I want to know why the order of quantifiers are important in a logic formula?
When I read books about logic programming, such points are mentioned, but did not say why.
Is there any one could explain with some examples?
Also, how can we determine order of quantifiers from a given logic formula?
Thanks in advance!
You would be well advised to read a book about first-order logic before the books about
logic programming.
Consider the true statement:
1. Everybody has a mother
Let's formalize it in FOL. To keep it simple, we'll say
that the universe of discourse is the set of people, i.e.
our individual variables x, y, z... range over people. Then
1 becomes:
1F. (x)(Ey)Mother(y,x)
which we can read as: For every person x there exists
some person y such that y is the mother of x.
Now let's swap the order of the universal quantifier (x) and existential
quantifier (Ey):
2F. (Ey)(x)Mother(y,x)
That reads: There is some person y such that for every person x,
y is the mother of x. Or in plain English:
2. There is somebody who is the mother of everybody
You see that swapping the quantifiers changes the meaning of the statement,
taking us from the true statement 1 to the false statement 2. Indeed, to the absurdly false statement
2, which entails that somebody is their own mother.
That's why the order of quantifiers matters.
how can we determine order of quantifiers from a given logic formula?
Well, in 1F and 2F, for example, all the variables are already bound by quantifiers,
so there's nothing to determine. The order of the quantifiers is what you see,
left to right.
Suppose one of the variables was free (not bound), e.g.
3F. (Ey)Mother(y,x)
You might read that as: There is someone who is the mother of x, for variable person x.
But this formula really doesn't express any statement. It expresses a unary predicate of persons, the predicate Someone is the mother of x. If you free up the remaining variable:
4F. Mother(x,y)
then you have the binary predicate, or relation: x is the mother of y.
A formula with 1,2,...,n free variables expresses a unary, binary,...,n-ary predicate.
Given a predicate, you can make a statement by binding free variables with quantifiers and/or substituting individual constants for the free variables. From 4F you can make:
(x)(y)Mother(x,y) (Everybody is everybody's mother)
(Ex)(y)Mother(x,y) (Somebody is everybody's mother)
(Ex)(Ey)Mother(x,y) (Somebody is somebody's mother)
(x)Mother(x,Arnold) (Everybody is the mother of Arnold)
(x)Mother(Bernice,x) (Bernice is the mother of everybody)
Mother(Arnold,Bernice) (Arnold is the mother of Bernice)
...
...
and so on ad nauseam.
What this should make clear is that if a formula has free variables, and therefore expresses
a predicate, the formula as such does not imply any particular way of quantifying
the free variables, or that they should be quantified at all.
Related
What is the purpose of this:
route(Town1,Town2,Distance):-
road(Town1,Town2,Distance).
Also, what does ! mean in this context? If someone knows exactly what it means, because all I hear is "cut" without any other explanation.
%TRAVELLING SALESMAN PROBLEM
DOMAINS
town = symbol
distance = integer
PREDICATES
nondeterm road(town,town,distance)
nondeterm route(town,town,distance)
CLAUSES
road("tampa","houston",200).
road("gordon","tampa",300).
road("houston","gordon",100).
road("houston","kansas_city",120).
road("gordon","kansas_city",130).
route(Town1,Town2,Distance):-
road(Town1,Town2,Distance).
route(Town1,Town2,Distance):-
road(Town1,X,Dist1),
route(X,Town2,Dist2),
Distance=Dist1+Dist2, !.
GOAL
route("tampa", "kansas_city", X),
write("Distance from Tampa to Kansas City is ",X),nl.
Prolog programs use recursion heavily, which is why you need a base case for the recursion to stop. Here:
route(Town1,Town2,Distance):-
road(Town1,Town2,Distance).
We're saying "If there's a direct road between Town1 and Town2, then we're done, no need to check for a route through any other town because this is a route, just give me the distance." In other words, we are saying "If there is a defined clause that matches the first two arguments, what must Distance be to fully match it?"
The ! is the cut/1 predicate. It stops backtracking. Here it is, described more thoroughly:
Sometimes it is desirable to selectively turn off backtracking. Prolog provides a predicate that performs this function. It is called the cut/1, represented by an exclamation point (!).
The cut/1 effectively tells Prolog to freeze all the decisions made so far in this predicate. That is, if required to backtrack, it will automatically fail without trying other alternatives.
In the case of your code:
route(Town1,Town2,Distance):-
road(Town1,X,Dist1),
route(X,Town2,Dist2),
Distance=Dist1+Dist2, !.
