I would like to prove some basic facts about a datatype_new and a codatatype: the first does not have an infinite element, and that the latter does have one.
theory Co
imports BNF
begin
datatype_new natural = Zero | Successor natural
lemma "¬ (∃ x. x = Successor x)"
oops
codatatype conat = CoZero | CoSucc conat
lemma "∃ x. x = CoSucc x"
oops
The problem was that I could not come up with a pen-and-paper proof, let alone a proof script.
An idea for the first was to use the size function, which has a theorem
size (Successor ?natural) = size ?natural + Suc 0
and somehow using that size is a function, applying it to the two sides of the original equation one cannot have a natural number equal to its successor. But I do not see how I could formalise this.
For the latter I did not even have an idea how to derive this theorem from the facts that the codatatype package proves.
How can I prove these?
Personally, I don't know the first thing about codatatypes. But let me try to help you nevertheless.
The first lemma you posted can be proven automatically by sledgehammer. It finds a proof using the size function, effectively reducing the problem on natural to the same problem on nat:
by (metis Scratch.natural.size(2) n_not_Suc_n nat.size(4) size_nat)
If you want a very basic, step-by-step version of this proof, you could write it like this:
lemma "¬(∃x. x = Successor x)"
proof clarify
fix x assume "x = Successor x"
hence "size x = size (Successor x)" by (rule subst) (rule refl)
also have "... = size x + Suc 0" by (rule natural.size)
finally have "0 = Suc 0" by (subst (asm) add_0_iff) (rule sym)
moreover have "0 ≠ Suc 0" by (rule nat.distinct(1))
ultimately show False by contradiction
qed
If you want a more “elementary” proof, without the use of HOL natural numbers, you can do a proof by contradiction using induction on your natural:
lemma "¬(∃x. x = Successor x)"
proof clarify
fix x assume "x = Successor x"
thus False by (induction x) simp_all
qed
You basically get the two cases in the induction:
Zero = Successor Zero ⟹ False
⋀x. (x = Successor x ⟹ False) ⟹
Successor x = Successor (Successor x) ⟹ False
The first subgoal is a direct consequence of natural.distinct(1), the second one can be reduced to the induction hypothesis using natural.inject. Since these rules are in the simpset, simp_all can solve it automatically.
As for the second lemma, the only solution I can think of is to explicitly construct the infinite element using primcorec:
primcorec infinity :: conat where
"infinity = CoSucc infinity"
Then you can prove your second lemma simply by unfolding the definition:
lemma "∃x. x = CoSucc x"
proof
show "infinity = CoSucc infinity" by (rule infinity.ctr)
qed
Caveat: these proofs work, but I am not sure whether they are the easiest and/or most elegant solution to this problem. I have virtually no knowledge of codatatypes or the new datatype package.
Related
I want to prove the following
lemma
fixes pi :: "'a path" and T :: "'a ts"
shows "valid_path T pi s ⟹ ∀ op ∈ set pi. valid_operator T op"
by induction on pi where
fun valid_path :: "'a ts ⇒ 'a path ⇒ 'a state ⇒ bool" where
"valid_path T [] s = True" |
"valid_path T (op#ops) s = (valid_operator T op ∧ valid_path T ops (effect op s))
and path is just a type synonym for an operator list.
The other definitions should not play a role for the proof.
The base case works fine.
The problem is that, informally, for the inductive step where pi = (x # xs) I'm assuming that
if valid_path T xs s
then ∀ op ∈ set xs. valid_operator T op
and I must show that this implies
if valid_path T (x#xs) s
then ∀ op ∈ set (x#xs). valid_operator T op
I can use the definition of valid_path here, so this last expression is equivalent to
if valid_path T (xs) (effect x s)
then ∀ op ∈ set (x#xs). valid_operator T op
If I could be able to use the induction hypothesis on valid_path T (xs) (effect x s) I would be done.
