Informatics 1 Lecture 8 Resolution (continued) Michael Fourman - PDF document

In this lecture we consider formal descriptions of the relationships between a finite number of individuals. We may have different types of individual Informatics 1 Lecture 8 Resolution (continued) Michael Fourman "I am never really satisfied that I understand anything, because, understand it well as I may, my comprehension can only be an infinitesimal fraction of all I want to understand." Ada Lovelace, the world's first programmer, 
 student of de Morgan, who taught her mathematics. 1 2 3 Again, [the Analytical Engine] might act upon other things besides number, were objects found whose mutual fundamental relations could be expressed by those of the abstract science of operations, and which should be also susceptible of adaptations to the action of the operating notation and mechanism of the engine . . . Supposing, for instance, that the fundamental relations of pitched sounds in the science of harmony and of musical composition were susceptible of such expression and adaptations, the engine might compose elaborate and scientific pieces of music of any degree of complexity or extent. In 1852, when only 37 years of age, Ada died of cancer.

4 Premises X ⋁ ¬D Y ⋁ D X ⋁ Y Conclusion Any assignment of truth values that A valid makes all the premises true inference will make the conclusion true. The conclusion follows from the premises 5 Premises X ⋁ ¬D Y ⋁ D X ⋁ Y Conclusion Any assignment of truth values that For any valid makes the conclusion false will make inference at least one of the premises false. Premises where D does 
 not occur in X or Y X ⋁ ¬D Y ⋁ D X ⋁ Y Conclusion A special property If some assignment of this inference abc of values for ABC makes the conclusion false then the assignments abc D and abc D ̅ each make one or other of the two premises false.

7 Resolution U ⋁ V ⋁ W ⋁ X ⋁ ¬C X ⋁ Y ⋁ Z ⋁ C U ⋁ V ⋁ W ⋁ X ⋁ Y ⋁ Z Resolution gives the same 8 E, B ¬B, D ¬E, B D C information as our earlier E, D ¬E, D graphical analysis. B A D B E ¬E A, E ¬A, C ¬E, A C, E ¬E, C ¬A ¬B C ¬C ¬D A Using sets builds in Clausal Form idempotence, associativity Resolution uses CNF and commutativity. a conjunction of disjunctions of literals (¬A ⋁ C) ⋀ (¬B ⋁ D) ⋀ (¬E ⋁ B) ⋀ (¬E ⋁ A) ⋀ (A ⋁ E) ⋀ (E ⋁ B) ⋀ (¬B ⋁ ¬C ⋁ ¬D) Clausal form is a set of sets of literals { {¬A,C}, {¬B,D}, {¬E,B}, {¬E,A}, {A,E}, {E,B}, {¬B, ¬C, ¬D} } Each set of literals represents the disjunction of its literals. An empty disjunction {} represents false ⊥ . The clausal form represents the conjunction of these disjunctions (an empty conjunction {} represents true ⊤ ). 9

� Clausal Form Clausal form is a set of sets of literals { {¬A,C}, {¬B,D}, {¬E,B}, {¬E,A}, {A,E}, {E,B},{¬B, ¬C, ¬D} } A (partial) truth assignment makes a clause true iff it makes at least one of its literals true (so it can never make the empty clause {} true) A (partial) truth assignment makes a clausal form true iff it makes all of its clauses true ( so the empty clausal form { } is always true ). 10 If we have derived {} by Clausal form is a set of sets of literals { X 0 , X 1 , … , X n-1 } resolution, then, for any where x i = { L 0 ,…,L m i -1 } Resolution rule for clauses valuation we are given, the X Y special property lets us find a where ¬A ∈ X , A ∈ Y (X ⋃ Y) \ { ¬A, A } constraint that it violates. So If a valuation makes everything in the conclusion false then that valuation must make everything in one or there are no valuations other of the premises false. satisfying all the constraints. If it makes A true, then it makes everything in X false If it makes A false, then it makes everything in Y false 11 On this slide, indentation Davis Putnam indicates grouping. So, for Take a collection � of clauses. For each propositional letter, A each atom, we resolve all pairs For each pair ���������� ∊ �� ⋀ �� ∊ �� ⋀ � A ∊ �� ⋀ �� A ∊ � if �������� A ������� return UNSAT satisfying A ∊ X ⋀ ¬A ∊ Y. Once if �������� A �� is contingent ��������������������� ���� remove any clauses containing A or � A all the A-resolvants have been return SAT produced we can forget about Where �������� A �������������� A, � A � , and 
 a clause is contingent if does not contain any clauses containing A or ¬A. complementary pair of literals Heuristic: start with variables that occur seldom. Removing clauses that contain A or ¬A will not prevent us

