I am struggling to convert this sentence to CNF:
(A ∨ B) ⇔ (C ∧ D).
I have already tried to use the Biconditional elimination logic rule to eliminate the ⇔.
(A ∨ B) → (C ∧ D) ∧ (C ∧ D) → (A ∨ B).
Then I eliminated the → with the Implication elimination logic rule. Now I have
¬(A ∨ B) ∨ (C ∧ D) ∧ ¬(C ∧ D) ∨ (A ∨ B).
I am pretty much stuck here. My professor says I should use Distributivity rule to reduce the sentence. I can't seem to find anything that matches the requirements of Distributivity rule. So, I can't seem to use Distributivity rule before doing some logical rule that I do not know of.
What am I missing here? Can Stack Overflow help me to resume the conversion to CNF?
You began with the expression:
(A ∨ B) ⇔ (C ∧ D).
You tried to perform the first few steps. Here I added brackets to be clear and correct:
[(A ∨ B) → (C ∧ D)] ∧ [(C ∧ D) → (A ∨ B)]. (by definition of ⇔)
[¬(A ∨ B) ∨ (C ∧ D)] ∧ [¬(C ∧ D) ∨ (A ∨ B)]. (by definition of →)
Apply the De Morgan negation law to ¬(A ∨ B) and ¬(C ∧ D):
[(¬A ∧ ¬B) ∨ (C ∧ D)] ∧ [(¬C ∨ ¬D) ∨ (A ∨ B)].
Simplify the right half:
[(¬A ∧ ¬B) ∨ (C ∧ D)] ∧ [¬C ∨ ¬D ∨ A ∨ B].
The distributive law for ∨ over ∧ states that: X ∨ (Y ∧ Z) ⇔ (X ∨ Y) ∧ (X ∨ Z).
We apply the law to the left half, with X = (¬A ∧ ¬B), Y = C, Z = D:
[((¬A ∧ ¬B) ∨ C) ∧ ((¬A ∧ ¬B) ∨ D)] ∧ [¬C ∨ ¬D ∨ A ∨ B].
Apply the distributive law to two subexpressions in the left half:
[[(¬A ∨ C) ∧ (¬B ∨ C)] ∧ [(¬A ∨ D) ∧ (¬B ∨ D)]] ∧ [¬C ∨ ¬D ∨ A ∨ B].
Remove the extra brackets because ∧ is associative and commutative:
(¬A ∨ C) ∧ (¬B ∨ C) ∧ (¬A ∨ D) ∧ (¬B ∨ D) ∧ [¬C ∨ ¬D ∨ A ∨ B].
Rearrange the variables, and we have our final formula in conjunctive normal form (CNF):
(¬A ∨ C) ∧ (¬A ∨ D) ∧ (¬B ∨ C) ∧ (¬B ∨ D) ∧ (A ∨ B ∨ ¬C ∨ ¬D).
Related
I know that Isabelle can do case analysis by constructors (e.g. of a list), but
Is there a way to split into cases based on whether a condition is true or false?
For example, in proving the following lemma, my logic (as indicated by the following invalid proof in invalid syntax), is that if the condition "x ∈ A" is true, the proof simplifies to something trivial; it also simplifies when the condition is false (i.e. "x ∉ A"):
lemma "(x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C) ⟹ x ∈ A ∨ (x ∈ B ∧ x ∈ C)"
proof (case "x ∈ A")
(* ... case true *)
show "x ∈ A ∨ (x ∈ B ∧ x ∈ C)" by (rule disjI1)
next (* ... case false *)
have "x ∈ B ∧ x ∈ C" by simp
show "x ∈ A ∨ (x ∈ B ∧ x ∈ C)" by (rule disjI2)
But I don't know how to translate this "case analysis" in English into Isabelle.
Is there way in Isabelle/HOL to express this kind of case analysis by the true or false of a condition? (as of Isabelle 2021)
(Or does it require additional axioms such as the law of excluded middle?)
You've almost correctly guessed the syntax, you can write a proof by cases for any predicate with the syntax proof (cases "<pred>").
For the example you provided:
lemma "(x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C) ⟹ x ∈ A ∨ (x ∈ B ∧ x ∈ C)"
proof (cases "x ∈ A")
(* ... case true *)
case True
then show "x ∈ A ∨ (x ∈ B ∧ x ∈ C)" by (rule disjI1)
next (* ... case false *)
case False
then have "x ∈ B ∧ x ∈ C" sorry (* by simp*)
then show "x ∈ A ∨ (x ∈ B ∧ x ∈ C)" by (rule disjI2)
I'm writing an Isar proof inside the Isabelle2020 and working with the locales.
