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L={w|w€{a,b}, number of a is divisible by 2 }is the language. Can someone help me with the regular grammer of this?
The language is the set of all strings of a and b with an even number of a. This is a regular language and the goal is to produce a regular grammar for it.
Unless the regular grammar you're going to need is trivial, I would recommend always writing down the finite automaton first, and then converting it into a grammar. Converting a finite automaton into a grammar is very easy, and solving this problem is easy with a finite automaton. We will have two states: one will correspond to having seen an even number of a, the other an odd number. The state corresponding to having seen an even number of a will be accepting, and seeing b will not cause us to change states. The DFA is therefore:
b b
/-\ /-\
| V | V
----->(q0)--a-->(q1)
^ |
| a |
\---------/
A regular grammar for this can be formed by writing the transitions down as productions, using the states as nonterminal symbols, and including an empty production for the accepting state:
(q0) -> b(q0) | a(q1) | e
(q1) -> b(q1) | a(q0)
For the sake of completeness, you could run some other algorithms on the grammar or automaton and get a regular expression, maybe like this: b*(ab*ab*)* (just wrote that down, not sure if it's right or not, left as an exercise).
In the car industry you have thousand of different variants of components available to choose from when you buy a car. Not every component is combinable, so for each car there exist a lot of rules that are expressed in propositional logic. In my case each car has between 2000 and 4000 rules.
They look like this:
A → B ∨ C ∨ D
C → ¬F
F ∧ G → D
...
where "∧" = "and" / "∨" = "or" / "¬" = "not" / "→" = "implication".
With the tool "limboole" (http://fmv.jku.at/limboole/) I am able to to convert the propositional logic expressions into conjunctive normal form (CNF). This is needed in case I have to use a SAT solver.
Now, I would like to check the buildability feasibility for specific components within the rule set. For example, for each of the following expressions or combinations, I would like to check if the are feasible within the rule set.
(A) ∧ (B)
(A) ∧ (C ∨ F)
(B ∨ G)
...
My question is how to solve this problem. I asked a similar questions before (Tool to solve propositional logic / boolean expressions (SAT Solver?)), but with a different focus and now I am stuck again. Or I just do not understand it.
One option is to calculate all solutions with an ALLSAT approach of the rule set. Then I could check if each combination is part of any solution. If yes, I can derive that this specific combination is feasible.
Another option would be, that I add the combination to the rule set and then run a normal SAT solver. But I would have to do it for each expression I want to check.
What do you think is the most elegant or rather easiest way to solve this problem?
The best method which is known to me is to use "incremental solving under assumptions" technique. It was motivated by the same problem you have: multiple SAT instances (CNF formulae) which have some common subformulae.
Formally, you have some core Boolean formula C in CNF. And you have a set of assumptions {A_i}, i=1..n, where A_i is a Boolean formula in CNF also.
On the step 0 you provide to the solver your core formula C. It tries to solve it, says a result to you and save its state (lets call this state as core-state). If formula C is satisfiable, on the step i you provide assumption A_i to the solver and it continues its execution from the core-state. Actually, it tries to solve a formula C ∧ A_i but not from the beginning.
You can find a bunch of papers related to this topic easily, where much information is located. Also, you can check you favorite SAT-solver for the support of this technique.
~(P /\ Q) |- Q -> ~P
I don't know where to start.
Negation confuses me.
I have to solve this in Isabelle (a program), but if someone explains how to solve using natural deduction, it will be enough help.
This is an example that the quality of an SO question is many times determined by an answer, not the question. I kind of give this answer to thank M.Eberl for another useful answer, since I can't make comments.
As to a comment above, that you may be asking a homework question, the comment is valid, but if you're confused by negation, then you're mostly doomed anyway, until you make progress, so even one complete answer wouldn't help you, and here, there's no right answer.
The formula is so basic, except through applying step-by-step rules, it would be hard for anyone to prove that they understand what they're proved, without going through the multitude of tedious steps to do so.
For example:
lemma "~(P ∧ Q) ==> Q --> ~P"
by auto
Surely that gets you nothing, if the requirement is that you demonstrate understanding.
I've largely made progress "by the method of absorption over time", and in his answer, M.Eberl gave a significant outline of the basics of natural deduction. My interest in it was to mess around and see if I could absorb a little more.
As to rule and erule, there is the cheat sheet:
http://www.phil.cmu.edu/~avigad/formal/FormalCheatSheet.pdf
As to the proof of logic by means of Isabelle, Isabelle/HOL is so big and involved, that a little help, once, doesn't get you much, though collectively, it's all important.
A basic, logic equivalency
I learned long ago the equivalent statement of an implication. It's even in HOL.thy, line 998:
lemma disj_not2: "(P | ~Q) = (Q --> P)"
From that, it's easy to see, along with DeMorgon's laws (line 993 of HOL.thy), that you stated an equivalency in your question.
