How does a Resolution algorithm work for propositional logic? - algorithm

I haven't been able to understand what the resolution rule is in propositional logic. Does resolution simply state some rules by which a sentence can be expanded and written in another form?
Following is a simple resolution algorithm for propositional logic. The function returns the set of all possible clauses obtained by resolving it's 2 input. I can't understand the working of the algorithm, could someone explain it to me?
function PL-RESOLUTION(KB,α) returns true or false
inputs: KB, the knowledge base, a sentence α in propositional logic, the query, a
sentence in propositional logic
clauses <--- the set of clauses in the CNF representation of KB ∧ ¬α
new <--- {}
loop do
for each Ci, Cj in clauses do
resolvents <----- PL-RESOLVE(Ci, Cj)
if resolvents contains the empty clause then return true
new <--- new ∪ resolvents
if new ⊆ clauses then return false
clauses <---- clauses ∪ new

It's a whole topic of discussion but I'll try to explain you one simple example.
Input of your algorithm is KB - set of rules to perform resolution. It easy to understand that as set of facts like:
Apple is red
If smth is red Then this smth is sweet
We introduce two predicates R(x) - (x is red) and S(x) - (x is sweet). Than we can written our facts in formal language:
R('apple')
R(X) -> S(X)
We can substitute 2nd fact as ¬R v S to be eligible for resolution rule.
Caluclating resolvents step in your programs delete two opposite facts:
Examples: 1) a & ¬a -> empty. 2) a('b') & ¬a(x) v s(x) -> S('b')
Note that in second example variable x substituted with actual value 'b'.
The goal of our program to determine if sentence apple is sweet is true. We write this sentence also in formal language as S('apple') and ask it in inverted state. Then formal definition of problem is:
CLAUSE1 = R('apple')
CLAUSE2 = ¬R(X) v S(X)
Goal? = ¬S('apple')
Algorithm works as follows:
Take clause c1 and c2
calculate resolvents for c1 and c2 gives new clause c3 = S('apple')
calculate resolvents for c3 and goal gives us empty set.
That means our sentence is true. If you can't get empty set with such resolutions that means sentence is false (but for most cases in practical applications it's a lack of KB facts).

Consider clauses X and Y, with X = {a, x1, x2, ..., xm} and Y = {~a, y1, y2, ..., yn}, where a is a variable, ~a is its negation, and the xi and yi are literals (i.e., possibly-negated variables).
The interpretation of X is the proposition (a \/ x1 \/ x2 \/ ... \/ xm) -- that is, at least one of a or one of the xi must be true, assuming X is true. Likewise for Y.
We assume that X and Y are true.
We also know that (a \/ ~a) is always true, regardless of the value of a.
If ~a is true, then a is false, so ~a /\ X => {x1, x2, ..., xm}.
If a is true, then ~a is false. In this case a /\ Y => {y1, y2, ..., yn}.
We know, therefore, that {x1, x2, ..., xm, y1, y2, ..., yn} must be true, assuming X and Y are true. Observe that the new clause does not refer to variable a.
This kind of deduction is known as resolution.
How does this work in a resolution based theorem prover? Simple: we use proof by contradiction. That is, we start by turning our "facts" into clauses and add the clauses corresponding to the negation of our "goal". Then, if we can eventually resolve to the empty clause, {}, we will have reached a contradiction since the empty clause is equivalent to falsity. Because the facts are given, this means that our negated goal must be wrong, hence the (unnegated) goal must be true.

resolution is a procedure used in proving that argument which are expressible in predicate logic are correct
resolution lead to refute theorem proving technique for sentences in propositional logic.
resolution provides proof by refutation. i.e. to show that it is valid,resolution attempts to show that the negation of the statement produces a contradiction with a known statement
algorithm:
1). convert all the propositions of axioms to clause form
2). negate propositions & convert result to clause form
3)resolve them
4)if the resolvent is the empty clause, then contradiction has been found

Related

How to write it in skolem form?(Prolog)

