Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 3 years ago.
Improve this question
It's a homework question. Can we consider the following as a CNF?
(¬x ∧ y) ∧ (x ∨ z)
I thinks it's a CNF just like A ∧ B, but my friend doesn't agree with me.
According to https://en.wikipedia.org/wiki/Conjunctive_normal_form
(¬x ∧ y) ∧ (x ∨ z)
itself is not strictly a CNF, which requires a disjunction of conjunctive clauses.
It is however equivalent to the following, according to the associative rule ( (a ∧ b) ∧ c is equivalent to a ∧ b ∧ c ).
¬x ∧ y ∧ (x ∨ z)
And that can be considered CNF, as being a disjunction of cunjunctive clauses, two of which only contain a single literal, one of which is negated.
Note that CNF does NOT require each conjunctive clause to contain ALL literals.
So in short: Strictly it is not, practically it only requires dropping redundant ().
Related
The universal quantifier in first-order logic (the symbol is ∀) and meta-universal quantifier in meta-logic (the symbol is ⋀) what is the main difference?
For the following two lemmas, the first example proves successful using universal quantifiers, while the meta-universal quantifiers do not.
lemma "∀ x. P x ⟹ P 0"
apply simp
done
lemma "⋀ x. P x ⟹ P 0"
oops
Isabelle is a generic framework for interactive theorem proving. Its meta-logic Isabelle/Pure allows to define a broad range of object-logics, one of them being Isabelle/HOL. As you already hinted, the symbol ∀ is Isabelle/HOL's universal quantifier and the symbol ⋀ is Isabelle/Pure's universal quantifier. Also, the symbol ⟹ is Isabelle/Pure's implication. The operator precedence rules state that ⋀ has lower precedence than ⟹, and ∀ has higher precedence than ⋀ and ⟹. Therefore ⋀ x. P x ⟹ P 0 is actually parsed as ⋀ x. (P x ⟹ P 0) (which clearly doesn't hold) instead of (⋀ x. P x) ⟹ P 0, so you need to explicitly parenthesize the proposition ⋀ x. P x. Then, your lemma can be trivially proved using the usual elimination rule for ⋀ in natural deduction as follows:
lemma "(⋀ x. P x) ⟹ P 0"
by (rule meta_spec)
In order to try to avoid this kind of nuances, I'd suggest you to adopt the Isabelle/Isar style for stating your lemmas, namely the following:
lemma
assumes "⋀ x. P x"
shows "P 0"
using assms by (rule meta_spec)
Please refer to Programming and Proving in Isabelle/HOL and The Isabelle/Isar Reference Manual for more information.
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.
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 5 years ago.
Improve this question
I can easily understand how to proof a simple big o notation like n5 + 3n3 ∈ O(n5).
But how can I proof something more complex like 3n or 2n ∉ O(nk)?
Use a proof by contradiction.
Let's prove that 2n ∉ O(n2). We assume the opposite, and deduce a contradiction from a consequence.
So: assumption: there exists M and n0 such that 2n < M n2 for all n >= n0.
Let x be an number such that x > 5, and x > n0 and 2x > 4 M. Do you agree that such a number must exist?
Finish off the proof by deducing a contradiction based on the inequality that 22x < 4 M x2 by assumption.
Now do the analogous proof for k = 3. Then do it for k = 4. Then generalize your result for all k.
I have a problem problem such as:
¬P ∨ (Q ∨ R)
I used the law that :
¬P ∨ (Q ∨ R) ≡ ¬P ∨ Q ∨ R
But I do not remember the name of the law. Can anybody help me?
You are thinking of the associative law, but there is a subtlety:
What the associative law says is that
P v (Q v R) = (P v Q) v R
Depending on your formal system of logic, there is a good chance that
P v Q v R
isn't officially a wff (well-formed formula) since it is syntactically ambiguous. The associative law guarantees that both ways of parsing it are equivalent, hence it is a common abbreviation for (P v Q) v R.
Thus, I would tend to regard
P v (Q v R) = P v Q v R
as giving an abbreviation which is underwritten by the associative law rather than as an application of the associative law per se.
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.