# Re: [isabelle] Finite_Set comp_fun_commute

Awesome! Thanks so much Peter, that's really kind of you.

John

On 20 Feb 2013, at 01:30, Peter Lammich wrote:

> 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
> prove your stronger fold_image rule.
>
> 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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