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First Order Predicate Logic Limitation of Propositional Logic The facts: “peter is a man”, “paul is a man”, “john is a man” can be symbolized by P, Q and R, respectively in PL. But we would not be able to draw any conclusions about similarities between P, Q and R.
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First Order Predicate Logic Limitation of Propositional Logic • The facts: “peter is a man”, “paul is a man”, “john is a man” can be symbolized by P, Q and R, respectively in PL. • But we would not be able to draw any conclusions about similarities between P, Q and R. • It would be much better to represent these facts as MAN(peter), MAN(paul) and MAN(john). • Further, we are even in more difficulty, if we try to represent sentences like “All men are mortal” in PL.
In Predicate Logic, such sentences can be easily represented and the limitations of propositional logic are removed to great extent. • The predicate logic is logical extension of propositional logic. The first order predicate logic is one where the quantification is over simple variables. Predicate Calculus • It has three more logical notions as compared to propositional calculus. • Terms, • Predicates • Quantifiers (universal or existential quantifiers i.e. “for all' and “there exists”)
A term is a constant (single individual or concept i.e.,5, john etc.), a variable that stands for different individuals and n-place function f(t1, …, tn) where t1, …, tn are terms. • A function is a mapping that maps ‘n’ terms to a term. • A predicate is a relation that maps ‘n’ terms to a truth value true (T) or false (F). Examples: • A statement x is greater than y is represented in predicate calculus as GREATER(x, y). It is defined as follows: GREATER( x, y) = T , if x y = F , otherwise
A statement x loves y is represented as LOVE(x, y) which maps it to true or false when x and y get instantiated to actual values. • A statement john’s father loves john is represented as LOVE(father(john), john). Here father is a function that maps john to his father. • The predicate names GREATER and LOVE take two terms and map to T or F depending upon the values of their terms. Quantifiers: Variables are used in conjunction with quantifiers. There are two types of quantifiers viz.., “there exist” () and “for all” ().
Example: Translate the text "Every man is mortal. John is a man. Therefore, John is mortal" into Predicate Calculus formula. Solution: Let MAN(x), MORTAL(x) represent ‘x is a man’ and ‘x is mortal’ respectively. • Every man is mortal :(x) (MAN(x) MORTAL(x)) • John is a man : MAN(john) • John is mortal : MORTAL(john) The whole text can be represented by the following formula. (x) ((MAN(x) MORTAL(x)) MAN(john)) MORTAL(john)
First Order Predicate Logic (FOL) • First order predicate calculus becomes First Order Predicate Logic if inference rules are added to it. • Using inference rules one can derive new formula using the existing ones. Interpretations of Formulae in FOL • In PL, an interpretation is simply an assignment of truth values to the atoms. In FOL, there are variables, so we have to do more than that. • An interpretation of a formula in FOL consists of a non empty domain D and an assignment of values to each constant, function symbol and predicate symbol.
We use notation I[] = T (or F), which means that is evaluated to be true or false under interpretation I over a domain D. • Let and are formulae and I be an interpretation over any domain D. The following holds true. • I[] = I[] I[] • I[ V ] = I[] V I[] • I[] = I[] I[] • I[~] = ~ I[] = T if and only if I[] = F
For any interpretation I and formulae using (x) and ( x), the following results holds true. • I[(x)P(x)] = T iff I[P(x)] = T, x D = F, otherwise • I[(x) P(x)] = T iff cD such that I[P(c)] = T = F, otherwise Example: Let : (x) ( y) P(x, y) be a formula. Evaluate it under the following interpretation I. (Ans: True) D = {1, 2} I I[P(1, 1)] = F, I[P(1, 2)] = T, I[P(2, 1)] = T, I[P(2, 2)] = F
A formula is said to be consistent (satisfiable) if and only if there exists an interpretation I such that I[] = T. Alternatively we say that I is a model of or I satisfies . • A formula is said to be inconsistent(unsatisfiable) if and only if no interpretation that satisfies or there exists no model for . • A formula is valid if and only if for every interpretation I, I[] = T. • A formula is a logical consequence of a set of formulae {1, 2, ..., n } if and only if for every interpretation I, if I[1 …n ] = T, then I[] = T.
