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Logic seminar 5

Logic seminar 5. The resolution principle 12.12.2005. Slobodan Petrović. The resolution principle. Herbrand’s procedure has one major drawback: It requires the generation of sets S 1 ’, S 2 ’,… of ground instances of clauses. For most cases, this sequence grows exponentially. Example:

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Logic seminar 5

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  1. Logic seminar 5 The resolution principle 12.12.2005. Slobodan Petrović

  2. The resolution principle • Herbrand’s procedure has one major drawback: • It requires the generation of sets S1’, S2’,… of ground instances of clauses. • For most cases, this sequence grows exponentially. • Example: • S={P(x,g(x),y,h(x,y),z,k(x,y,z)), P(u,v,e(v),w,f(v,w),x)}

  3. The resolution principle • Example (cont.): • H0={a} • H1={a,g(a),h(a,a),k(a,a,a),e(a),f(a,a)} • … • The earliest set that is unsatisfiable is S5’. • S5’ has the order of 10256 elements. • It is impossible even to store S5’ in a computer.

  4. The resolution principle • In order to avoid the generation of sets of ground instances of clauses, the resolution principle was introduced by Robinson in 1965. • The resolution principle can be applied directly to any set S of clauses (not necessarily ground clauses) to test the unsatisfiability of S.

  5. The resolution principle • The essential idea of the resolution principle: • To check whether S contains the empty clause □. • If S contains □, then S is unsatisfiable. • If S does not contain □, then it should be checked whether □ can be derived from S. • By Herbrand’s theorem, checking for the presence of □ is equivalent to counting the number of nodes of a closed semantic tree for S.

  6. The resolution principle • S is unsatisfiable if and only if there is a finite closed semantic tree T for S. • S contains □ if and only if T consists of only one node – the root node. • If S does not contain □, then T must contain more than one node. • If we can reduce the number of nodes in T to one, □ may be forced to appear.

  7. The resolution principle • We can view the resolution principle as an inference rule that can be used to generate new clauses for S. • If we put these new clauses into S, some nodes of the original T can be forced to become failure nodes. • Thus the number of nodes in T can be reduced and the empty clause □ will be eventually derived.

  8. The resolution principle • The resolution principle for the propositional logic • The resolution principle is essentially an extension of the one-literal rule (Davis and Putnam). • Example. The clauses: C1: P C2: PQ. • Using the one-literal rule, from C1 and C2 we can obtain a clause: C3: Q

  9. The resolution principle • The one-literal rule requires the examination whether there is a complementary pair of a literal (in this example P) in C1 and a literal (in this example P) in C2. • Then the complementary pair is deleted from C1 and C2 to obtain C3.

  10. The resolution principle • By extending the one-literal rule and applying it to any pair of clauses (not necessarily unit clauses) we get the resolution principle: • For any two clauses C1 and C2, if there is a literal L1 in C1 that is complementary to a literal L2 in C2, then delete L1 and L2 from C1 and C2, respectively, and construct the disjunction of the remaining clauses. • The constructed clause is a resolvent of C1 and C2.

  11. The resolution principle • Example. The clauses: C1: PR C2: PQ. • Clause C1 has the literal P, which is complementary to P in C2. • By deleting P and P from C1 and C2, respectively, and constructing the disjunction of the remaining clauses R and Q, we obtain a resolvent: RQ.

  12. The resolution principle • Example. The clauses: C1: PQR C2: QS. • The resolvent of C1 and C2 is PRS. • Example. The clauses: C1: PQ C2: PR • Since there is no literal in C1 that is complementary to any literal in C2, there is no resolvent of C1 and C2.

  13. The resolution principle • Theorem: • Given two clauses C1 and C2, a resolvent C of C1 and C2 is a logical consequence of C1 and C2. • If we have two unit clauses, then the resolvent of them, if there is one, is the empty clause □. • If a set S of clauses is unsatisfiable, we can use the resolution principle to generate □ from S.

  14. The resolution principle • Definition: • Given a set S of clauses, a (resolution) deduction of C from S is a finite sequence C1, C2,…, Ck of clauses such that each Ci either is a clause in S or a resolvent of clauses preceding Ci, and Ck=C. • A deduction of □ from S is called a refutation, or a proof of S. • A clause C can be deduced or derived from S if there is a deduction of C from S.

