CS 4700: Foundations of Artificial Intelligence

CS 4700:Foundations of Artificial Intelligence Carla P. Gomes gomes@cs.cornell.edu Module: Propositional Logic: Inference (Reading R&N: Chapter 7)

Proof Methods

Current module we’ve talked about this approach Next module Proof methods • Proof methods divide into (roughly) two kinds: • Application of inference rules • Legitimate (sound) generation of new sentences from old • Proof = a sequence of inference rule applications Can use inference rules as operators in a standard search algorithm • Different types of proofs • Model checking • truth table enumeration (always exponential in n) • improved backtracking, e.g., Davis--Putnam-Logemann-Loveland (DPLL) (including some inference rules) • heuristic search in model space (sound but incomplete) e.g., min-conflicts-like hill-climbing algorithms

Proof • The sequence of wffs (w1, w2, …, wn) is called a proof (or deduction) of wn from • a set of wffs Δ iff each wi in the sequence is either in Δ or can be inferred from a • wff (or wffs) earlier in the sequence by using a valid rule of inference. • If there is a proof of wn from Δ, we say that wn is a theorem of the set Δ. • Δ├ wn • (read: wn can be proved or inferred from Δ) • The concept of proof is relative to a particular set of inference rules used. If we • denote the set of inference rules used by R, we can write the fact that wn can be • derived fromΔ using the set of inference rules in R: • Δ├ R wn • (read: wn can be proved from Δ using the inference rules in R)

Propositional logic: Rules of Inference or Methods of Proof • How to produce additional wffs (sentences) from other ones? What steps can we • perform to show that a conclusion follows logically from a set of hypotheses? • Example • Modus Ponens P P  Q ______________  Q The hypotheses (premises) are written in a column and the conclusions below the bar The symbol  denotes “therefore”. Given the hypotheses, the conclusion follows. The basis for this rule of inference is the tautology (P  (P  Q))  Q) [aside: check tautology with truth table to make sure] • In words: when P and P  Q are True, then Q must be True also. (meaning of • second implication)

Propositional logic: Rules of Inference or Methods of Proof • Example • Modus Ponens If you study the CS 4700 material  You will pass You study the CS4700material ______________  you will pass • Nothing “deep”, but again remember the formal reason is that • ((P ^ (P  Q))  Q is a tautology.

Propositional logic: Rules of Inference or Method of Proof

Valid Arguments • An argument is a sequence of propositions. The final proposition is called the conclusion of the argument while the other propositions are called the premises or hypotheses of the argument. • An argument is valid whenever the truth of all its premises implies the truth of its conclusion. • How to show that q logically follows from the hypotheses (p1 p2  …pn)? Show that (p1  p2  …pn)  q is a tautology One can use the rules of inference to show the validity of an argument.

Proof Tree • Proofs can also be based on partial orders – we can represent them using a tree structure: • Each node in the proof tree is labeled by a wff, corresponding to a wff in the original set of hypotheses or be inferable from its parents in the tree using one of the rules of inference; • The labeled tree is a proof of the label of the root node. • Example: • Given the set of wffs: • P, R, PQ • Give a proof of Q  R

P PQ R Q Q  R Tree Proof Given: P, P Q, R; Prove: Q  R MP Conj. What rules of inference did we use?

Length of Proofs • Why bother with inference rules? We could always use a truth table • to check the validity of a conclusion from a set of premises. But, resulting proof can be much shorter than truth table method. Consider premises: p_1, p_1  p_2, p_2  p_3 … p_(n-1)  p_n To prove conclusion: p_n Inference rules: Truth table: n-1 P steps 2n Key open question: Is there always a short proof for any valid conclusion? Probably not. The NP vs. co-NP question.

Resolution in Propositional Logic

P  Q P  R  Q  R Soundness of rule (validity of rule): [(P  Q)  (P  R)]  (Q  R) is valid Resolution (for CNF) Very important inference rule – several other inference rules can be seen as special cases of resolution. Resolution for CNF – applied to a special type of wffs: conjunction of clauses. Literal – either an atom (e.g., P) or its negation (P). Clause – disjunction of of literals (e.g., (P  Q  R)). Note: Sometimes we use the notation of a set for a clause: e.g. {P,Q,R} corresponds to the clause (PQ R); the empty clause (sometimes written as Nil or {}) is equivalent to False;

P  Q P  R  Q  R Soundness of Resolution:Validity of the Resolution Inference Rule resolving on P • Validity (Tautology): (P  Q) (P  R) (Q  R) ;

