What are finitely failed derivations? Are refutations the same as contradictions in the mathematical sense? What's the difference between general logic programs and definite logic programs?
There are no finitely failed derivations. Only failed derivations and finitely failed derivations trees. A failed derivations is a derivation that ends in failure. For example:
p :- q.
p :- p.
q :- fail.
The derivation that consists of the first rule of p and then the only rule of q is a failed derivation. Derivations might not only fail because an undefined predicate such as fail, but also because some head unification does not completely succeed.
Now what is a finitely failed derivation tree. Well if you look at all the derivations you get a tree. In a finitely failed derivation tree, the tree is finite and each derivation is failed. Finitely failed derivation trees have the following nice property:
- The interpreter terminates.
- The interpreter does not produce any answer substitution.
In practical Prolog systems this means that after posing your question you will get a No after a while (in some Prolog systems a false is displayed). Interestingly the above program will not terminate for the query p. It is an instance of a infinite derivation tree where each derivation is failed. The derivations are:
p - q - fail
p - p - q - fail
p - p - p - q - fail
Etc..
The notion of finitely failed derivation trees is defined for definite Prolog programs. One can now extend the notion of a Prolog program into general Prolog programs. In a general Prolog program the body might contain negative literals. And the idea is that the interpreter regresses into checking for finitely failed derivation trees for these literals.
One important question is how finitely failed derivation trees relate to mathematical derivations. Under what mathematical semantic should a goal fail? And how could we build an interpreter that implements this semantic? A particular class of semantics is based on the refutation method. Here we explain a derivation as establishing a contradiction:
P, ~G |= f => P |- G
This more or less implies double negation elimination and thus classical logic. But also other logics can be instrumental. As a start you might want to lookup the following book:
Logic for Applications
Anil Nerode, Richard A. Shore
2nd. Edition, 1997, Springer
Bye
Related
If I have a subset of logic programming which contains only one function symbol, am I able to do everything?
I think that I cannot but I am not sure at all.
A programming language can do anything user wants if it is a Turing-complete language. I was taught that this means it has to be able to execute if..then..else commands, recursion and that natural numbers should be defined.
Any help and opinions would be appreciated!
In classical predicate logic, there is a distinction between the formula level and the term level. Since an n-ary function can be represented as an (n+1)-ary predicate, restricting only the number of function symbols does not lessen the expressivity.
In prolog, there is no difference between the formula and the term level. You might pick an n-ary symbol p and try to encode turing machines or an equivalent notion(e.g. recursive functions) via nestings of p.
From my intution I would assume this is not possible: you can basically describe n-ary trees with variables as leaves, but then you can always unify these trees. This means that every rule head will match during recursive derivations and therefore you are unable to express any case distinction. Still, this is just an informal argument, not a proof.
P.S. you might also be interested in monadic logic, where only unary predicates are allowed. This fragment of first-order logic is decidable.
So I have the following working code in Prolog that produces the factorial of a given value of A:
factorial(0,1).
factorial(A,B) :- A>0, C is A-1, factorial(C,D), B is A*D.
I am looking for an explanation as to how this code works. I.e, what exactly happens when you ask the query: factorial(4, Answer).
Firstly,
factorial(0, 1).
I know the above is the "base case" of the recursive definition. What I am not sure of why/how it is the base case. My guess is that factorial(0, 1) inserts some structure containing (0, 1) as a member of "factorial". If so, what does the structure look like? I know if we say something like "rainy(seattle).", this means that Seattle is rainy. But "factorial(0, 1)"... 0, 1 is factorial? I realize it means factorial of 0 is 1, but how is this being used in the long run? (Writing this is helping me understand more as I go along, but I would like some feedback to make sure my thinking is correct.)
factorial(A,B) :- A>0, C is A-1, factorial(C,D), B is A*D.
Now, what exactly does the above code mean. How should I read it?
I am reading it as: factorial of (A, B) is true if A>0, C is A-1, factorial(C, D), B is A*D. That does not sound quite right to me... Is it?
"A > 0". So if A is equal to 0, what happens? It must not return at this point, or else the base case would never be used. So my guess is that A > 0 returns false, but the other functions are executed one last time. Did recursion stop because it reached the base case, or because A was not greater than 0? Or a combination of both? At what point is the base case used?
I guess that boils down to the question: What is the purpose of having both a base case and A > 0?
Sorry for the badly formed questions, thank you.
EDIT: In fact, I removed "A > 0" from the procedure and the code still works. So I guess my questions were not stupid at least. (And that code was taken from a tutorial.)
It is counterproductive to think of Prolog facts and rules in terms of data structures. When you write factorial(0, 1). you assert a fact to the Prolog interpreter that is assumed to be universally true. With this fact alone Prolog can answer questions of three types:
What is the factorial of 0? (i.e. factorial(0, X); the answer is X=1)
A factorial of what number is 1? (i.e. factorial(X,1); the answer is X=0)
Is it true that a factorial of 0 is 1? (i.e. factorial(0,1); the answer is "Yes")
As far as the rest of your Prolog program is concerned, only the first question is important. That is the question that the second clause of your factorial/2 rule will be asking at the end of evaluating a factorial.