  15. The resolution principle • Example. Given a set S of clauses: (1) PQ (2) Q (3) P • From (1) and (2), we can obtain a resolvent (4) P. • From (3) and (4), we can obtain a resolvent (5) □. • Since □ is derived from S by resolution, □ is a logical consequence of S. Hence S is unsatisfiable.

  16. The resolution principle • Example. Given a set S of clauses: (1) PQ (2) PQ (3) PQ (4) PQ • We generate the following resolvents: (5) Q (from (1) and (2)) (6) Q (from (3) and (4)) (7) □ (from (5) and (6)). • Since □ is derived, S is unsatisfiable.

  17. The resolution principle • Example (cont.) • The deduction can be represented by a tree – deduction tree.

  18. The resolution principle • Substitution and unification • The most important part of applying the resolution principle – finding a literal in a clause that is complementary to a literal in another clause. • For clauses containing no variables, this is very simple. • For clauses containing variables, it is more complicated.

  19. The resolution principle • Example. Consider the clauses: C1: P(x)Q(x) C2: P(f(x))R(x). • There is no literal in C1 that is complementary to any literal in C2. • However, if we substitute f(a) for x in C1 and a for x in C2, we obtain: C1’: P(f(a))Q(f(a)) C2’: P(f(a))R(a).

  20. The resolution principle • Example (cont.) • C1’and C2’are ground instances of C1 and C2, respectively. • P(f(a)) and P(f(a)) are complementary to each other. • Therefore, from C1’ and C2’ we can obtain a resolvent C3’: Q(f(a))R(a).

  21. The resolution principle • Example (cont.) • More generally, if we substitute f(x) for x in C1, we obtain: C1*: P(f(x))Q(f(x)) • C1* is an instance of C1. • The literal P(f(x)) in C1* is complementary to the literal P(f(x)) in C2. • The resolvent from C1* and C2 is C3: Q(f(x))R(x).

  22. The resolution principle • Example (cont.) • C3’ is an instance of C3. • By substituting appropriate terms for the variables in C1 and C2, we can generate new clauses from C1 and C2. • The clause C3 is the most general clause in the sense that all other clauses that can be generated by this process are instances of C3. • C3 is also called a resolvent of C1 and C2.

  23. The resolution principle • Definition • A substitution is a finite set of the form {t1/v1,…,tn/vn}, where every vi is a variable, every ti is a term different from vi, and no two elements in the set have the same variable after the stroke symbol. • When t1,…,tn are ground terms, the substitution is called a ground substitution. • The substitution that consists of no elements is called the empty substitution (denoted by ).

  24. The resolution principle • Example {f(z)/x,y/z} {a/x,g(y)/y,f(g(b))/z} are substitutions. • Definition • Let ={t1/v1,…,tn/vn} be a substitution and E be an expression. • Then E is an expression obtained from E by replacing simultaneously each occurrence of the variable vi, 1in, in E by the term ti. • E is called an instance of E.

  25. The resolution principle • Example. Let ={a/x,f(b)/y,c/z} and E=P(x,y,z}. • Then E=P(a,f(b),c).

  26. The resolution principle • Definition • Let ={t1/x1,…,tn/xn} and ={u1/y1,…,um/ym} be two substitutions. • Then the composition◦ of  and  is a substitution that is obtained from the set {t1/x1,…,tn/xn,u1/y1,…,um/ym} by deleting any element tj/xj for which tj=xj, and any element ui/yi such that yi is among {x1,x2,…,xn}.

  27. The resolution principle • Example. Let ={t1/x1,t2/x2}={f(y)/x,z/y} ={u1/y1,u2/y2,u3/y3}={a/x,b/y,y/z} • Then {t1/x1,t2/x2,u1/y1,u2/y2,u3/y3}={f(b)/x,y/y,a/x,b/y,y/z}. • Since t2=x2, t2/x2, i.e. y/y should be deleted from the set. • Since y1 and y2 are among {x1,x2}, u1/y1 and u2/y2, i.e. a/x and b/y should also be deleted. Thus: ◦={f(b)/x,y/z}.

  28. The resolution principle • The composition of substitutions is associative. • The empty substitution  is both a left and a right identity. • That is (◦)◦= ◦(◦) ◦=◦ for all , , and .

  29. The resolution principle • In the resolution proof procedure, in order to identify a complementary pair of literals, very often we have to unify two or more expressions. • We have to find a substitution that can make several expressions identical. • This process is called unification of expressions.