You are telling Prolog that, if you find a road between Town1 and Town X, and then from Town X there is a route to finish of the path between Town1 and Town2, then recurse; if there is a direct road between X and Town2, then the predicate at the top of your question holds. If not, it will recursively loop. Eventually, Dist2 will have a final value which will float back up to this predicate.
This is where ! comes in. Once you have a route that holds, it says "Stop, don't try and find another route, I only want 1 route. Don't backtrack for any other possible routes." Otherwise, you'd get multiple write("Distance from Tampa to Kansas City is ",X),nl. with different X values, which might not have made sense to the author of this program.
Also, as #DavidTonhofer says, the = should be replaced with is as the latter will force arithmetic evaluation. For example X = 2+4 compares X to the unevaluated expression 2+4, but with is you'll be comparing X to 6.
Translate the following statements into FOL sentances
1) Alex likes John
Likes(alex, john) - I know this is correct
2) Each person is either a man or woman
AxAy( man(x) v woman(y) )
EDIT: Is this better??: Az(Person(z) -> man(x) v woman(y))
OR EDIT: Is this better??: Ax(Person(x) -> man(x) v woman(x))
3) No one is both man and woman
Ex( (man(x) ^ ¬woman(x)) v (¬man(x) ^ woman(x)) )
4) Alex likes a man who likes a woman
AxEy(Likes( man(x), woman(y) ) -> Likes(alex, man(x) ))
Thanks
Here is a screenshot of the background info
EDIT: For number 3, I have found this online
"The exclusive disjunction of p and q asserts that either p is true or q is true but not both. The natural, but long-winded, way to express exclusive disjunction, then, is (p | q) & ~(p & q)."
If this can apply, then I assume the correct answer is Ax( (man(x) v woman(x)) ^ ¬(man(x) & woman(x)) )
But now I am getting confused as to how 2 and 3 are different...
Hey I just wanted to know if these were correct
1. Testing the Correctness of Translation
One of the ways to test if a first order sentence agrees with an informal specification is to use a model.
To perform a test you need:
To determine how many relations are there in your first order sentence and list them. This list would play the role of the logical signature of your model.
Now take a set of individuals large enough to have all interesting combinations of properties assigned to at least one individual.
Assign the properties to individuals according to your understanding of the meaning of relations in your first order sentence.
Finally for each universally quantified variable try different assignments of individuals to variables and check if the property holds.
Consider one of the examples from the post.
Informal specification: Alex likes a man who likes a woman
We have
one constant symbol: Alex
two unary relations: Man(x) and Woman(x)
a binary relation: Likes(x,y)
1.1 Structure No. 1
Now consider a structure where we have an individual for Alex, an individual for a man who is not Alex, and an individual for a woman who is not Alex.
Let's start with three: p1, p2, p3.
Alex is p1
Man(p2)
Woman(p3)
Woman(p1)
Likes (p2,p3)
Likes (p1,p2)
The informal specification says that Alex likes someone who is a Man and Likes someone else who is a Woman. Our model satisfies this specification.
Now consider the following statement taken from the OP:
AxEy(Likes( man(x), woman(y) ) -> Likes(alex, man(x) ))
This statement is from a different language. Here man(x) and woman(y) are unary functions instead of unary relations, so we cannot check this sentence.
I would not speculate any further on the subject of woman(y) or man(x) being a function (Pun intended). Instead I would consider a different sentence.
AxEy(Man(x) & Woman(y) & Likes( x, y ) -> Likes(alex, x ))
Let's see what happens if x=p2 and y=p1. In our model the premise
Man(p2) & Woman(p3) & Likes( p2, p3 )
holds, and the conclusion
Likes(p1,p2)
also holds, therefore the formula is true under this variable assignment.
1.2 Structure No. 2
Now consider a different model. This time with four individuals: p1,p2,p3,p4.
Alex is p1
Man(p2)
Man(p4)
Woman(p3)
Woman(p1)
Likes (p2,p3)
Likes (p1,p2)
Likes (p4,p3)
This model also satisfies our informal specification as Alex likes p2 who is a Man and likes p3 who is a Woman.
Consider the assignment x=p4 and y=p3.
Once again the premise holds in our model
Man(p4) & Woman(p3) & Likes( p4, p3 )
however the conclusion
Likes(p1,p4)
does not hold in our model. Since x is universally quantified this might be a problem as the formula does not hold under this assignment, however y is under an existential quantifier. Let's see if we can find an assignment for y that would turn our formula into a true statement with x=p4.
This is indeed possible. Let y=p2.