13 A A contradictory cycle Y X ¬A X Y A ¬A ∨ X {¬A, X} {¬A} ¬X ∨ ¬A {¬X, ¬A} {} {A, Y} A ∨ Y {A} {¬Y, A} ¬Y ∨ A 14 A A contradictory cycle Y X ¬A A X Y ¬A ∨ X {Y} {¬X, Y} {¬A, X} ¬X ∨ ¬A {¬X, ¬A} {X, Y} {} {¬Y} A ∨ Y {A, Y} {X, ¬Y} {Y, ¬Y} {¬Y, A} ¬Y ∨ A {¬X, ¬Y} By our analysis of the picture, 15 D C ¬A, C we know that any valuation ¬B, D satisfying the binary B A ¬E, B constraints must make A, B, C, ¬E, A E ¬E D all true. So adding this new A, E E, B constraint makes an ¬A ¬B ¬C, ¬D inconsistent set of constraints. ¬C ¬D Use resolution to show this directly.

By our analysis of the picture, A ¬A, C we know that any valuation ¬B, D satisfying the binary ¬E, B constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly. By our analysis of the picture, 17 A ¬A, C ¬E, C we know that any valuation ¬B, D C, E satisfying the binary ¬E, B constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly. By our analysis of the picture, 18 A B ¬A, C ¬E, C we know that any valuation ¬B, D C, E satisfying the binary ¬E, B constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly.

By our analysis of the picture, 19 A B ¬A, C ¬E, C ¬E, D we know that any valuation ¬B, D C, E E, D satisfying the binary ¬E, B constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly. By our analysis of the picture, 20 A B C ¬A, C ¬E, C ¬E, D we know that any valuation ¬B, D C, E E, D satisfying the binary ¬E, B constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly. By our analysis of the picture, 21 A B C ¬A, C ¬E, C ¬E, D ¬E, ¬D we know that any valuation ¬B, D C, E E, D ¬D, E satisfying the binary ¬E, B constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly.

By our analysis of the picture, 22 A B C D ¬A, C ¬E, C ¬E, D ¬E, ¬D we know that any valuation ¬B, D C, E E, D ¬D, E satisfying the binary ¬E, B constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly. By our analysis of the picture, 23 A B C D ¬A, C ¬E, C ¬E, D ¬E, ¬D ¬E, E we know that any valuation E, E ¬B, D C, E E, D ¬D, E satisfying the binary ¬E, ¬E ¬E, B ¬E, E constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly. By our analysis of the picture, 24 A B C D E ¬A, C ¬E, C ¬E, D ¬E, ¬D ¬E, E we know that any valuation E , E ¬B, D C, E E, D ¬D, E satisfying the binary ¬E , ¬E ¬E, B ¬E, E constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly.

By our analysis of the picture, 25 A B C D E ¬A, C ¬E, C ¬E, D ¬E, ¬D ¬E, E we know that any valuation E , E ¬B, D C, E E, D ¬D, E {} satisfying the binary ¬E , ¬E ¬E, B ¬E, E constraints must make A, B, C, ¬E, A D all true. So adding this new A, E constraint makes an E, B ¬C, ¬D inconsistent set of constraints. Use resolution to show this directly. Resolution A complete proof procedure for propositional logic that works on formulas expressed in conjunctive normal form. (Robinson 1965) Conjunctive Normal Form (CNF) Literal: a propositional variable p or its negation ¬p Clause: a disjunction of (a set of) literals. CNF: a conjunction of clauses. 26 Resolution From two clauses C 1 = (X ∪ {A}), C 2 = (Y ∪ {¬A}) the resolution rule generates the new clause (X ∪ Y) = R(C 1 ,C 2 ) where X and Y are sets of literals, not containing A or ¬A. (X ∪ Y) is the resolvant A is the variable resolved on 27