R  P P  Q  R  Q R  P P  Q  R Q P P  Q  Q P P  Q  Q can be re-written can be re-written Rule of Inference Chaining Rule of Inference: Modus Ponens Resolution:Special Cases 1 – Rule of Inference: Chaining 2 – Rule of Inference: Modus Ponens

P P  Q  Q P P  Q  Q Resolution:Special Cases 3 – Unit Resolution

P  Q  R P  W  Q  R  P  R  R  W P  Q  R P  W  Q  R  P  Q  Q  W P  Q  R P  W  Q  R  P  W Resolution:Caution! No duplications in the resolvent set only one instance of Q appears in the resolvent, which is a set! P  Q  R  S P  Q  W  Q  R  S  W Resolving one pair at a time DO NOT Resolve on Q and R Resolving on Q Resolving on R True

CNF Conjunctive Normal Form (CNF) A wff is in CNF format when it is a conjunction of disjunctions of literals. (P  Q  R) (S  P  T R) (Q  S) Resolution for CNF – applied to wffs in CNF format. {λ}  Σ1 { λ} Σ2  Σ1 Σ2 Σi- sets of literals i =1 ,2 λ – atom; Resolution Resolvent of the two clauses atom resolved upon

Conversion to CNF • P  (Q  R) • 1.Eliminate , replacing α  β with (α  β)(β  α). (P  (Q  R))  ((Q R)  P) • 2. Eliminate , replacing α  β with α β. (P  Q  R)  ((Q R)  P) • 3. Move  inwards using de Morgan's rules and double-negation: (P Q R)  ((QR)  P) • 4. Apply distributivity law ( over ) and flatten: (P  Q  R)  (Q P)  (R  P)

Converting DNF (Disjunctions of conjunctions) into CNF • 1 – create a table – each row corresponds to the literals in each conjunct; • 2 - Select a literal in each row and make a disjunction of these literals; Example: (PQ R ) (S R P) (Q S  P) (P  S  Q)  (P  R  Q) (P  P  Q) (P  S  S) (P  R  S) (P  P  S) (P  P  Q)… How many clauses?

Resolution:Wumpus World • P31P2,2, • P2,2 •  P31 P? P?

Resolution Refutation • Resolution is sound – but resolution is not complete – e.g., (P R) ╞ (P  R) but • we cannot infer (P  R) using resolution  • we cannot use resolution directly to decide all logical entailments. • Resolution is Refutation Complete: • We can show that a particular wff W is entailed from a given KB, how? • Proof by contradiction: • Write the negation of what we are trying to prove (W) as a conjunction of clauses; • Add those clauses (W) to the KB (also a set of clauses), obtaining KB’; prove inconsistency for KB’, i.e., • Apply resolution to the KB’ until: • No more resolvents can be added • Empty clause is obtained • To show that (P  R) ╞Res (P  R) do: (1) negate (P  R), i.e.: (P) (R) ; (2) prove that • (P  R)  (P) (R) is inconsistent ! !

Propositional Logic:Proof by refutation or contradiction: • Satisfiability is connected to inference via the following: • KB ╞ α if and only if (KBα) is unsatisfiable • One assumes α and shows that this leads to a contradiction with the facts in KB

Resolution:Robot Domain • Example: • BatIsOk • RobotMoves • BatIsOk  BlockLiftable RobotMoves Show that KB ╞ BlockLiftable KB • BatIsOk  BlockLiftable  RobotMoves BlockLiftable • BatIsOk • RobotMoves • BatIsOk  BlockLiftable  RobotMoves BlockLiftable RobotMoves KB’ • BatIsOk  RobotMoves BatIsOk BatIsOk Nil

Resolution • Resolution is refutation complete (Completeness of resolution refutation): • If KB╞ W, the resolution refutation procedure, i.e., applying resolution on KB’, will produce the empty clause. • Decidability of propositional calculus by resolution refutation: • If KB is a set of finite clauses and if KB╞ W, then the resolution refutation procedure will terminate without producing the empty clause. • Ground Resolution Theorem • If a set of clauses is not satisfiable, then resolution closure of those clauses contains the empty clause. In general, resolution for propositional logic is exponential ! The resolution closure of a set of clauses W in CNF, RC(W), is the set of all clauses derivable by repeated application of the resolution rule to clauses in W or their derivatives.