On some point I want to use the axiom from the locale, which states
((A.par x y) ∧ (F x ≃ F y)) → (x ≃ y)
where "A" is another locale given as the information for the one where the above axiom is stored, "A.par" is simply a predicate then.
All the predicates and functions in the Axiom use polymorphic types, the free variable "x" itself is of type 'a, for example, the function "F" is of type 'a => 'b and "G" is of type 'b => 'c:
F :: "'a => 'b"
G :: "'b => 'c"
x :: "'a"
Generally, I have to axioms of this form: one for the locale "F" (showed above) and one for the locale "G" (which is exactly the same but with other types here):
((A.par x y) ∧ (G x ≃ G y)) → (x ≃ y)
The problem is that inside the proof I want to use this axiom but applied as
(A.par (F x) (F y)) ∧ (G(F x) ≃ G(F y)) → (F x) ≃ (F y)
Here the types coincide (as I see), because "F x" is of type 'b and "G" is applied exactly on this type.
And it seems I have to explicitly show the prover that "(F x)" here should be considered as "x" in the axiom and so on.
The proof step itself is just to apply the lemma to the conclusion I have already got:
(A.par (F x) (F y)) ∧ (G(F x) ≃ G(F y))
to get
(F x) ≃ (F y)
If we take a look on the output of the Isabelle we will see these:
USING 1)
((E (domain' (F x)) ∧ ¬ (¬ (E (domain' (F y))) ∨ domain' (F x) ≠ domain'
(F y))) ∧ E (codomain' (F x)) ∧ ¬ (¬ (E (codomain' (F y))) ∨ codomain'
(F x) ≠ codomain' (F y))) ∧ E (G (F x)) ∧ ¬ (¬ (E (G (F y))) ∨ G (F x) ≠ G
(F y))
AND 2)
(E ?x ∧ ¬ (¬ (E ?y) ∨ ?x ≠ ?y)) ← (((E (domain' ?x) ∧ ¬ (¬ (E (domain' ?y
)) ∨ domain' ?x ≠ domain' ?y)) ∧ E (codomain' ?x) ∧ ¬ (¬ (E (codomain' ?y
)) ∨ codomain' ?x ≠ codomain' ?y)) ∧ E (G ?x) ∧ ¬ (¬ (E (G ?y)) ∨ G ?x ≠ G
?y))
GET THIS
E (F x) ∧ ¬ (¬ (E (F y)) ∨ F x ≠ F y)
So, here one should merely use the 2)nd fact with substituted schematic variables by "F x" and "F y".
Will be glad to see any suggestions.
I am looking for a method to do rewriting, but with implications instead of equalities.
For example I know that x = 3 ∧ y = 4 implies Q x y and now I want to replace a positive occurrence of Q x y in my current subgoal with x = 3 ∧ y = 4.
Is there an existing method in Isabelle to do this?
For example I would like to do somthing like this (where implication_subst is the name of the method I am looking for):
lemma
assumes a1: "⋀x y. x = 3 ∧ y = 4 ⟹ Q x y"
shows "(∃x y. A x ∧ Q x y ∧ B y)"
proof (implication_subst a1)
show "∃x y. A x ∧ (x = 3 ∧ y = 4) ∧ B y"
sorry
qed
Below is my (incomplete) attempt to implement such a method using Eisbach, maybe this gives a better idea of what I am looking for:
named_theorems pos_cong
lemma implication_subst_exists[pos_cong]:
assumes "⋀x. P x ⟹ Q x"
and "∃x. P x"
shows "∃x. Q x"
using assms by blast
lemma implication_subst_conjl[pos_cong]:
assumes "P ⟹ Q"
and "P ∧ A"
shows "Q ∧ A"
using assms by blast
lemma implication_subst_conjr[pos_cong]:
assumes "P ⟹ Q"
and "A ∧ P"
shows "A ∧ Q"
using assms by blast
lemma implication_subst_neg[pos_cong]:
assumes "P ⟹ Q"
and "P"
shows "¬¬Q"
using assms by auto
lemma implication_subst_impl[pos_cong]:
assumes "P ⟹ ¬Q"
and "¬P ⟶ A"
shows "Q ⟶ A"
using assms by auto
lemma implication_subst_impr[pos_cong]:
assumes "P ⟹ Q"
and "A ⟶ P"
shows "A ⟶ Q"
using assms by auto
lemma implication_subst_neg_disj_l[pos_cong]:
assumes "P ⟹ ¬Q"
and "¬(¬P ∨ A)"
shows "¬(Q ∨ A)"
using assms by auto
lemma implication_subst_neg_disj_r[pos_cong]:
assumes "P ⟹ ¬Q"
and "¬(A ∨ ¬P)"
shows "¬(A ∨ Q)"
using assms by auto
method implication_subst_h uses r declares pos_cong = (
rule r
| (rule pos_cong, implication_subst_h r: r, assumption))
method implication_subst uses r declares pos_cong =
(implication_subst_h r: r pos_cong: pos_cong, (unfold not_not)?)