Man(p4) & Woman(p2) & Likes( p4, p2 )
does not hold, therefore the formula
Man(p4) & Woman(p2) & Likes( p4, p2 ) -> Likes(p1,p4)
is true. This sounds better yet to satisfy the formula we took an assignment that is far from the intended meaning of the informal specification. The original specification clearly was not speaking about a man who likes another man. So we are not done yet.
1.3 Structure No. 3
Consider a model with just two individuals: p1 and p2.
Alex is p1
Man(p2)
Woman(p1)
The informal specification does not hold in this model, so our sentence also should be false. There are four possible assignments of variables
x=p1,y=p2
x=p1,y=p1
x=p2,y=p1
x=p2,y=p2
Since Like(x,y) is always false in our model, the premise fails therefore according to the rules of implication, the formula is true. So our sentence is also true in a model where it should not hold. Once again, our first-order formalization does not hold against the informal specification.
This seems to be a very complex process that we have taken to test the translation. It assumes a certain skill of finding counterexamples and always keeping in mind the intended meaning of the informal specification.
2. How to Come up with a Correct Translation
Let's look back at our informal specification
Informal specification: Alex likes a man who likes a woman
and reformulate it in a way that is easier to translate
There exists a man whom Alex likes, and this man likes some woman
There are two parts in this sentence connected with an and
Ex (Man(x) & Likes(Alex,x))
Ey (Woman(y) & Likes(x,y))
So we have
Ex (Man(x) & Likes(Alex,x) & Ey (Woman(y) & Likes(x,y))).
You like prenex normal forms where all quantifiers are collected in one quantifier prefix. We can apply logical equivalence to get
ExEy (Man(x) & Likes(Alex,x) & Woman(y) & Likes(x,y)).
Now let's check if this statement agrees with the specification in each of the structures from the previous section.
2.1 Structure 1
Since the variables are existentially quantified and the specification holds, we need to find only one satisfying assignment for the formula.
Consider
Alex is p1
x = p2
y = p3
The conjunction holds under this assignment.
2.2 Structure 2
The same assignment as in the previous subsection can be used. In fact
Structure 1 is an substructure of Structure 2. For an existentially quantified statement we know that if it is true in a substructure, it is also true in the whole structure.
2.3 Structure 3
Since there is no pair of elements (x,y) such that Likes(x,y) in our structure, the conjunction
Man(x) & Likes(Alex,x) & Woman(y) & Likes(x,y)
cannot be true, so the statement is false. We also know that Structure 3 does not satisfy our informal specification, so our formula has passed our tests.
Our testing procedure is by no means complete. However it gives us some assurance that the translation is indeed correct.
I was reading Learn Prolog Now! and I'm confused about their use of logic and I was hoping someone can clarify the ambiguity of arguments in rules.
For example, on Chapter 1, knowledge base 4 they reference the following rule:
loves(vincent,mia).
They then provide a query:
loves(marsellus,X), woman(X).
and then the make the English statement:
"Now, remember that , means "and", so this query says: is there any individual X such that Marsellus loves X and X is a woman ?"
The key word being, Marsellus loves X.
This is worded in a way that makes me think there is some "logic" tied to the arrangement of arguments, such as to say,
Is it true that loves(vincent,mia). "logically" means:
vincent loves mia?
BUT is it false that
mia loves vincent?
That's what I originally thought but then on Chapter 3, example 2: Descendant they have the following example of:
child(bridget,caroline).
Then the English statement: "That is, Caroline is a child of Bridget"
But if #1 above is true (vincent loves mia), then shouldn't this be read as Bridget is a child of Caroline?
Or does the order of arguments not matter and it's your additional programming that will determine the logic of the arguments?
loves(X, Y) intuitively would mean X loves Y but not Y loves X. But as a programmer, you could assume it's symmetrical. It just wouldn't be as intuitive. It just depends upon how you choose to define it. Prolog Now chose the most intuitive meaning, so loves(X, Y) means X loves Y, but not Y loves X.
Whether child(X, Y) means X is a child of Y or Y is a child of X is up to the programmer. child used by itself as the name of the fact/predicate in this context is a bit ambiguous here (it's not a verb and isn't part of a phrase) and you just have to roll with whatever the Prolog Now site chose as their meaning. When you view the functor as a verb, the arguments are interpreted left to right as expected. So, better, would be to use child_of(X, Y) to mean X is a child of Y so that it's clearer.
Prolog as a language wouldn't care whether you define love(X,Y) as X loves Y or Y loves X. It's probably just a convention that we tend to follow in first order logic.