Resolution example:Wumpus World • KB = (B1,1 (P1,2 P2,1))  B1,1 α = P1,2

Resolution example:Wumpus World • KB = (B1,1 (P1,2 P2,1))  B1,1 α = P1,2 KB = (B11  (P1,2 P2,1)) ^ ((P1,2 P2,1)  B11)  B1,1 =(B11  P1,2 P2,1) ^ ((P1,2 P2,1)  B11)  B1,1 =(B11  P1,2 P2,1) ^(( P1,2 ^  P2,1)  B11))  B1,1 =(B11  P1,2 P2,1) ^( P1,2 B11) ^ ( P2,1  B11)  B1,1

Resolution example:Wumpus World • KB = (B1,1 (P1,2 P2,1))  B1,1 α = P1,2 KB = (B11  (P1,2 P2,1)) ^ ((P1,2 P2,1)  B11)  B1,1 =(B11  P1,2 P2,1) ^ ((P1,2 P2,1)  B11)  B1,1 =(B11  P1,2 P2,1) ^(( P1,2 ^  P2,1)  B11))  B1,1 =(B11  P1,2 P2,1) ^( P1,2 B11) ^ ( P2,1  B11)  B1,1

New resolvents added at the end Resolution algorithm • Proof by contradiction, i.e., show KBα unsatisfiable

New resolvents added at the end Resolution algorithm • Proof by contradiction, i.e., show KBα unsatisfiable Any complete search algorithm applying only the resolution rule, can derive any conclusion entailed by any knowledge base in propositional logic – resolution can always be used to either confirm or refute a sentence – refutation completeness (Given A, it’s true we cannot use resolution to derive A OR B; but we can use resolution to answer the question of whether A OR B is true.)

Resolution Closure • Definition The resolution closure RC(f) of a formula f in CNF is the set of all clauses derivable by repeated application of the resolution rule to clauses in f or their derivatives.

Resolution • Theorem (Ground Resolution Theorem) • If f ( in CNF) is unsatisfiable then RC(f) contains the empty clause (that evolves from P¬P for some variable P). • Proof (by contraposition) • If RC(f) does not contain the empty clause then it is satisfiable. • Let the variables in f be P1,..,Pn, and let us assume that RC(f) does not contain the empty clause. • Therefore we can construct a model for m for f with suitable truth • values for P1,..,Pn.

Resolution (contd.) • Construction procedure: • For i=1,..,n, if there exists a clause in RC(f) containing the literal ¬Pi such that all its other literals are false under the assignment chosen for P1,..,Pi-1, then assign false to Pi. . Otherwise we set Pi=True. • Claim: RC(f)|m is true. I.e., m is a model for f.

Resolution (contd.) • Claim, RC(f)|m is true. I.e., m is a model for f. • Assume the opposite. Then, there exists a clause that evaluates to false • under m. • Consider a non-satisfied clause in RC(f) over variables P1,..Pi that • minimizes i. • Wlog, we may write this clause as (ab), where b is a literal over Pi and • a is a formula over variables P1,.., Pi-1. • Note that there cannot exist another such clause where b is negated • (c¬b) as otherwise i was not minimal [consider (ac)  RC(f)].

Resolution (contd.) • So we write the clause as (ab), with b a literal over Pi • However, then Pi is set such that b evaluates to True, which contradicts the • assumption that the clause was not satisfied. • Therefore all the clauses are satisfied and m is indeed a model for RC(f). • QED

BatIsOk  BlockLiftable  RobotMoves BlockLiftable RobotMoves • BatIsOk  RobotMoves BatIsOk BatIsOk 0th level resolvents • BatIsOk • RobotMoves • BatIsOk  BlockLiftable  RobotMoves BlockLiftable Nil Resolution Refutation – Ordering Search Strategies • Original clauses – 0th level resolvents • Breadth first strategy • Generate all 1st level resolvents, then all 2nd level resolvents, etc. • Depth first strategy • Produce a 1st level resolvent; • Resolve the 1st level resolvent with a 0th level resolvent to produce a 2nd level resolvent, etc. • With a depth bound, we can use a backtrack search strategy; Depth first strategy

BatIsOk  BlockLiftable  RobotMoves BlockLiftable RobotMoves • BatIsOk  RobotMoves BatIsOk BatIsOk Nil Refinement Resolution Strategies Set-of-support Resolution Strategy • Definitions: • A clause γ2 is a descendant of a clause γ1 iif: • Is a resolvent of γ1 with some other clause • Or is a resolvent of a descendant of γ1 with some other clause; • If γ2 is a descendant of γ1, γ1 is an ancestor ofγ2; • Set-of-support – set of clauses that are either clauses coming from the negation of the theorem to be proved or descendants of those clauses. • Set-of-support Strategy – it allows only refutations in which one of the clauses being resolved is in the set of support. • Set-of-support Strategy is refutation complete. Set-of-support Strategy

Refinement Strategies • Ancestry-filtered strategy – allows only resolutions in which at least one member of the clauses being resolved either is a member of the original set of clauses or is an ancestor of the other clause being resolved; • The ancestry-filtered strategy is refutation complete.