Translate the following formula into a horn formula in Skolem form:
∀w¬∀x∃z(H(w)∧(¬G(x,x)∨¬H(z)))
it's translated from german to english, how to write it in horn form and then in skolem form, i didn't find anything on internet...plz help me
I will always use the satisfiability preserving version of skolemization, i.e. the one where those are replaced which would become existential quantifiers when moved to the head of the formula.
To make life a bit simpler, let's push the negations to the atoms. We can also see that w doesn't occur in ¬G(x,x)∨¬H(z) and that x,z don't occur in H(w), allowing us to distribute the quantifiers a bit inside.
Then we obtain the formula ∀w¬H(w) ∨ ∃x∀z (G(x,x)∧ H(z)) .
If we want to refute the formula:
We skolemize ∃x and delete ∀w, ∀z and obtain:
¬H(w) ∨ (G(c,c)∧ H(z))
after CNF transformation, we have:
(¬H(w) ∨ G(c,c)) ∧ (¬H(w) ∨ H(z))
both clauses have exactly one positive literal, so they are horn clauses. Translated to Prolog syntax we get:
g(c,c) :- h(W).
h(Z) :- h(W).
If we want to prove the formula:
We have to negate before we skolemize, leading to:
∃w H(w) ∧ ∀x∃z (¬G(x,x) ∨ ¬H(z))
after skolemizing ∃w and ∃z, deleting ∀x and CNF transformation, we obtain:
H(c) ∧ (¬G(x,x) ∨ ¬H(f(x)))
That could be interpreted as a fact h(c) and a query ?- g(X,X), h(f(X)).
To be honest, both variants don't make much sense - the first does not terminate for any input and in the second version, the query will fail because g/2 is not defined.
does this page help?
6.3 Convert first-order logic expressions to normal form
A horn clause consists of various goals that all have to be satisfied in order for the whole clause to be true.
∀w¬∀x∃z(H(w)∧(¬G(x,x)∨¬H(z)))
First you want to translate the whole statement to human language for clarity. ¬ means NOT, ∧ means AND and ∨ means OR. The () are used to group goals.
∀w¬∀x∃z
For all w, all NOT x, at least 1 Z. If a w is true, x must be false and there must be at least 1 z.
H(w)
Is w a H? There must be a fact that says H(w) is true in your knowledge base.
¬G(x,x)
Is there a fact G(x,x)? If yes, return false.
¬H(z)
Is there a fact H(z)? If yes, return false.
z(H(w)∧(¬G(x,x)∨¬H(z)))
What this says that z is only true if H(w) is true AND either G(x,x) OR H(z) is false.
In Prolog you'd write this as
factCheck(W,X,Z) :- h(W), not(g(X,X);not(checkZ(Z)).
where Z is a list with at least 1 entry in it. If ANY element in list Z is true, fail.
%is the list empty?
checkZ([])
%is h true for the first element of the list?
checkZ([Head|Tail]) :- h(Head), !.
%remove the first element of the list
checkZ([Head|Tail]) :- checkZ(Tail).

prolog and translating propositional logic

My ultimate goal is to load a set of propositional formulas in to Prolog from a file in order to deduce some facts. Suppose I have the propositional formula:
p implies not(q).
In Prolog this would be:
not(q) :- p
Prolog does not seem to like the not operator in the head of the rule. I get the following error:
'$record_clause'/2: No permission to redefine built-in predicate `not/1'
Use :- redefine_system_predicate(+Head) if redefinition is intended
I know two ways to rewrite the general formula for p implies q. First, use the fact that the contrapositive is logically equivalent.
p implies q iff not(q) implies not(p)
Second, use the fact that p implies q is logically equivalent to not(p) or q (the truth tables are the same).
The first method leads me to my current problem. The second method is then just a conjunction or disjunction. You cannot write only conjunctions and disjunctions in Prolog as they are not facts or rules.
What is the best way around my problem so that I can express p implies not(q)?
Is it possible to write all propositional formulas in Prolog?
EDIT: Now I wish to connect my results with other propositional formulae. Suppose I have the following rule:
something :- formula(P, Q).
How does this connect? If I enter formula(false, true) (which evaluates to true) into the interpreter, this does not automatically make something true. Which is what I want.
p => ~q === ~p \/ ~q === ~( p /\ q )
So we can try to model this with a Prolog program,
formula(P,Q) :- P, Q, !, fail.
formula(_,_).
Or you can use the built-in \+ i.e. "not", to define it as formula(P,Q) :- \+( (P, Q) ).
This just checks the compliance of the passed values to the formula. If we combine this with domain generation first, we can "deduce" i.e. generate the compliant values:
13 ?- member(Q,[true, false]), formula(true, Q). %// true => ~Q, what is Q?
Q = false.
14 ?- member(Q,[true, false]), formula(false, Q). %// false => ~Q, what is Q?
Q = true ;
Q = false.
You are using the wrong tool. Try Answer Set Programming.

Resolution Inference Rule Algorithm

I have the following doubt in the Resolution Inference Rule.
1* In for each Ci, Cj in clauses do , does each Ci and Cj necessarily contain complimentary symbols (ex. one contains A and the other contains ~A) ?
2* In the above example , what if both clauses have the same symbol (ex A and A). Should I consider it for inference? If so , what result does it return?
3* when does the if new ⊆ clauses then return false run? After all the clauses have been explored?
4* What is the use of if new ⊆ clauses then return false ?
5* What is the use of if new ⊆ clauses then return false ?
function PL-RESOLUTION(KB,α) returns true or false
inputs: KB, the knowledge base, a sentence α in propositional logic,
the query, a sentence in propositional logic
clauses <--- the set of clauses in the CNF representation of KB ∧ ¬α
new <--- {}
loop do
for each Ci, Cj in clauses do
resolvents <----- PL-RESOLVE(Ci, Cj)
if resolvents contains the empty clause then return true
new <--- new ∪ resolvents
if new ⊆ clauses then return false
clauses <---- clauses ∪ new
Yes.
No, you can't apply resolution in that case.
Well, why shouldn't it happen?
I guess it's to avoid infinite loops (in that case you return to the conditions at the previous iteration)
Same question?
More about 1 and 2. The idea for resolution is that if you have (A v B) ^ (not A v C) then you can safely infer that B or C are true (informally because A is either true or false). If you have (A v B) ^ (A v C) you can't apply the same reasoning.