As to why we'd choose the ordering convention we usually follow, It's probably because the predicate defines a relation between two terms. So the order would depend on how you define the relation.
A predicate r(a,b) means that (a,b) is in the relation r:A->B.
The way i would clear my doubts of the case you've given in the question is:
loves(a,?) ? should anything which a loves. And
child(a,?) ? should be anything which is a child of a.
It would depend on how you define the relation 'child'.
As far as i know ( which isn't far ), There's no hard rule in logic that stops you from defining predicates the way you want. You could define love(X,Y) as 'X loves Y' or 'X is loved by Y' as long as you're consistent.
I read the question asked in Herbrand universe, Herbrand Base and Herbrand Model of binary tree (prolog) and the answers given, but I have a slightly different question more like a confirmation and hopefully my confusion will be clarified.
Let P be a program such that we have the following facts and rule:
q(a, g(b)).
q(b, g(b)).
q(X, g(X)) :- q(X, g(g(g(X)))).
From the above program, the Herbrand Universe
Up = {a, b, g(a), g(b), q(a, g(a)), q(a, g(b)), q(b, g(a)), q(b, g(b)), g(g(a)), g(g(b))...e.t.c}
Herbrand base:
Bp = {q(s, t) | s, t E Up}
Now come to my question(forgive me for my ignorance), i included q(a, g(a)) as an element in my Herbrand Universe but from the fact, it states q(a, g(b)). Does that mean that q(a, g(a)) does not suppose to be there?
Also since the Herbrand models are subset of the Herbrand base, how do i determine the least Herbrand model by induction?
Note: I have done a lot of research on this, and some parts are well clear to me but still i have this doubt in me thats why i want to seek the communities opinion. Thank you.
From having the fact q(a,g(b)) you cannot conclude whether or not q(a,g(a)) is in the model. You will have to generate the model first.
For determining the model, start with the facts {q(a,g(b)), q(b,g(b))} and now try to apply your rules to extend it. In your case, however, there is no way to match the right-hand side of the rule q(X,g(X)) :- q(X,g(g(g(X)))). to above facts. Therefore, you are done.
Now imagine the rule
q(a,g(Y)) :- q(b,Y).
This rule could be used to extend our set. In fact, the instance
q(a,g(g(b))) :- q(b,g(b)).
is used: If q(b,g(b)) is present, conclude q(a,g(g(b))). Note that we are using here the rule right-to-left. So we obtain
{q(a,g(b)), q(b,g(b)), q(a,g(g(b)))}
thereby reaching a fixpoint.
Now take as another example you suggested the rule
q(X, g(g(g(X)))) :- q(X, g(X)).
Which permits (I will no longer show the instantiated rule) to generate in one step:
{q(a,g(b)), q(b,g(b)), q(a,g(g(g(b)))), q(b, g(g(g(b))))}
But this is not the end, since, again, the rule can be applied to produce even more! In fact, you have now an infinite model!
{g(a,gn+1(b)), g(b, gn+1(b))}
This right-to-left reading is often very helpful when you are trying to understand recursive rules in Prolog. The top-down reading (left-to-right) is often quite difficult, in particular, since you have to take into account backtracking and general unification.
Concerning your question:
"Also since the Herbrand models are subset of the Herbrand base, how do i determine the least Herbrand model by induction?"
If you have a set P of horn clauses, the definite program, then you can define
a program operator:
T_P(M) := { H S | S is ground substitution, (H :- B) in P and B S in M }
The least model is:
inf(P) := intersect { M | M |= P }
Please note that not all models of a definite program are fixpoints of the
program operator. For example the full herbrand model is always a model of
the program P, which shows that definite programs are always consistent, but
it is not necessarily a fixpoint.
On the other hand each fixpoint of the program operator is a model of the
definite program. Namely if you have T_P(M) = M, then one can conclude
M |= P. So that after some further mathematical reasoning(*) one finds that
the least fixpoint is also the least model:
lfp(T_P) = inf(P)
But we need some further considerations so that we can say that we can determine
the least model by a kind of computation. Namely one easily observes that the
program operator is contiguous, i.e. preserves infinite unions of chains, since
horn clauses do not have forall quantifiers in their body:
union_i T_P(M_i) = T_P(union_i M_i)
So that again after some further mathematical reasoning(*) one finds that we can
compute the least fixpoint via iteration, witch can be used for simple
induction. Every element of the least model has a simple derivation of finite
depth:
union_i T_P^i({}) = lpf(T_P)
Bye
(*)
Most likely you find further hints on the exact mathematical reasoning
needed in this book, but unfortunately I can't recall which sections
are relevant:
Foundations of Logic Programming, John Wylie Lloyd, 1984
http://www.amazon.de/Foundations-Programming-Computation-Artificial-Intelligence/dp/3642968287
Because Prolog uses chronological backtracking(from the Prolog Wikipedia page) even after an answer is found(in this example where there can only be one solution), would this justify Prolog as using eager evaluation?
mother_child(trude, sally).
father_child(tom, sally).
father_child(tom, erica).
father_child(mike, tom).
sibling(X, Y) :- parent_child(Z, X), parent_child(Z, Y).
parent_child(X, Y) :- father_child(X, Y).
parent_child(X, Y) :- mother_child(X, Y).