Refinement Strategies • Linear Input Resolution Strategy – at least one of the clauses being resolved is a member of the original set of clauses (including the theorem being proved). • Linear Input Resolution Strategy is not refutation complete. Example: (P Q) (P Q) (P Q) (P Q) This set of clauses is inconsistent; but there is no linear-input refutation strategy; but there is a resolution refutation strategy; (P Q) (P Q) (P Q) (P Q) Q Q This is NOT Linear Input Resolution Strategy Nil

: Resolution and Pigeon Hole Principle • Consider a formula to represent the Pigeon Hole principle: (n+1) pigeons • in n holes. • Pij – pigeon i goes in hole j Pij  {0,1} i= 1, 2, …, n+1; j = 1, 2, …n • Each pigeon has to go in a hole: • Starting with pigeon 1 • P11  P12  …  P1n • … • Pn+1,1  Pn+1,2  …  Pn+1,n • .

Resolution Proofs of PH • And? • Two pigeons cannot go in the same hole • P11 ~P21; P11 ~P31; … P11 ~Pn+1,1; • ~P11 ~P21; ~P11 ~P31; … ~P11~Pn+1,1; • …. • ~Pn+1,n ~P2,n; ~Pn+1,n ~P3n; … ~Pn+1,n~Pn,n; Resolution proofs of PH take exponentially many steps: Resolution proofs of inconsistency of PH require an exponential number of clauses, no matter in what order we resolve the clauses (Armin Haken 85). Related to NP vs. Co-NP questions

Horn Clauses

Inference with propositional Horn clauses can be done in linear time ! Horn Clauses • Definition: • A Horn clause is a clause that has at most one positive literal. • Examples: • P; P  Q;  P  Q;  P  Q  R; Types of Horn Clauses: Fact – single atom – e.g., P; Rule – implication, whose antecendent is a conjunction of positive literals and whose consequent consists of a single positive literal – e.g., PQ  R; Head is R; Tail is (PQ ) Set of negative literals - in implication form, the antecedent is a conjunction of positive literals and the consequent is empty. e.g., PQ  ; equivalent to  P  Q.

P P  Q  Q P P  Q  Q Unit Resolution for Horn Clauses • Theorem • Unit resolution is refutation complete for HF, i.e. if kb¬a is a HF then unit propagation shows that kb¬a is unsatisfiable iff kb╞ a. • Proof • If kb¬a is not satisfiable then kb¬a must contain a clause that contains one non-negated variable only:This clause must be a unit clause. (otherwise consider the model that sets all variables to false). • Using unit resolution of this variable with all other clauses essentially eliminates the variable from the formula while preserving the Horn property. • Continuing this process, unit propagation hits P¬P iff the formula is not satisfiable.

Deciding entailment with Horn clauses can be done in linear time, in the size of the KB ! Forward chainingHORN (Expert Systems and Logic Programming) • Horn Form (restricted) KB = conjunction of Horn clauses • Horn clause = • proposition symbol; or • (conjunction of symbols)  symbol • E.g., C  (B  A)  (C  D  B) • Modus Ponens (for Horn Form): complete for Horn KBs • α1, … ,αn, α1 …  αnβ • β

Forward Chaining:Diagnosis systems • Example: diagnostic system IF the engine is getting gas and the engine turns over THEN the problem is spark plugs IF the engine does not turn over and the lights do not come onTHEN the problem is battery or cables IF the engine does not turn over and the lights come onTHEN the problem is starter motor IF there is gas in the fuel tank and there is gas in the carburator THEN the engine is getting gas

Forward chaining(Data driven reasoning) • Idea: fire any rule whose premises are satisfied in the KB, • add its conclusion to the KB, until query is found AND-OR graph

Count Agenda Inferred A B P => Q 1 L and M => P 2 B and L => M 2 A and P => L 2 A and B => L 2 P F L F M F B F A F Forward Chaining Algorithm • Forward chaining is • sound and complete • for Horn KB

Forward chaining example

CS 4700: Foundations of Artificial Intelligence