Well, of course, not quite, and that's where all the hassle comes in. Rearranging things, based on trivial equivalencies, to finally prove the equivalency. (While also knowing what the notation means, such as that your |- will be ==>. I use ASCII because I don't trust the graphical in browsers.)
M.Eberl mentioned structured proofs. Consider this one:
lemma "~(P ∧ Q) ==> Q --> ~P"
proof-
fix P Q :: bool
assume "~(P ∧ Q)"
hence "~P ∨ ~Q" by simp
hence "~Q ∨ ~P" by metis
thus "Q --> ~P" by metis
qed
What would I deserve in points, for homework? Nothing much. It's actually a testimony that metis knows how to use basic first-order logic. Otherwise, how did it know to make the jump from ~Q ∨ ~P to Q --> ~P?
Assuming you are talking about Isabelle/HOL, you can use ‘single-step tactics’ like rule, erule, assumption with the basic natural deduction rules. The ones you will probably need for your proposition are:
introduction rules notI, conjI, disjE impI
elimination rules like notE, conjE, disjE, impE
destruction rules like mp (modus ponens), conjunct1, conjunct2
If you want to find out what a particular rule means, just write e.g. thm notI and Isabelle will display the statement of the theorem.
You can set up a goal like
lemma "¬(P ∧ Q) ⟹ Q ⟶ ¬P"
and then write e.g.
apply (rule impI)
to apply the introduction rules for implication, which leaves you with the updated goal state
goal (1 subgoal):
1. ¬ (P ∧ Q) ⟹ Q ⟹ ¬ P
Now you find the next appropriate rule and apply that one etc. until all subgoals are solved. Then you can write done and your proof is complete.
As for assumption and erule: if you end up with a goal that has some P to prove and P is already in the assumptions, you can use apply assumption to solve it. (erule is like rule with assumption chained directly after it and is often convenient for applying elimination rules)
However, this kind of proof is very tedious to do. A better way would be to do the whole proof in Isar, Isabelle's structured proof language. For an introduction to Isar, you can have a look at chapter 5 of Concrete Semantics.
Similar to a JIT compiler, this is a LJEFAATGAA answer (learn just enough from another answer to give an answer).
This is more about what I learned than about what others may learn, which may help others learn; the Isabelle learning curve is quite brutal. I'd think the time has come and gone for the OP's need for help.
Except for M.Eberl's answer to exI and refl behavior explanation needed, it would still be a mystery to me about why rule thm produces specific goal states, for a particular thm, and why rule thm produces the message Failed to apply initial proof method.
Except for the precise outline given by Chris, with the rules and braces, I wouldn't have been able to fill in the precise details. An example that if a person has the time to learn, it's better for them to be given a partial answer, to make them do a little work, than to give them the complete answer.
Two main driving points
After a few comments, I show my proof from the outline. It got me the following understanding from having to work out the details, where I talk as if I know I'm right:
The use of apply(rule thm) is being applied to a combination of the chained facts and the goal statement, where this, in the output, displays one or more chained facts.
Braces start and end a local context/scope. Variables inside and outside the context/scope work as we would expect them to work, that is, as scope normally works in programming languages. So if I state fix P Q :: bool at the beginning of a proof, then state fix Q :: bool in a sub-context, the Q in the sub-context refers to a different variable than the parent-context Q.
Having properly credited Chris, I insert his outline here, so it can easily be compared to the Isar source:
1. ~(P & Q) premise
2. { Q assumption
3. { P assumption
4. P & Q and-introduction 3, 2
5. _|_ negation-elimination 1, 4 }
6. ~P negation-introduction 3-5 }
7. Q -> ~P implication-introduction 2-6
The source:
lemma "~(P & Q) ==> Q --> ~P"
proof-
fix P Q :: bool
assume a1: "~(P & Q)"
{
assume a2: "Q"
{
assume a3: "P"
have a4: "P & Q"
apply(rule conjI) apply(rule a3) by(rule a2)
(* NOTE: have to set 'notE' up right. Next 'hence False' doesn't do it. *)
(* hence False apply(rule) by(metis a1) *)
(* 'rule' applies the default of 'conjI', because there is the fact
'this: P & Q', which comes from 'hence'; 'rule notE' gives an error.*)
from a1 a4
have False (* From 'this' and 'goal': ~(P & Q) ⟹ P & Q ==> False *)
by(rule notE) (* notE: ~P ==> P ==> R *)
}
hence "~P"
apply(rule notI) by(assumption)
}
thus "Q --> ~P"
apply(rule impI) by(assumption)
qed
Understanding scope better
Thinking was being required at my statement have False. I wasn't setting things up right for notE. Just to see if I was on the right track, I would execute sledgehammer, but it wasn't able to prove False within that context.