With the following output:
?- sibling(sally, erica).
true ;
false.
To summarize the discussion with #WillNess below, yes, Prolog is strict. However, Prolog's execution model and semantics are substantially different from the languages that are usually labelled strict or non-strict. For more about this, see below.
I'm not sure the question really applies to Prolog, because it doesn't really have the kind of implicit evaluation ordering that other languages have. Where this really comes into play in a language like Haskell, you might have an expression like:
f (g x) (h y)
In a strict language like ML, there is a defined evaluation order: g x will be evaluated, then h y, and f (g x) (h y) last. In a language like Haskell, g x and h y will only be evaluated as required ("non-strict" is more accurate than "lazy"). But in Prolog,
f(g(X), h(Y))
does not have the same meaning, because it isn't using a function notation. The query would be broken down into three parts, g(X, A), h(Y, B), and f(A,B,C), and those constituents can be placed in any order. The evaluation strategy is strict in the sense that what comes earlier in a sequence will be evaluated before what comes next, but it is non-strict in the sense that there is no requirement that variables be instantiated to ground terms before evaluation can proceed. Unification is perfectly content to complete without having given you values for every variable. I am bringing this up because you have to break down a complex, nested expression in another language into several expressions in Prolog.
Backtracking has nothing to do with it, as far as I can tell. I don't think backtracking to the nearest choice point and resuming from there precludes a non-strict evaluation method, it just happens that Prolog's is strict.
That Prolog pauses after giving each of the several correct answers to a problem has nothing to do with laziness; it is a part of its user interaction protocol. Each answer is calculated eagerly.
Sometimes there will be only one answer but Prolog doesn't know that in advance, so it waits for us to press ; to continue search, in hopes of finding another solution. Sometimes it is able to deduce it in advance and will just stop right away, but only sometimes.
update:
Prolog does no evaluation on its own. All terms are unevaluated, as if "quoted" in Lisp.
Prolog will unfold your predicate definitions as written and is perfectly happy to keep your data structures full of unevaluated uninstantiated holes, if so entailed by your predicate definitions.
Haskell does not need any values, a user does, when requesting an output.
Similarly, Prolog produces solutions one-by-one, as per the user requests.
Prolog can even be seen to be lazier than Haskell where all arithmetic is strict, i.e. immediate, whereas in Prolog you have to explicitly request the arithmetic evaluation, with is/2.
So perhaps the question is ill-posed. Prolog's operations model is just too different. There are no "results" nor "functions", for one; but viewed from another angle, everything is a result, and predicates are "multi"-functions.
As it stands, the question is not correct in what it states. Chronological backtracking does not mean that Prolog will necessarily backtrack "in an example where there can be only one solution".
Consider this:
foo(a, 1).
foo(b, 2).
foo(c, 3).
?- foo(b, X).
X = 2.
?- foo(X, 2).
X = b.
So this is an example that does have only one solution and Prolog recognizes that, and does not attempt to backtrack. There are cases in which you can implement a solution to a problem in a way that Prolog will not recognize that there is only one logical solution, but this is due to the implementation and is not inherent to Prolog's execution model.
You should read up on Prolog's execution model. From the Wikipedia article which you seem to cite, "Operationally, Prolog's execution strategy can be thought of as a generalization of function calls in other languages, one difference being that multiple clause heads can match a given call. In that case, [emphasis mine] the system creates a choice-point, unifies the goal with the clause head of the first alternative, and continues with the goals of that first alternative." Read Sterling and Shapiro's "The Art of Prolog" for a far more complete discussion of the subject.
from Wikipedia I got
In eager evaluation, an expression is evaluated as soon as it is bound to a variable.
Then I think there are 2 levels - at user level (our predicates) Prolog is not eager.
But it is at 'system' level, because variables are implemented as efficiently as possible.
Indeed, attributed variables are implemented to be lazy, and are rather 'orthogonal' to 'logic' Prolog variables.