  30. The resolution principle • Definition • A substitution  is called a unifier for a set {E1,…,Ek} if and only if E1=E2=…=Ek. • The set {E1,…,Ek} is said to be unifiable if there is a unifier for it. • Definition • A unifier  for a set {E1,…,Ek} of expressions is a most general unifier if and only if for each unifier  for the set there is a substitution  such that =◦.

  31. The resolution principle • Example • The set {P(a,y),P(x,f(b))} is unifiable since the substitution ={a/x,f(b)/y} is a unifier for the set.

  32. The resolution principle • Unification algorithm • We have to find the most general unifier for a finite unifiable set of nonempty expressions – by means of a unification algorithm. • When the given set is not unifiable, the algorithm has also to detect that fact. • Example • Given P(a) and P(x), we have to find the disagreement between these two expressions, and then try to eliminate it.

  33. The resolution principle • Example (cont.) • For P(a) and P(x), the disagreement is {a,x}. • Since x is a variable, x can be replaced by a, and thus the disagreement can be eliminated. • Definition • The disagreement set of a nonempty set W of expressions is obtained by locating the first symbol (counting from the left) at which not all the expressions in W have exactly the same symbol, and then extracting from each expression in W the subexpression that begins with the symbol occupying that position. • The set of these respective subexpressions is the disagreement set of W.

  34. The resolution principle • Example W={P(x,f(y,z)),P(x,a),P(x,g(h(k(x))))} • The first symbol position at which not all atoms in W are exactly the same is the fifth, since they all have the first four symbols P(x, in common. • Thus the disagreement set consists of the respective subexpressions (underlined terms) that begin in symbol position number five. • It is the set {f(y,z),a,g(h(k(x)))}.

  35. The resolution principle • Unification algorithm (1) Set k=0, Wk=W, and k=. (2) If Wk is a singleton, stop; k is the most general unifier for W. Otherwise, find the disagreement set Dk of Wk. (3) If there exist elements vk and tk in Dk such that vk is a variable that does not occur in tk, go to Step 4. Otherwise, stop; W is not unifiable. (4) Let k+1=k◦{tk/vk} and Wk+1=Wk{tk/vk}. (Note that Wk+1=Wk+1.) (5) Set k=k+1 and go to Step 2.

  36. The resolution principle • Example • Find the most general unifier for W={P(a,x,f(g(y))),P(z,f(z),f(u))}. (1) 0= and W0=W. Since W0 is not a singleton, 0 is not the most general unifier for W. (2) The disagreement set D0={a,z}. In D0, there is a variable v0=z that does not occur in t0=a. (3) 1=0◦{t0/v0}=◦{a/z}={a/z}; W1=W0{t0/v0}={P(a,x,f(g(y))),P(z,f(z),f(u))}{a/z}={P(a,x,f(g(y))),P(a,f(a),f(u))}.

  37. The resolution principle • Example (cont.) (4) W1 is not a singleton, hence we find the disagreement set D1 of W1: D1={x,f(a)}. (5) From D1, we find that v1=x and t1=f(a). (6) 2=1◦{t1/v1}={a/z}◦{f(a)/x}={a/z,f(a)/x}; W2=W1{t1/v1}= ={P(a,x,f(g(y))),P(a,f(a),f(u))}{f(a)/x}= ={P(a,f(a),f(g(y))),P(a,f(a),f(u))}. (7) W2 is not a singleton, hence we find the disagreement D2 of W2: D2={g(y),u}. From D2, we find that v2=u and t2=g(y).

  38. The resolution principle • Example (cont.) (8) 3=2◦{t2/v2}={a/z,f(a)/x}◦{g(y)/u}= ={a/z,f(a)/x,g(y)/u}; W3=W2{t2/v2}= ={P(a,f(a),f(g(y))),P(a,f(a),f(u))}{g(y)/u}= ={P(a,f(a),f(g(y))),P(a,f(a),f(g(y)))}= ={P(a,f(a),f(g(y)))} (9) W3 is a singleton, hence 3={a/z,f(a)/x,g(y)/u} is the most general unifier for W.