I can't since the hypothesis only holds for valid_path T (xs) s instead of valid_path T xs (effect x s).
But this does not really matter since the predicate of that if statement does not depend on s at all!
But Isabelle does not know that so it complains.
How can I make it such that I can apply the inductive hypothesis on valid_path T (xs) (effect x s)?
I have a feeling that I have to make the claim more general, so that I can use the hypothesis on the proof, but I don't know how.
It is very common that you have to generalize some terms in an induction. Use the keyword arbitrary in the induct method.
proof (induct pi arbitrary: s)
This is explained in Chapter 2.4 of Programming and Proving in Isabelle/HOL.
I'm trying to learn how to use Isabelle/Isar with HOL, and I decided a good way to do that would be to develop some elementary number theory. I defined my own plus and times operations so that proof methods wouldn't be doing all the work for me as there's already plenty proved about +, * in Main. My versions are defined as:
fun p:: "nat ⇒ nat ⇒ nat" (infix "⊕" 80) where
p_0: "0 ⊕ n =n" |
p_rec: " (Suc m) ⊕ n = Suc (m ⊕ n)"
fun t:: "nat ⇒ nat ⇒ nat" (infix "⊗" 90) where
t_0: "0 ⊗ n= 0" |
t_rec: "Suc m ⊗ n = n + m ⊗ n"
and I've already shown that multiplication and addition are commutative the distributive law holds. Then I tried to show the following:
lemma euclidean_division_existence: "∃q r. n=q⊗m⊕r"
proof (induction n)
case 0
have "0= 0 ⊗ m" by auto
hence "∃q. 0 =q ⊗ m" by auto
but it's telling me it can't finish the proof of the last step. I've tried various things, but I can't figure out how to tell Isabelle I just gave it a witness for the existence statement I'm trying to prove. How can I make Isabelle recognize that?
Edit:
xanonec helped me get past this step, but I immediately got stuck on the next step by a seemingly similar problem. Ultimately I want to show:
"∃ q r. 0 = q ⊗ m⊕r"
but I can't figure out how to simultaneously introduce the two existentially quantified variable from
"0 = 0 ⊗ m ⊕ 0"
A suitable strategy for the solution could be a direct rule application (in jEdit you can cntrl+LMB or cmd+LMB on exI to navigate to its statement):
lemma euclidean_division_existence: "∃q r. n=q⊗m⊕r"
proof(induction n)
case 0
have "0 = 0 ⊗ m" by auto
hence "∃q. 0 = q ⊗ m" by (rule exI)
qed
More generally, in many similar cases sledgehammer can find a suitable (but, often, suboptimal) proof. A tutorial on the use of sledgehammer is a part of the official documentation of Isabelle. Also, I would like to suggest the following resources: "Concrete Semantics with Isabelle/HOL" by Tobias Nipkow and Gerwin Klein and "A Proof Assistant for Higher-Order Logic" by Tobias Nipkow et al.
An update following an amendment made to the statement of the question
The following listing presents a proof that relies only on the most basic methods and direct rule application:
lemma euclidean_division_existence: "∃q r. n=q⊗m⊕r"
proof (induction n)
case 0 show ?case
proof-
have "0 = 0 ⊗ m ⊕ 0" by simp
then have "∃r. 0 = 0 ⊗ m ⊕ r" by (rule exI)
then show "∃q r. 0 = q ⊗ m ⊕ r" by (rule exI)
qed
case (Suc n) show ?case sorry
qed
However, if you can afford to rely more on proof automation, then you can use metis to prove the entire theorem:
lemma euclidean_division_existence: "∃q r. n=q⊗m⊕r" by (metis p_0 t_0)
I need to prove the following:
lemma "m = min_list(x#xs) ⟹ m ∈ set (x#xs)"
In plain English, I need to prove that the return value from "min_list (x#xs)" is always a member of (x#xs)
I tried:
apply(induct xs)
apply(auto)
I also tried to reuse existing lemmas for the min_list by using:
find_theorems min_list
The sub-goal at this point is so long that I do not know how to proceed.