lemma example1:
assumes a1: "⋀x y. x = 3 ∧ y = 4 ⟹ Q x y"
shows "∃x y. A x ∧ Q x y ∧ B y"
proof (implication_subst r: a1)
show "∃x y. A x ∧ (x = 3 ∧ y = 4) ∧ B y"
sorry
qed
lemma example2:
assumes a1: "⋀x y. x = 3 ∧ y = 4 ⟹ Q x y"
shows "(∃x y. ¬(¬A x ∨ ¬Q x y ∨ ¬B y))"
proof (implication_subst r: a1)
show "∃x y. ¬ (¬ A x ∨ ¬ (x = 3 ∧ y = 4) ∨ ¬ B y)"
sorry
qed
Having gone through most exercises and also solved/proved in LEAN the first five propositional validities/properties at the end of chapter 3 in the LEAN manual, I still have trouble with the following implication (one of the implications needed for the proof of property 6):
theorem Distr_or_L (p q r : Prop) : (p ∨ q) ∧ (p ∨ r) → p ∨ (q ∧ r) :=
begin
intros pqpr,
have porq : p ∨ q, from pqpr.left,
have porr : p ∨ r, from pqpr.right,
sorry
end
The difficulty I face is mainly due to the case when p is not true, as I don't know how to combine, using LEAN tools, the two sides of the and in the hypothesis to obtain the fact that both q and r must hold under that scenario. I would greatly appreciate any help here; please help me understand how to construct this proof in the above setting without importing any other tactics except those in standard LEAN. For completeness, here is my proof of the other direction:
theorem Distr_or_R (p q r : Prop) : p ∨ (q ∧ r) → (p ∨ q) ∧ (p ∨ r) :=
begin
intros pqr,
exact or.elim pqr
( assume hp: p, show (p ∨ q) ∧ (p ∨ r), from
and.intro (or.intro_left q hp) (or.intro_left r hp) )
( assume hqr : (q ∧ r), show (p ∨ q) ∧ (p ∨ r), from
and.intro (or.intro_right p hqr.left) (or.intro_right p hqr.right) )
end
Hint. Try case splitting on both porq and porr.
Here's a solution
theorem Distr_or_L (p q r : Prop) : (p ∨ q) ∧ (p ∨ r) → p ∨ (q ∧ r) :=
begin
intros pqpr,
have porq : p ∨ q, from pqpr.left,
have porr : p ∨ r, from pqpr.right,
{ cases porq with hp hq,
{ exact or.inl hp },
{ cases porr with hp hr,
{ exact or.inl hp },
{ exact or.inr ⟨hq, hr⟩ } } }
end
Here is an asnwer for say problem without using cases. (explained later in tbe mentioned book)
example : p ∨ (q ∧ r) ↔ (p ∨ q) ∧ (p ∨ r) :=
Iff.intro
(fun hpqr => Or.elim hpqr
(fun hp: p =>
show (p ∨ q) ∧ (p ∨ r) from And.intro (Or.inl hp) (Or.inl hp))
(fun hqr: q ∧ r =>
show (p ∨ q) ∧ (p ∨ r) from And.intro (Or.inr hqr.left) (Or.inr hqr.right)))
(fun hpqpr: (p ∨ q) ∧ (p ∨ r) =>
have hpq: (p ∨ q) := hpqpr.left;
have hpr: (p ∨ r) := hpqpr.right;
Or.elim hpq
(fun hp: p => show p ∨ (q ∧ r) from Or.inl hp)
(fun hq: q => Or.elim hpr
(fun hp: p => show p ∨ (q ∧ r) from Or.inl hp)
(fun hr: r => show p ∨ (q ∧ r) from Or.inr (And.intro hq hr))))
You're using the information in the right part of the conjunction, otherwise, the left part doesn't give any info regarding r.
I'm trying to prove the following simple theorem I've come up with, that:
A point is on the boundary iff any small enough ball around that point contains points both in S and out of S.
Below I've managed to do the forward direction but I'm stuck on the backwards direction.