can any one help me to understand this recursive prolog example?

here is the plus code that i don't understand
plus(0,X,X):-natural_number(X).
plus(s(X),Y,s(Z)) :- plus(X,Y,Z).
while given :
natural_number(0).
natural_number(s(X)) :- natural_number(X).
I don't understand this recursion. If I have plus(s(0),s(s(s(0))),Z) how can i get the answer of 1+3=4?
I need some explanation for the first code. I try that plus(0,X,X) will stop the recursion but I think that I do it wrong.
So, let's start with natural_number(P). Read this as "P is a natural number". We're given natural_number(0)., which tells us that 0 is always a natural number (i.e. there are no conditions that must be met for it to be a fact). natural_number(s(X)) :- natural_number(X). tells us that s(X) is a natural number if X is a natural number. This is the normal inductive definition of natural numbers, but written "backwards" as we read Prolog "Q := P" as "Q is true if P is true".
Now we can look at plus(P, Q, R). Read this as "plus is true if P plus Q equals R". We then look at the cases we're given:
plus(0,X,X) :- natural_number(X).. Read as Adding 0 to X results in X if X is a natural number. This is our inductive base case, and is the natural definition of addition.
plus(s(X),Y,s(Z)) :- plus(X,Y,Z). Read as "Adding the successor of X to Y results in the successor Z if adding X to Y is Z'. If we change the notation, we can read it algebraically as "X + 1 + Y = Z + 1 if X + Y = Z", which is very natural again.
So, to answer you direct question "If I have plus(s(0),s(s(s(0))),z), how can i get the answer of 1+3=4?", let's consider how we can unify something with z each step of the induction
Apply the second definition of plus, as it's the only one that unifies with the query. plus(s(0),s(s(s(0))), s(z')) is true if plus(0, s(s(s(0))), z') is true for some z
Now apply the first definition of plus, as it's the only unifying definition: plus(0, s(s(s(0))), z') if z' is s(s(s(0))) and s(s(s(0))) is a natural number.
Unwind the definition of natural_number a few times on s(s(s(0))) to see that is true.
So the overall statement is true, if s(s(s(0))) is unified with z' and s(z') is unified with z.
So the interpreter returns true, with z' = s(s(s(0))) and z = s(z'), i.e. z = s(s(s(s(0)))). So, z is 4.
That code is a straightforward implementation of addition in Peano arithmetic.
In Peano arithmetic, natural numbers are represented using the constant 0 and the unary function s. So s(0) is a representation of 1, s(s(s(0))) is representation of 3. And plus(s(0),s(s(s(0))),Z) will give you Z = s(s(s(s(0)))), which is a representation of 4.
You won't get numerical terms like 1+3=4, all you get is the term s/1 which can embed itself to any depth and thus can represent any natural number. You can combine such terms (using plus/3) and thereby achieve summing.
Note that your definition of plus/3 has nothing to do with SWI-Prolog's built-in plus/3 (which works with integers and not with the s/1 terms):
?- help(plus).
plus(?Int1, ?Int2, ?Int3)
True if Int3 = Int1 + Int2.
At least two of the three arguments must be instantiated to integers.

prolog unification resolution

Why does this work:
power(_,0,1) :- !.
power(X,Y,Z) :-
Y1 is Y - 1,
power(X,Y1,Z1),
Z is X * Z1.
And this gives a stack overflow exception?
power(_,0,1) :- !.
power(X,Y,Z) :-
power(X,Y - 1,Z1),
Z is X * Z1.
Because arithmetic operations are only performed on clauses through the is operator. In your first example, Y1 is bound to the result of calculating Y - 1. In the later, the system attempts to prove the clause power(X, Y - 1, Z1), which unifies with power(X', Y', Z') binding X' = X, Y' = Y - 1, Z' = Z. This then recurses again, so Y'' = Y - 1 - 1, etc for infinity, never actually performing the calculation.
Prolog is primarily just unification of terms - calculation, in the "common" sense, has to be asked for explicitly.
Both definitions do not work properly.
Consider
?- pow(1, 1, 2).
which loops for both definitions because the second clause can be applied regardless of the second argument. The cut in the first clause cannot undo this. The second clause needs a goal Y > 0 before the recursive goal. Using (is)/2 is still a good idea to get actual solutions.
The best (for beginners) is to start with successor-arithmetics or clpfd and to avoid prolog-cut altogether.
See e.g.: Prolog predicate - infinite loop

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