  39. The resolution principle • Example. Determine whether or not the set W={Q(f(a),g(x)),Q(y,y)} is unifiable. (1) 0=, W0=W. (2) W0 is not a singleton, hence we find the disagreement set D0 of W0: D0={f(a),y}. From D0, we know that v0=y and t0=f(a). (3) 1=0◦{t0/v0}=◦{f(a)/y}={f(a)/y}; W1=W0{t0/v0}={Q(f(a),g(x)),Q(y,y)}{f(a)/y}= ={Q(f(a),g(x)),Q(f(a),f(a))}.

  40. The resolution principle • Example (cont.) (4) W1 is not a singleton, hence we find the disagreement set D1 of W1: D1={g(x),f(a)}. (5) No element of D1 is a variable. Hence the unification algorithm is terminated and we conclude that W is not unifiable. • Theorem • If W is a finite nonempty unifiable set of expressions, the unification algorithm will always terminate at Step 2, and the last k is the most general unifier for W.

  41. The resolution principle • The resolution principle for the first-order logic • Definition • If two or more literals (with the same sign) of a clause C have the most general unifier , then C is called a factor of C. • If C is a unit clause, it is called a unit factor of C.

  42. The resolution principle • Example. C=P(x)P(f(y))Q(x) • The first and the second literals have the most general unifier ={f(y)/x}. • Hence C=P(f(y))Q(f(y)) is a factor of C. • Definition • Let C1 and C2 be two clauses (called parent clauses) with no variables in common. • Let L1 and L2 be two literals in C1 and C2, respectively.

  43. The resolution principle • Definition (cont.) • If L1 and L2 have the most general unifier , then the clause (C1-L1)(C2-L2) is called a binary resolvent of C1 and C2. • The literals L1 and L2 are called the literals resolved upon.

  44. The resolution principle • Example. • Let C1=P(x)Q(x) and C2=P(a)R(x). • Since x appears in both C1 and C2, we rename the variable in C2 and let C2=P(a)R(y). • We choose L1=P(x) and L2=P(a). • Since L2=P(a), L1 and L2 have the most general unifier ={a/x}. (C1-L1)(C2-L2)= =({P(a),Q(a)}-{P(a)})({P(a),R(y)}-{P(a)})= ={Q(a)}{R(y)}={Q(a),R(y)}=Q(a)R(y). • Thus Q(a)R(y) is a binary resolvent of C1 and C2. P(x) and P(a) are the literals resolved upon.

  45. The resolution principle • Definition • A resolvent of (parent) clauses C1 and C2 is one of the following binary resolvents: • a binary resolvent of C1 and C2, • a binary resolvent of C1 and a factor of C2, • a binary resolvent of a factor of C1 and C2, • a binary resolvent of a factor of C1 and a factor of C2.

  46. The resolution principle • Example • Let C1=P(x)P(f(y))R(g(y)) and C2=P(f(g(a)))Q(b). • A factor of C1 is C1’=P(f(y))R(g(y)). • A binary resolvent of C1’ and C2 is R(g(g(a)))Q(b). • Therefore, R(g(g(a)))Q(b) is a resolvent of C1 and C2.

  47. The resolution principle • Example • Show that alternate interior angles formed by a diagonal of a trapezoid are equal. • We first axiomatize the theorem. • Let T(x,y,u,v) mean that xyuv is a trapezoid with upper-left vertex x, upper-right vertex y, lower-right vertex u, and lower-left vertex v. • Let P(x,y,u,v) mean that the line segment xy is parallel to the line segment uv. • Let E(x,y,z,u,v,w) mean that the angle xyz is equal to the angle uvw.

  48. The resolution principle • Example (cont.) • Then the axioms are the following: • A1: (x)(y)(u)(v)[T(x,y,u,v)P(x,y,u,v)] (definition of a trapezoid) • A2: (x)(y)(u)(v)[P(x,y,u,v)E(x,y,v,u,v,y)] (alternate interior angles of parallel lines are equal) • A3: T(a,b,c,d).

  49. The resolution principle • Example (cont.) • From these axioms, we should be able to conclude that E(a,b,d,c,d,b) is true, i.e. A1A2A3E(a,b,d,c,d,b) is a valid formula. • Since we want to prove this by refutation, we negate the conclusion and prove that A1A2A3E(a,b,d,c,d,b) is unsatisfiable.

  50. The resolution principle • Example (cont.) • The standard form of the formula: S={T(x,y,u,v)P(x,y,u,v), P(x,y,u,v)E(x,y,v,u,v,y), T(a,b,c,d),E(a,b,d,c,d,b)}. • The standard form S is a set of four clauses.

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