Prenex Normal Form • In FOL, there are infinite number of domains and consequently infinite number of interpretations of a formula. • Therefore, unlike PL, it is not possible to verify a validity and inconsistency of a formula by evaluating it under all possible interpretations. • We will discuss the formalism for verifying inconsistency and validity in FOL. There is also a third type of normal form called Prenex Normal Form in FOL, . • A closed formula of FOL is said to be in Prenex Normal Form(PNF) if and only if is represented as (q1 x1) (q2 x2) … (qn xn) (M), where, qk , (1 k n ) quantifiers ( or ) and M is a formula free from quantifiers. • A list of quantifiers [(q1 x1) … (qn xn)] is called prefixand M is called the matrix of a formula . Here M is represented in CNF.
Transformation of Formula into Prenex Normal Form • A formula can be easily transformed into PNF using various equivalence laws. We use the following conventions: [x] - a formula , which contains a variable x. - a formula without a variable x. q - Quantifier ( or ).
Equivalence Laws: There are following pairs of logically equivalent formulae called equivalence laws in addition to the equivalence laws given for PL. 1. (q x) [x] V (q x) ( [x] V ) 2. V (q x) [x] (q x) ( V [x]) 3. (q x) [x] (q x) ( [x] ) 4. (q x) [x] (q x) ( [x]) 5. ~ ((x) [x]) (x) (~ [x]) 6. ~ ((x) [x]) (x) (~ [x])
There are few more equivalence laws: 7. (x) [x] (x) [x] (x) ( [x] [x] ) 8. (x) [x] V (x) [x] (x) ( [x] V [x]) • Quantifiers and can distribute over and V respectively, but cannot distribute over V and . i. (x) [x] V (x) [x] (x) ( [x] V [x] ) ii. (x) [x] (x) [x] (x) ( [x] [x])
Show through an example that (x) [x] V (x) [x] and (x) ([x] V [x] ) are not equivalent. Example: Consider : (x) P(x) V (x) Q(x) and an interpretation I as follows. D = {1, 2} I I[P(1)] = T, I[P(2)] = F I[Q(1)] = F, I[Q(2)] = T • We can easily show that is false under interpretation I whereas a formula (x) (P(x) V Q(x)) is true under I.
Renaming Bound Variables • For the cases mentioned below, we have to do something special. i. (x) [x] V (x) [x] (x) ( [x] V [x] ) ii. (x) [x] (x) [x] (x) ( [x] [x]) • Every bound variable x in [x] can be renamed by y such that y does not occur in and . Then, (x) [x] V (x) [x] (x) [x] V (y) [y] , (x) (y) ( [x] V [y] ) (using laws 1 and 2)
Thus by renaming all occurrence of x by y {, } in a formula , we get, two more equivalence laws as follows: 9. (x) [x] V (x) [x] (x) (y) ( [x] V [y] ) 10. (x) [x] (x) [x] (x) (y) ( [x] [y] ) • These equivalence laws are used for transforming a formula into PNF.
Skolemisation (Standard Form) • Prenex normal form of a formula is further transformed into special form called Skolemisation or Standard Form. • This form is used in resolution of clauses. • The process of eliminating existential quantifiers and replacing the corresponding variable by a constant or a function is called skolemisation. • A constant or a function is called skolem constant or function.
Convertion of PNF to its equivalent Standard Form • Scan prefix from left to right. If q1 is the first existential quantifier then choose a new constant c Matrix. Replace all occurrence of x1 appearing in Matrix by c and delete (q1 x1) from the prefix to obtain new prefix and matrix. • If qr is an existential quantifier and q1….qr-1 are universal quantifiers appearing before qr , then choose a new (r-1) place function symbol ’ f ' Matrix. Replace all occurrence of xr in Matrix by f(x1, …,xr-1 ) and remove (qr xr ). • Repeat the process till all existential quantifiers are removed from matrix.
Any formula in FOL can be transformed into its standard form. • Matrix of standard formula is in CNF and prefix is free from existential quantifiers. • Formula in standard form is expressed as: (x1)… (xn) (C1… Cm), where Ck ,(1 k m) is formula in disjunctive normal form. • Since all the variables in prefix are universally quantified, we omit prefix part from the standard form for the sake of convenience and write standard form as (C1… Cm).