I am not looking for a full answer just hints on how to approach this lemma. Moreover, is this proof an easy one or significantly difficult one for someone just learning Isabelle?
Spoiler: it is possible to use the standard list induction and auto to prove the theorem, i.e. something similar to by (induct xs ...) (auto simp: ...). I deliberately left out sections in the proof for you to fill in on your own. You will need to think about if any variables (i.e. m or x) need to be specified as arbitrary and also understand what information the simplifier may need (look for clues in the specification of min_list in the theory List).
With regard to your question about the difficulty of the problem, I believe, that difficulty is a function of experience. Most certainly, when I started learning Isabelle, I was finding it difficult to formalise proofs similar to the one in your question. After a certain time spent coding in Isabelle (by the time of answering this question, I must have accrued an equivalent of 4-5 months of full-time coding in Isabelle), such problems no longer seem to present a significant challenge for me. Of course, there are other factors that need to be taken into account, e.g. previous training in mathematics or logic and previous coding experience.
General advice from someone who is learning Isabelle on his own (the advice may not be consistent with the approach that is normally recommended by professional instructors)
I believe, when proving similar results, it is important to understand that Isabelle is, primarily, a tool for formalisation of 'pen-and-paper' proofs. Therefore, it is important to have the 'pen-and-paper' proof at hand before trying to formalise it. I would suggest the following general approach when attacking similar problems:
Write the proof on paper.
Formalise the proof using Isar, providing as many details as possible and not caring too much about the length of the proof. Also, try not to rely on the tools for automated reasoning (i.e. auto, blast, meson, metis, fastforce) and use direct methods like rule and intro as much as you can.
Once your Isar proof is complete, apply tools for automated reasoning (e.g. auto, blast) to your Isar proof to simplify your proof as much as possible.
Of course, eventually, it will become increasingly easy to omit 1 and 2 as you make progress in learning Isabelle.
I can provide further details, e.g. the complete short proof and the long Isar version of the proof.
UPDATE
As per your request in the comments, I provide an informal proof.
Lemma. m = min_list (x # xs) ⟹ m ∈ set (x # xs).
Remarks. For completeness, I also provide the definition of min_list and some comments about the const set. The definition of min_list can be found in the theory List:
fun min_list :: "'a::ord list ⇒ 'a" where
"min_list (x # xs) = (case xs of [] ⇒ x | _ ⇒ min x (min_list xs))"
The const set is defined implicitly and constitutes a part of the datatype infrastructure for list (see the document "Defining (Co)datatypes and Primitively (Co)recursive Functions in Isabelle/HOL" in the standard documentation if Isabelle). In particular, it is called the 'set function' of the datatype. Many basic properties of the const set can be found by inspection/search, e.g. find_theorems list.set. I believe that the theorem thm list.set is representative of the main properties of the const set (I took the liberty to rename the schematic variables in the theorem):
set [] = {}
set (?x # ?xs) = insert ?x (set ?xs)
Proof. The proof is by structural induction on the list xs. The induction principle is stated as an unnamed lemma at the beginning of the theory List. For completeness, I restate the induction principle below:
"P [] ⟹ (⋀a list. P list ⟹ P (a # list)) ⟹ P list"
Base case: assume xs = [], show m = min_list (x # xs) ⟹ m ∈ set (x # xs) for all x. From the definition of min_list, it is trivial to see that min_list (x # []) = x. Similarly, set (x # []) = {x} can be shown directly from the properties of the const set. Substituting into the predicate above, it remains to show that m = x ⟹ m ∈ {x} for all x. This follows from basic set theory.