The second rule uses comma operator, which is Prolog's way of saying "and". Your interpretation can be rewritten in terms of variables A and B like this:
B is a factorial of A when A>0, and C is set to A-1, and D is set to the factorial of C, and B is set to A times D
This rule covers all As above zero. The reference to factorial(C,D) will use the same rule again and again, until C arrives to zero. This is when this rule stops being applicable, so Prolog would grab the "base case" rule, and use 1 as its output. At this point, the chain of evaluating factorial(C, D) starts unwrapping, until it goes all the way to the initial invocation of the rule. This is when Prolog computes the final answer, and factorial/2 returns "Yes" and produces the desired output value.
In response to your edit, removing the A>0 is not dangerous only for getting the first result. Generally, you can ask Prolog to find you more results. This is when the factorial/2 with A>0 removed would fail spectacularly, because it would start going down the invocation chain of the second clause with negative numbers - a chain of calls that will end in numeric overflow or stack overflow, whichever comes first.
If you come from a procedural language background, the following C++ code might help. It mirrors pretty accurately the way the Prolog code executes (at least for the common case that A is given and B is uninstantiated):
bool fac(int a, int &b)
{
int c,d;
return
a==0 && (b=1,true)
||
a>0 && (c=a-1,true) && fac(c,d) && (b=a*d,true);
}
The Prolog comma operates like the sequential &&, and multiple clauses like a sequential ||.
My mental model for how prolog works is a tree traversal.
The facts and predicates in a prolog database form a forest of trees. When you ask the Prolog engine to evaluate a predicate:
?- factorial(6,N).
the Prolog engine looks for the tree rooted with the specified functor and arity (factorial/2 in this case). The Prolog engine then performs a depth-first traversal of that tree trying to find a solution using unification and pattern matching. Facts are evaluated as they are; For predicates, the right-hand side of the :- operator is evaluated, walking further into the tree, guided by the various logical operators.
Evaluation stops with the first successful evaluation of a leaf node in the tree, with the prolog engine remembering its state in the tree traversal. On backtracking, the tree traversal continues from where it left off. Execution is finally complete when the tree traversal is completed and there are no more paths to follow.
That's why Prolog is a descriptive language rather than an imperative language: you describe what constitutes truth (or falsity) and let the Prolog engine figure out how to get there.
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
I have been trying to acclimate to Prolog and Horn clauses, but the transition from formal logic still feels awkward and forced. I understand there are advantages to having everything in a standard form, but:
What is the best way to define the material conditional operator --> in Prolog, where A --> B succeeds when either A = true and B = true OR B = false? That is, an if->then statement that doesn't fail when if is false without an else.
Also, what exactly are the non-obvious advantages of Horn clauses?
What is the best way to define the material conditional operator --> in Prolog
When A and B are just variables to be bound to the atoms true and false, this is easy:
cond(false, _).
cond(_, true).
But in general, there is no best way because Prolog doesn't offer proper negation, only negation as failure, which is non-monotonic. The closest you can come with actual propositions A and B is often
(\+ A ; B)
which tries to prove A, then goes on to B if A cannot be proven (which does not mean that it is false due to the closed-world assumption).
Negation, however, should be used with care in Prolog.
Also, what exactly are the non-obvious advantages of Horn clauses?
That they have a straightforward procedural reading. Prolog is a programming language, not a theorem prover. It's possible to write programs that have a clear logical meaning, but they're still programs.
To see the difference, consider the classical problem of sorting. If L is a list of numbers without duplicates, then
sort(L, S) :-
permutation(L, S),
sorted(S).
sorted([]).
sorted([_]).
sorted([X,Y|L]) :-
X < Y,
sorted([Y|L]).
is a logical specification of what it means for S to contain the elements of L in sorted order. However, it also has a procedural meaning, which is: try all the permutations of L until you have one that it sorted. This procedure, in the worst case, runs through all n! permutations, even though sorting can be done in O(n lg n) time, making it a very poor sorting program.
See also this question.
I wan to use the destruct tactic to prove a statement by cases. I have read a couple of examples online and I'm confused. Could someone explain it better?
Here is a small example (there are other ways to solve it but try using destruct):
Inductive three := zero
| one
| two.
Lemma has2b2: forall a:three, a<>zero /\ a<>one -> a=two.
Now some examples online suggest doing the following:
intros. destruct a.
In which case I get:
3 subgoals H : zero <> zero /\ zero <> one
______________________________________(1/3)
zero = two
______________________________________(2/3)
one = two
______________________________________(3/3)
two = two
So, I want to prove that the first two cases are impossible. But the machine lists them as subgoals and wants me to PROVE them... which is impossible.
Summary:
How to exactly discard the impossible cases?
I have seen some examples using inversion but I don't understand the procedure.
Finally, what happens if my lemma depends on several inductive types and I still want to cover ALL cases?