A clause is a closed formula of the form (x1)… (xn) (L1V … V Lm), where each Lk is a literal and x1,…,xn are variables occurring in L1, …, Lm. • Since all the variables in a clause are universally quantified, we omit them and write clause as (L1V … V Lm ) for the sake of simplicity. • The standard form of any formula is of the form (C1… Cm), where each Ci , (1 i m) is a clause. • Define S = { C1, … ,Cm } to be a set of clauses that represents a standard form of a formula .
S is said to be unsatisfiable (inconsistent) if and only if there no interpretation that satisfies all the clauses of S simultaneously. • A formula is unsatisfiable (inconsistent) if and only if its corresponding set S is unsatisfiable. • S is said to be satisfiable (consistent) if and only if each clause is consistent under at least one interpretation i.e., an interpretation that satisfies all the clauses of S simultaneously. • Alternatively, an interpretation I models S if and only if I models each clause of S.
Resolution in Predicate Logic • Resolution method is used to test unsatisfiability of a set S of clauses in Predicate Logic. It is an extension of resolution method for propositional logic. • The resolution principle basically checks whether empty clause is contained or derived from S. • A clause is a closed formula of the form (x1)… (xn) (L1V …V Lm), where Lk ,(1 k m) is a literal and xj, (1 j n) are all the variables occurring in L1, …, Lm.
Resolution for the clauses containing no variables is very simple and is similar to PL. It becomes complicated when clauses contain variables. • In such case, two complementary literals are resolved after proper substitutions so that both the literals have same arguments. Example: Consider two clauses C1 and C2 as follows: C1 = P(x) V Q(x) C2 = ~ P(f(x)) V R(x)
Substitute 'f(a)' for 'x' in C1 and 'a' for 'x' in C2 , where 'a' is a new constant from the domain, then C3 = P(f(a)) V Q(f(a)) C4 = ~ P(f(a)) V R(a) • Resolvent of C3 and C4 is [Q(f(a)) V R(a)] • Here C3 and C4 do not have variables. They are called ground instances of C1 and C2 . • In general, if we substitute 'f(x)' for 'x' in C1 , then C’1 = P(f(x)) V Q(f(x)) • Resolvent of C'1 and C2 is [Q(f(x)) V R(x)] • We notice that [Q(f(a)) V R(a)] is an instance of [Q(f(x)) V R(x)].
Before we discuss a systematic method of generating resolvent from the parents, we discuss substitutions and unification of clauses . Substitution and Unification • A substitution = {x1 / t1, …., xn / tn}, is a finite set of elements, where each xk is a distinct variable and each tk is a term different from xk , ( 1 k n). Each element xk / tk of a set is called a binding for xk .
A termin predicate logicis either a constant, a variable or n-place function symbol with t1, …, tn as arguments, where each ti (1 ti n) is a term. • A ground term is one which is free from variables. • A substitution is a ground if t1, t2 ,…, tn are all ground terms. • A substitution is called a pure variable substitution if t1, t2 ,…, tn are all variables. • Let = {x1 / t1, …., xn / tn} be a substitution and F be any formula in predicate logic.
The formula F is obtained by replacing each occurrence of the variable xk in F by the corresponding term tk . • F is called an instance of F. If F is ground, then it is called ground instance of F. • Let = {x1 / t1,…, xn / tn} and = {y1 / s1,…, ym / sm} be two substitutions. • Then the composition is obtained from the set { x1 / (t 1) , …., xn / (tn ), y1 / s1, …., ym / sm } after deleting any binding xk / (tk ) for which xk = tk and binding yj / sj for which yj {x1, …., xn}
Example: Consider two substitutions and defined as follows: ={x / f(y), y / z} and = {x / a, y / b, z / y}. Show that = {x / f(b), z / y}. Solution: = {x / f(y), y / z} = {x / (f(y) ) , y / (z ) , x / a, y / b, z / y } = {x / f(b), y / y, x / a , y / b , z / y} According to the definition, delete y / y, x / a and y / b from above set, we get, {x / f(b), z / y}. Hence = {x / f(b), z / y}
The substitution given by an empty set is called identity substitution and denoted by . It satisfies the condition F = F, formulae F. • Let F1 and F2 be two formulae. F1 and F2 are variants of each other, if there exist substitutions and such that F1 = F2 and F2 = F1 . Example: Show that P(f(x, y), g(z), a) and P(f(y, x), g(u), a) are variant of each other. Solution:Consider two substitutions ={x / y, y / x, z / u} and = {y / x, x / y, u / z} • Let F1= P(f(x, y), g(z), a) and F2 = P(f(y, x), g(u), a) • Then, F1 = F2 and F2 = F1
Let F be a formula and U be a set of variables occurring in F. A renamingsubstitution for F is made a pure variable substitution such as {x1 / y1 , …, xn / yn } if following conditions are satisfied. • {x1 , …, xn } U and yk (1 k n) are distinct • { U – {x1 , …, xn }} {y1 , …, yn } = • If = {F1, …, Fn} is a finite set of formulae and is any substitution, then = {F1, …, Fn}. • A substitution is called a unifier for a set if and only if is a singleton set (a set containing single element) i.e., F1 = ….= Fn .