Inductive step: assume ⋀x. m = min_list (x # xs) ⟹ m ∈ set (x # xs), show m = min_list (a # x # xs) ⟹ m ∈ set (a # x # xs) for all a, x and xs. Fix a, x and xs. Assume m = min_list (a # x # xs). Then it remains to show that m ∈ set (a # x # xs). Given m = min_list (a # x # xs), from the definition of min_list, it is easy to infer that either m = a or m = min_list (x # xs). Consider these cases explicitly:
Case I: m = a. a ∈ set (a # x # xs) follows from the definitions. Then, m ∈ set (a # x # xs) by substitution.
Case II: m = min_list (x # xs). Then, from the assumption ⋀x. m = min_list (x # xs) ⟹ m ∈ set (x # xs) it follows that m ∈ set (x # xs). Thus, m ∈ set (a # x # xs) follows from the properties of set.
In all possible cases m ∈ set (a # x # xs), which is what was required to prove.
Thus, the proof is concluded.
Concluding thoughts. Try converting this informal proof to an Isar proof. Also, please note that the proof may not be ideal - I might make edits to the proof later.
I noticed that Isabelle automatically simplifies ¬ (a ∈ (- A)) and ¬ (x = y) to a ∉ A and x ≠ y, respectively.
Here is a simple pen-and-paper proof in natural deduction but fails in Isabelle. In the 2nd line, ¬ (a ∈ (- A)) is simplified to a ∉ - A. From the latter, we cannot apply ComplD, but why?
lemma "- (- A) ⊆ (A::'a set)"
proof
fix a assume "a ∈ - (- A)"
hence "¬ (a ∈ (- A))" by (rule ComplD)
hence "¬ (¬ (a ∈ A))" by (rule ComplD) (* fail! *)
thus "a ∈ A" by (rule notnotD)
qed
Is there a way to go back to the non-simplified expression?
Of course, the lemma can be proved in one line by simp. But my purpose is to explicitly use natural deduction rules (for teaching).
These are not simplifications. If you look at the definitions of a ≠ b and x ∉ A in Isabelle (e.g. by ctrl-clicking on the symbols), you will find that they are simply abbreviations for ¬(a = b) and ¬(x ∈ A). The statements are internally represented exactly the way you wrote them above, they are just printed differently for increased readability.
The reason why you cannot apply the rule ComplD is that it simply does not match. ComplD says that ?c ∈ - ?A ⟹ ?c ∉ ?A. However, in the failing step, your assumption is a ∉ -A, and that cannot be unified to the premise ?c ∈ -?A of ComplD, and therefore rule fails.
I am relatively certain that you will need classical reasoning for this proof since your statement does not hold intuitionistically. This means you will have to do a proof by contradiction, e.g. like this:
lemma "- (- A) ⊆ (A::'a set)"
proof
fix a assume a: "a ∈ - (- A)"
show "a ∈ A"
proof (rule ccontr)
assume "a ∉ A"
have "a ∈ -A"
proof (rule ComplI)
assume "a ∈ A"
with ‹a ∉ A› show False by contradiction
qed
moreover from a have "a ∉ -A" by (rule ComplD)
ultimately show False by contradiction
qed
qed
The rule ccontr in there starts the proof by contradiction; the proof method contradiction is merely a nice way to derive anything when one has proven a fact and its negation.
Here is a proof:
theory Example
imports Main
begin
datatype natural = Zero | Succ natural
lemma "⋀ n. n = Succ m ⟹ n ≠ Zero"
proof -
fix n
assume "n = Succ m"
from this show "n ≠ Zero" by (metis natural.distinct(2))
qed
end
The depth value is 0 throughout the proof but after
show "n ≠ Zero"
it changes to
proof (prove): depth 1
What does depth mean here? Is it any important when carrying out a proof?
In a nutshell, it refers to the current proof nesting level. In your case, it's 1, because show opens a new proof inside the proof.
To answer your second question: No, it's not important at all. Some people use it to measure how complicated a proof is, but to the system, it makes no difference.