How to discard impossible cases? Well, it's true that the first two obligations are impossible to prove, but note they have contradicting assumptions (zero <> zero and one <> one, respectively). So you will be able to prove those goals with tauto (there are also more primitive tactics that will do the trick, if you are interested).
inversion is a more advanced version of destruct. Additional to 'destructing' the inductive, it will sometimes generate some equalities (that you may need). It itself is a simple version of induction, which will additionally generate an induction hypothesis for you.
If you have several inductive types in your goal, you can destruct/invert them one by one.
More detailed walk-through:
Inductive three := zero | one | two .
Lemma test : forall a, a <> zero /\ a <> one -> a = two.
Proof.
intros a H.
destruct H. (* to get two parts of conjunction *)
destruct a. (* case analysis on 'a' *)
(* low-level proof *)
compute in H. (* to see through the '<>' notation *)
elimtype False. (* meaning: assumptions are contradictory, I can prove False from them *)
apply H.
reflexivity.
(* can as well be handled with more high-level tactics *)
firstorder.
(* the "proper" case *)
reflexivity.
Qed.
If you see an impossible goal, there are two possibilities: either you made a mistake in your proof strategy (perhaps your lemma is wrong), or the hypotheses are contradictory.
If you think the hypotheses are contradictory, you can set the goal to False, to get a little complexity out of the way. elimtype False achieves this. Often, you prove False by proving a proposition P and its negation ~P; the tactic absurd P deduces any goal from P and ~P. If there's a particular hypothesis which is contradictory, contradict H will set the goal to ~H, or if the hypothesis is a negation ~A then the goal will be A (stronger than ~ ~A but usually more convenient). If one particular hypothesis is obviously contradictory, contradiction H or just contradiction will prove any goal.
There are many tactics involving hypotheses of inductive types. Figuring out which one to use is mostly a matter of experience. Here are the main ones (but you will run into cases not covered here soon):
destruct simply breaks down the hypothesis into several parts. It loses information about dependencies and recursion. A typical example is destruct H where H is a conjunction H : A /\ B, which splits H into two independent hypotheses of types A and B; or dually destruct H where H is a disjunction H : A \/ B, which splits the proof into two different subproofs, one with the hypothesis A and one with the hypothesis B.
case_eq is similar to destruct, but retains the connections that the hypothesis has with other hypotheses. For example, destruct n where n : nat breaks the proof into two subproofs, one for n = 0 and one for n = S m. If n is used in other hypotheses (i.e. you have a H : P n), you may need to remember that the n you've destructed is the same n used in these hypotheses: case_eq n does this.
inversion performs a case analysis on the type of a hypothesis. It is particularly useful when there are dependencies in the type of the hypothesis that destruct would forget. You would typically use case_eq on hypotheses in Set (where equality is relevant) and inversion on hypotheses in Prop (which have very dependent types). The inversion tactic leaves a lot of equalities behind, and it's often followed by subst to simplify the hypotheses. The inversion_clear tactic is a simple alternative to inversion; subst but loses a little information.
induction means that you are going to prove the goal by induction (= recursion) on the given hypothesis. For example, induction n where n : nat means that you'll perform integer induction and prove the base case (n replaced by 0) and the inductive case (n replaced by m+1).
Your example is simple enough that you can prove it as “obvious by case analysis on a”.
Lemma has2b2: forall a:three, a<>zero/\a<>one ->a=two.
Proof. destruct a; tauto. Qed.
But let's look at the cases generated by the destruct tactic, i.e. after just intros; destruct a.. (The case where a is one is symmetric; the last case, where a is two, is obvious by reflexivity.)
H : zero <> zero /\ zero <> one
============================
zero = two
The goal looks impossible. We can tell this to Coq, and here it can spot the contradiction automatically (zero=zero is obvious, and the rest is a first-order tautology handled by the tauto tactic).
elimtype False. tauto.
In fact tauto works even if you don't start by telling Coq not to worry about the goal and wrote tauto without the elimtype False first (IIRC it didn't in older versions of Coq). You can see what Coq is doing with the tauto tactic by writing info tauto. Coq will tell you what proof script the tauto tactic generated. It's not very easy to follow, so let's look at a manual proof of this case. First, let's split the hypothesis (which is a conjunction) into two.
destruct H as [H0 H1].
We now have two hypotheses, one of which is zero <> zero. This is clearly false, because it's the negation of zero = zero which is clearly true.
contradiction H0. reflexivity.
We can look in even more detail at what the contradiction tactic does. (info contradiction would reveal what happens under the scene, but again it's not novice-friendly). We claim that the goal is true because the hypotheses are contradictory so we can prove anything. So let's set the intermediate goal to False.
assert (F : False).
Run red in H0. to see that zero <> zero is really notation for ~(zero=zero) which in turn is defined as meaning zero=zero -> False. So False is the conclusion of H0:
apply H0.
And now we need to prove that zero=zero, which is
reflexivity.
Now we've proved our assertion of False. What remains is to prove that False implies our goal. Well, False implies any goal, that's its definition (False is defined as an inductive type with 0 case).
destruct F.