Example: Consider = {P(f(x), z), P(y, a)}. Find an unifier of , if it exist Solution: Consider a substitution = {y / f(a) , x / a , z / a}. = { P(f(a), a), P(f(a), a)} = { P(f(a), a)} Since is a singleton set, = {y / f(a) , x / a , z / a} is an unifier of . • An unifier for a set ={ F1, …, Fn } is most general unifier (mgu), if for each unifier of , there exists a substitution such that = . • The most general unifier may not be unique.
Example: Consider = {P(f(x), z), P(y, a)}. Find the most general unifier. Solution: • The most general unifier for is = {y/f(x),z/a}, because for any unifier, say, = {y/f(a), x/a, z/a}, a substitution = {x / a} such that = • Consider = {y / f(x), z / a} {x / a} = {y/f(a), x/a, z/a} = • Hence = and therefore = {y/f(x), z/a} is most general unifier.
Disagreement Set • Consider two formulae F1 = P(a) and F2 = P(x) which are not identical. The disagreement is that 'a' occurs in F1 but variable 'x' occurs in F2 . • In order to unify F1 and F2 , find out the disagreement set and then try to eliminate it. In this case disagreement set is {a, x}. In general disagreement set of S is obtained as follows: • Locate the leftmost symbol position at which all the formulae of are different. • Extract those sub formulae from beginning at the position located above. This set is called disagreement set of denoted by D( ) at the particular position located above.
Method: • Initialize k = 0, k = (identity substitution), • Sk = Sk • If Sk is a singleton set then k is mgu of S else find disagreement set of Sk and denote it by Dk . • If a variable 'x' and a term 't' in Dk such that 'x' does not occur in 't', then compute k+1 = k {x / t}; k = k+1; Sk = Sk-1k ; else S is not unifiable. • Repeat the process • In unification algorithm, we try to find 'x' and 't'in Dk such that 'x' is a variable that does not occur in term 't'. • This is called occurs check. We say that a variable 'x' satisfies occurs check with respect to a term 't'.
Example: Apply unification algorithm and find mgu of a set S = {P(a, x, f(g(y))) , P(z, f(z), f(u)) }. Solution:0 = ; S0 = {P(a, x, f(g(y))) , P(z, f(z), f(u)) } D0 = {a, z}, 1 = {z / a }; S1 = S1 = { P(a, x, f(g(y))), P(a, f(a), f(u)) } D1 = {x, f(a)}; 2 = 1 {x / f(a) } = {z / a, x / f(a) }, S2 = S12 = { P(a, f(a), f(g(y))), P(a, f(a), f(u)) } D2 = {g(y), u}; 3 = 2 {u / g(y) }= {z / a, x / f(a), u / g(y) }, S3 = S23 = {P(a, f(a), f(g(y))),P(a, f(a), f(g(y))) } Since S3 is a singleton set, therefore, 3 = {z / a, x / f(a), u / g(y) }is the mgu of S.
Logical Consequence of a Set • A clause L is defined to be a logical consequenceof a set of clauses S if and only if every interpretation I that satisfies S also satisfies L. Lemma :If S = {C1,...,Cn} is a set of clauses, then L is a logical consequence of S if and only if ( C1… Cn L ) is valid. Lemma: Alternatively, we define L to be a logical consequence of S if and only if ~ ( ~ ( C1 … Cn ) V L ) is unsatisfiable or inconsistent.