It was because I was on auto-pilot, and was using fix Q :: bool and fix P :: bool in the two local contexts. I took them out and sledgehammer easily found proofs.
That's an example of a person knowing some logic, but not knowing how to implement the logic correctly in the languages Isabelle/HOL and Isabelle/Isar.
Next, I had to learn how to set up things correctly for apply(rule notE).
Belaboring a point
Part of my understanding about the above source comes from seeing the phrase chained facts in isar-ref.pdf. Minor exposure to the natural deduction rules, along with M.Eberl's explanation about unification, instantiation, and resolution finally helped make sense of what happens in the output panel.
Above, I have hence False commented out, and then use have False. Fortunately, in Isabelle, there are multiple ways to do things, but the goal was to apply notE. Even with that, there are different Isar keywords that can be used to set things up.
Anyway, seeing how chained facts are used with rule was a light-bulb moment. I guess here's effectively what's involved at the statement have False, as it relates to notE:
term "~P ==> P ==> R" (* notE: line 376 of HOL.thy *)
lemma "~(P & Q) ⟹ P & Q ==> False"
by(rule notE)
If I had an account, I would upvote the question to get rid of that -1.
Thanks to Chris for giving the precise outline.
Since you explicitly mentioned natural deduction. In a specific flavour of natural deduction -- where lines are numbered and the scope of assumptions is explicitly marked by boxes (denoted by curly braces below) -- one way of proving your statement is the following:
1. ~(P & Q) premise
2. { Q assumption
3. { P assumption
4. P & Q and-introduction 3, 2
5. _|_ negation-elimination 1, 4 }
6. ~P negation-introduction 3-5 }
7. Q -> ~P implication-introduction 2-6
Actually, since your goal is to prove an implication, you only have one choice at the start, namely implication introduction.
It would be a good exercise to translate the above proof as faithfully as possible into a structured Isar proof (e.g., using what is called "raw proof blocks" and incidentally also denoted by curly braces in Isabelle).
I have a Relation f defined as f: A -> B × C. I would like to write a firsr-order formula to constrain this relation to be a bijective function from A to B × C?
To be more precise, I would like the first order counter part of the following formula (actually conjunction of the three):
∀a: A, ∃! bc : B × C, f(a)=bc -- f is function
∀a1,a2: A, f(a1)=f(a2) → a1=a2 -- f is injective
∀(b, c) : B × C, ∃ a : A, f(a)=bc -- f is surjective
As you see the above formulae are in Higher Order Logic as I quantified over the relations. What is the first-order logic equivalent of these formulae if it is ever possible?
PS:
This is more general (math) question, rather than being more specific to any theorem prover, but for getting help from these communities --as I think there are mature understanding of mathematics in these communities-- I put the theorem provers tag on this question.
(Update: Someone's unhappy with my answer, and SO gets me fired up in general, so I say what I want here, and will probably delete it later, I suppose.
I understand that SO is not a place for debates and soapboxes. On the other hand, the OP, qartal, whom I assume is the unhappy one, wants to apply the answer from math.stackexchange.com, where ZFC sets dominates, to a question here which is tagged, at this moment, with isabelle and logic.
First, notation is important, and sloppy notation can result in a question that's ambiguous to the point of being meaningless.
Second, having a B.S. in math, I have full appreciation for the logic of ZFC sets, so I have full appreciation for math.stackexchange.com.
I make the argument here that the answer given on math.stackexchange.com, linked to below, is wrong in the context of Isabelle/HOL. (First hmmm, me making claims under ill-defined circumstances can be annoying to people.)
If I'm wrong, and someone teaches me something, the situation here will be redeemed.
The answerer says this:
First of all in logic B x C is just another set.
There's not just one logic. My immediate reaction when I see the symbol x is to think of a type, not a set. Consider this, which kind of looks like your f: A -> BxC:
definition foo :: "nat => int × real" where "foo x = (x,x)"
I guess I should be prolific in going back and forth between sets and types, and reading minds, but I did learn something by entering this term:
term "B × C" (* shows it's of type "('a × 'b) set" *)
Feeling paranoid, I did this to see if had fallen into a major gotcha:
term "f : A -> B × C"
It gives a syntax error. Here I am, getting all pedantic, and our discussion is ill-defined because the notation is ill-defined.
The crux: the formula in the other answer is not first-order in this context
(Another hmmm, after writing what I say below, I'm full circle. Saying things about stuff when the context of the stuff is ill-defined.)
Context is everything. The context of the other site is generally ZFC sets. Here, it's HOL. That answerer says to assume these for his formula, wich I give below:
Ax is true iff x∈A
Bx is true iff x∈B×C
Rxy is true iff f(x)=y
Syntax. No one has defined it here, but the tag here is isabelle, so I take it to mean that I can substitute the left-hand side of the iff for the right-hand side.
