qed
next
case goal2
from goal2(2,1) show ?case
proof (induct rule: fold_graph.induct)
case emptyI thus ?case by (auto intro: fold_graph.emptyI)
next
case (insertI x A r)
hence FG: "fold_graph f a (gA) r" by simp
from x\<notin>A insertI.prems have N: "g x \<notin> gA" by auto
from fold_graph.insertI[OF N FG] show ?case by simp
qed
qed

lemma fold_graph_image':
"inj_on g s \<Longrightarrow> fold_graph f a (gs) = fold_graph (f o
g) a s"
by (rule ext) (rule fold_graph_image)

theorem fold_image_even_stronger:
assumes I: "inj_on g A"
shows "Finite_Set.fold f x (gA) = Finite_Set.fold (f o g) x A"
unfolding Finite_Set.fold_def
by (simp add: fold_graph_image'[OF I, of f x])

Best,
Peter

On Di, 2013-02-19 at 19:34 +0100, John Wickerson wrote:
> Mm, yes, my mistake. The domain of f should be "g  A" rather than just "A". So the theorem should be:
>
> > lemma fold_image_stronger:
> > assumes "⋀x y. ⟦ x ∈ g  A ; y ∈ g  A ⟧ ⟹ f x ∘ f y = f y ∘ f x"
> > assumes "finite A" and "inj_on g A"
> > shows "fold f x (g  A) = fold (f ∘ g) x A"
>
> I think *this* one is true ... (but how to prove it?)
>
> Thanks,
> john
>
> On 19 Feb 2013, at 18:01, Lawrence Paulson wrote:
>
> > Is your desired theorem true?
> >
> > I would find it easier to believe if it also assumed "x : A" and "g  A <= A".
> >
> > Larry Paulson
> >
> >
> > On 19 Feb 2013, at 16:43, John Wickerson <jpw48 at cam.ac.uk> wrote:
> >
> >> Hi Peter, thanks very much for this. Forgive me if I'm mistaken, but I don't understand how either of these approaches would help. I think I would still need to reason about terms like
> >>
> >>> fold f s (insert a A)
> >>
> >>
> >> in order to complete the induction, and I can't reason about such terms without knowing that f satisfies the "comp_fun_commute" property.
> >>
> >> Let me state my problem more concretely...
> >>
> >> Finite_Set provides the following lemma (the first assumption comes from the context "comp_fun_commute"):
> >>
> >>> lemma fold_image:
> >>> assumes "⋀x y. f x ∘ f y = f y ∘ f x"
> >>> assumes "finite A" and "inj_on g A"
> >>> shows "fold f x (g  A) = fold (f ∘ g) x A"
> >>
> >> But I want the following lemma:
> >>
> >>> lemma fold_image_stronger:
> >>> assumes "⋀x y. ⟦ x ∈ A ; y ∈ A ⟧ ⟹ f x ∘ f y = f y ∘ f x"
> >>> assumes "finite A" and "inj_on g A"
> >>> shows "fold f x (g  A) = fold (f ∘ g) x A"
> >>
> >>
> >> How might I prove it? It's tricky because all the other lemmas about Finite_Set.fold are in the "comp_fun_commute" context where
> >>
> >>> ⋀x y. f x ∘ f y = f y ∘ f x
> >>
> >> holds, whereas I only have the weaker property
> >>
> >>> ⋀x y. ⟦ x ∈ A ; y ∈ A ⟧ ⟹ f x ∘ f y = f y ∘ f x
> >>
> >> available to me.
> >>
> >> Thanks very much,
> >>
> >> john
> >>
> >>
> >>
> >>
> >> On 19 Feb 2013, at 16:38, Peter Lammich wrote:
> >>
> >>> Hi.
> >>>
> >>> An alternative is to use an invariant rule, i.e., something like:
> >>>
> >>>
> >>> I s a0   !!x s a. [| I s a; x\in s |] ==> I (s-{x}) (f x a)
> >>> ------------------------------------------------------------ if finite s
> >>> I {} (fold f s a0)
> >>>
> >>>
> >>> or, alternatively, show that your proposition holds for folding over any
> >>> distinct list representing the set:
> >>>
> >>>
> >>> !!l. [| distinct l; set l = s |] ==> P (foldl f l a0)
> >>> --------------------------------------------------------  if finite s
> >>> P (fold f s a0)
> >>>
> >>>
> >>> Both rules (modulo my typos) should be provable by induction over the
> >>> finite set s.
> >>>
> >>> --
> >>> Peter
> >>>
> >>>
> >>>
> >>> On Di, 2013-02-19 at 16:01 +0100, John Wickerson wrote:
> >>>> Dear Isabelle,
> >>>>
> >>>> This question is directed at anybody familiar with the Finite_Set theory...
> >>>>
> >>>> http://isabelle.in.tum.de/library/HOL/Finite_Set.html
> >>>>
> >>>> ... in particular, the Finite_Set.fold functional. Consider the term
> >>>>
> >>>> Finite_Set.fold f s A
> >>>>
> >>>> Various lemmas (e.g. Finite_Set.comp_fun_commute.fold_image) require me to show that f satisfies the "comp_fun_commute" property, i.e.
> >>>>
> >>>> (1)    f x o f y = f y o f x
> >>>>
> >>>> for all x and y. This is too strong a requirement for me. I can show that (1) holds for all x and y in A, but not for all x and y in general. Morally, I *should* only have to show that f commutes when given inputs drawn from A.
> >>>>
> >>>> It would be quite a bit of hassle for me to convert these lemmas to stronger versions. So I was wondering if anybody has come across this problem before, or knows how to easily strengthen these lemmas, or has any other advice on this topic?
> >>>>
> >>>> Thanks,
> >>>> john
> >>>
> >>>
> >>>
> >>
> >>
> >
>

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