Theorem: L is a logical consequence of S if and only if S { ~ L} = { C1, …, Cn , ~ L} is unsatisfiable. • A deduction of an empty clause from a set S of clauses is called a resolution refutation of S. Theorem: (Soundness and completeness of resolution): There is a resolution refutation of S if and only if S is unsatisfiable (inconsistent). Theorem: L is a logical consequence of S if and only if there is a resolution refutation of S {~L}.
We can summarize that in order to show L to be a logical consequence of the set of formulae {1, …, n }, follow the following steps: • Obtain a set S of all the clauses by converting each k to its corresponding standard form and then to the clauses. • Show that a set S { ~ L} is unsatisfiable i.e., the set S { ~ L} contains either empty clause or empty clause can be derived in finite steps using resolution method. • If so, then report 'Yes' and conclude that L is a logical consequence of S and subsequently of formulae 1, …, n otherwise report 'No'.
Example: Consider the following set of formulae: : (x) (P(x) (Q(x) R(x))) : (x) (P(x) U(x)) Show that L = (x) (U(x) R(x)) is a logical consequence of and . Solution: Convert and to its corresponding clauses by generating their standard forms. Standard form of : (x) (~ P(x) V (Q(x) R(x))) is (x) ((~ P(x) V Q(x)) (~ P(x) V R(x))) Clauses of are ~ P(x) V Q(x) and ~ P(x) V R(x) Similarly the clauses of are P(a) and U(a), where a is a skolem constant. Then, S = {~ P(x) V Q(x) , ~ P(x) V R(x), P(a), U(a) }
Negation of a goal: ~ L : ~ ((x) (U(x) R(x))) (x) (~ U(x) V ~ R(x) ) Clause of ~ L: ~ U(x) V ~ R(x) Therefore, S { ~ L} = {~ P(x) V Q(x) , ~ P(x) V R(x), P(a), U(a), ~ U(x) V ~ R(x) } Now resolve the following clauses of S { ~ L} to get empty clause. • ~ P(x) V Q(x) • ~ P(x) V R(x) • P(a) • U(a) • ~ U(x) V ~ R(x)
Inverted tree (resolution tree) is given below whose root is deduced as (empty clause) by resolving clauses in a set S { ~ L}. Hence L is a logical consequence of and . U(a) ~ U(x) V ~ R(x) {x / a} ~ R(a) ~ P(x) V R(x) {x / a} ~ P(a) P(a) • Hence L is a logical consequence of and .
If the choice of clauses to resolve at each step is made in a systematic ways, then resolution algorithm will find a contradiction if one exists. • There exist various strategies for making the right choice that can speed up the process considerably. Useful Tips: • Initially choose a clause from the negated goal clauses as one of the parents to be resolved ( This corresponds to intuition that the contradiction we are looking for must be because of the formula we are trying to prove) .
Choose a resolvent and some existing clause if both contain complementary literals. • If such clauses do not exists, then resolve any pairs of clauses that contain complementary literals. • Whenever possible, resolve with the clauses with single literal. Such resolution generate new clauses with fewer literals than the larger of their parent clauses and thus probably algorithm terminates faster. • Eliminate tautologies and clauses that are subsumed by other clauses as soon as they are generated.
Semantic Tableaux (FOL) • There are four more rules handling variables in a predicate formula in addition to one given for Propositional logic. • Let us denote a formula containing a variable x by [x]. Rule 10: ( x) [x] [t] for any ground term t, where t is a term free from variables.
Rule 11: ~ {(x) [x] } ~ [c] for any new constant c not occurring in Rule 12 : ( x) [x] [c] for any new constant c Rule 13: ~ {( x) [x] } ~ [t] for any ground term t
Example: Show that the formula (x) (P(x) ~ ( Q(x) P(x)) ) is inconsistent. Solutions: If we show that tableau for [(x) (P(x) ~ ( Q(x) P(x)))] as a root is a contradictory tableau, then the formula is inconsistent. {Tableau root} (x) (P(x) ~ ( Q(x) P(x)) (1) {Apply R10 on 1} P(t) ~ ( Q(t) P(t)) , (2) where t is any ground term {Apply R1 on 2} P(t) ~ ( Q(t) P(t)) (3) {Apply R7 on 3} Q(t) ~ P(t) Closed {P(t), ~ P(t)}