Using the same approach fails on the last step, the goal is close but not quite there, and I'm not sure what to do here:
lemma frontier_ball: "x ∈ frontier S ⟷
(∃r>0. (∀δ>0. δ<r ⟶ ((ball x δ) ∩ S ≠ {} ∧ (ball x δ) ∩ -S ≠ {})))"
(is "?lhs = ?rhs")
proof
{
assume "?lhs"
hence "x ∉ interior S ∧ x ∉ interior (-S)" by (auto simp: frontier_def interior_complement)
hence "∀δ>0. ((ball x δ) ∩ S ≠ {} ∧ (ball x δ) ∩ -S ≠ {})" by (auto simp: mem_interior)
then have "?rhs" by (simp add: Orderings.no_top_class.gt_ex)
}
{
assume "¬?lhs"
hence "x ∈ interior S ∨ x ∈ interior (-S)" by (auto simp: frontier_def interior_complement)
hence "∃δ>0. ball x δ ∩ S = {} ∨ ball x δ ∩ -S = {}" by (auto simp: mem_interior)
then have "¬?rhs" by (simp add: subset_ball)
}
qed
I tried to tell isabelle how to obtain such a delta but it's stuck on the obtain step:
lemma frontier_ball: "x ∈ frontier S ⟷
(∃r>0. (∀δ>0. δ<r ⟶ ((ball x δ) ∩ S ≠ {} ∧ (ball x δ) ∩ -S ≠ {})))"
(is "?lhs = ?rhs")
proof
{
assume "?lhs"
hence "x ∉ interior S ∧ x ∉ interior (-S)" by (auto simp: frontier_def interior_complement)
hence "∀δ>0. ((ball x δ) ∩ S ≠ {} ∧ (ball x δ) ∩ -S ≠ {})" by (auto simp: mem_interior)
then have "?rhs" by (simp add: Orderings.no_top_class.gt_ex)
}
{
fix r::real
assume "¬?lhs ∧ r>0"
hence "x ∈ interior S ∨ x ∈ interior (-S)" by (auto simp: frontier_def interior_complement)
then obtain r2 where "r2>0" and "ball x r2 ∩ S = {} ∨ ball x r2 ∩ -S = {}" by (auto simp: mem_interior)
then obtain δ where "δ>0 ∧ δ<r ∧ δ<r2" by auto
}
qed
Any pointers would be great!
Well, you can just construct such a δ. If you have r > 0 and r2 > 0 you want some δ that fulfils 0 < δ ≤ r2 and 0 < δ < r, why not just use min r2 (r/2)? You can define δ to be that and then you can prove the properties you want:
def δ ≡ "min r2 (r/2)"
with r2 A have δ: "δ > 0" "δ < r" "δ ≤ r2" by auto
with r2 have δ': "ball x δ ∩ S = {} ∨ ball x r2 ∩ -S = {}" using subset_ball[OF δ(3)] by auto
Or, a bit more direct:
lemma frontier_ball: "(x :: 'a :: {metric_space}) ∈ frontier S ⟷
(∃r>0. (∀δ>0. δ<r ⟶ ((ball x δ) ∩ S ≠ {} ∧ (ball x δ) ∩ -S ≠ {})))"
(is "?lhs = ?rhs")
proof -
{
assume "?lhs"
hence "x ∉ interior S ∧ x ∉ interior (-S)" by (auto simp: frontier_def interior_complement)
hence "∀δ>0. ((ball x δ) ∩ S ≠ {} ∧ (ball x δ) ∩ -S ≠ {})" by (auto simp: mem_interior)
then have "?rhs" by (simp add: Orderings.no_top_class.gt_ex)
}
moreover
{
assume lhs: "¬?lhs"
{
fix r :: real assume r: "r > 0"
from lhs have "x ∈ interior S ∨ x ∈ interior (-S)"
by (auto simp: frontier_def interior_complement)
then obtain δ where "δ > 0" "ball x δ ∩ S = {} ∨ ball x δ ∩ -S = {}"
by (auto simp: mem_interior)
with r have "min δ (r/2) > 0" "min δ (r/2) < r"
"ball x (min δ (r/2)) ∩ S = {} ∨ ball x (min δ (r/2)) ∩ -S = {}" using subset_ball by auto
hence "∃δ>0. δ < r ∧ (ball x δ ∩ S = {} ∨ ball x δ ∩ -S = {})" by blast
}
hence "¬?rhs" by blast
}
ultimately show ?thesis by blast
qed
For the record, I would avoid doing things like assume "A ∧ B". Do assume "A" "B" instead. That gives you two facts that you can work with directly, instead of having them wrapped up with a HOL conjunction in one fact.