Also, the expression x ∈ A is what would be in the formula in a typical set theory textbook, not Rxy. Therefore, for the answerer's formula to have meaning, I can rightfully insert f(x) = y into it.
This then is why I did a lot of hedging in my first answer. The variable f cannot be in the formula. If it's in the formula, then it's a free variable which is implicitly quantified. Here's the formula in Isar syntax:
term "∀x. (Ax --> (∃y. By ∧ Rxy ∧ (∀z. (Bz ∧ Rxz) --> y = z)))"
Here it is with the substitutions:
∀x. (x∈A --> (∃y. y∈B×C ∧ f(x)=y ∧ (∀z. (z∈B×C ∧ f(x)=z) --> y = z)))
In HOL, f(x) = f x, and so f is implicitly, universally quantified. If this is the case, then it's not first-order.
Really, I should dig deep to recall what I was taught, that f(x)=y means:
(x,f(x)) = (x,y) which means we have to have (x,y)∈(A, B×C)
which finally gets me:
∀x. (x∈A -->
(∃y. y∈B×C ∧ (x,y)∈(A,B×C) ∧ (∀z. (z∈B×C ∧ (x,z)∈(A,B×C)) --> y = z)))
Finally, I guess it turns out that in the context of math.stackexchange.com, it's 100% on.
Am I the only one who feels compulsive about questioning what this means in the context of Isabelle/HOL? I don't accept that everything here is defined well enough to show that it's first order.
Really, qartal, your notation should be specific to a particular logic.
First answer
With Isabelle, I answer the question based on my interpretation of your
f: A -> B x C, which I take as a ZFC set, in particular a subset of the
Cartesian product A x (B x C)
You're sort of mixing notation from the two logics, that of ZFC
sets and that of HOL. Consequently, I might be off on what I think you're
asking.
You don't define your relation, so I keep things simple.
I define a simple ZFC function, and prove the first
part of your first condition, that f is a function. The second part would be
proving uniqueness. It can be seen that f satisfies that, so once a
formula for uniqueness is stated correctly, auto might easily prove it.
Please notice that the
theorem is a first-order formula. The characters ! and ? are ASCII
equivalents for \<forall> and \<exists>.
(Clarifications must abound when
working with HOL. It's first-order logic if the variables are atomic. In this
case, the type of variables are numeral. The basic concept is there. That
I'm wrong in some detail is highly likely.)
definition "A = {1,2}"
definition "B = A"
definition "C = A"
definition "f = {(1,(1,1)), (2,(1,1))}"
theorem
"!a. a \<in> A --> (? z. z \<in> (B × C) & (a,z) \<in> f)"
by(auto simp add: A_def B_def C_def f_def)
(To completely give you an example of what you asked for, I would have to redefine my function so its bijective. Little examples can take a ton of work.)
That's the basic idea, and the rest of proving that f is a function will
follow that basic pattern.
If there's a problem, it's that your f is a ZFC set function/relation, and
the logical infrastructure of Isabelle/HOL is set up for functions as a type.
Functions as ordered pairs, ZFC style, can be formalized in Isabelle/HOL, but
it hasn't been done in a reasonably complete way.
Generalizing it all is where the work would be. For a particular relation, as
I defined above, I can limit myself to first-order formulas, if I ignore that
the foundation, Isabelle/HOL, is, of course, higher-order logic.
I am studying membership algorithms and I am working on this particular problem which says the following:
Exhibit an algorithm that, given any regular language L, determines whether or not L = L*
So, my first thought was, we have L* which is Kleene star of L and to determine if L = L*, well couldn't we just say that since L is regular, we know L* is by definition which states that the family of regular languages is closed under star-closure.
Therefore L will always be equal to L*?
I feel like there is definitely a lot more to it, there is probably something I am missing. Any help would be appreciated. Thanks again.
since L is regular, we know L* is by definition which states that the family of regular languages is closed under star-closure. Therefore L will always be equal to L*?
No. Regular(L) --> Regular(L*), but that does not mean that L == L*. Just because two languages are both regular does not mean that they are the same regular language. For instance, a* and b* are both regular languages, but this does not make them the same language.
A example of L != L* would be the language L = a*b*, and thus L* = (a*b*)*. The string abab is part of L* but not part of L.
As far as an algorithm goes, let me remind you that the concept of a regular language is one that can be parsed by a DFA - and for any given DFA, there is a single optimal reduction of that DFA.
The implication that you stated is wrong. Closedness under the Kleene star means only that L* is again regular, if L is regular.
One possibility to check whether L = L* is to compute the minimal automaton for both and then checking for equivalence.