Introduction to Logic for Artificial Intelligence Lecture 1 Erik Sandewall 2010

Uses of Formal Logic in A.I. In preconditions of actions In postconditions of actions Contents of knowledgebases/beliefbases Contents of messages Queries Embedded expressions in action scripts Methods for planning and plan revision Obtaining conclusions from knowledgebase Many more

Uses of Formal Logic in A.I. To summarize: Expressions for evaluation, and use of their value Expressions for matching, obtaining variable values as the result Drawing conclusions from logic expressions In order to master these techniques, it is necessary to be precise about the character of logic formulas and the operations on them. This requires: Basic concepts Equivalence rules Inference rules

Propositional logic Expressions and Evaluation (not p) T -> F, F -> T (and p q) TT -> T, TF -> F, FT -> F, FF -> F (or p q) TT -> T, TF -> F, FT -> F, FF -> F (imp p q) TT -> T, TF -> F, FT -> T, FF -> T (eqv p q) TT -> T, TF -> F, FT -> F, FF -> T

Propositional logic Conventional Notations (not p) ¬p, ~p -p (and p q) p  q, p & q p * q (or p q) p  q p + q (imp p q) p → q, p  q (eqv p q) p ↔ q, p  q

Equivalence Relation between Formulas If p and q are two formulas, then we write p == q if and only if the value of p equals the value of q for each possible value of their elements. E.g. (and p q) == (and p q) (and p (or q r)) == (or (and p q)(and p r)) Notice that each of these lines contains a relation between two logic formulas, not a single logic formula It is important to write it as == and not as =, since = will be used for another purpose and within formulas

Equivalence rules for not, and and or (not (not p)) == p (and p p) == p (and p q) == (and q p) (and p (and q r)) == (and (and p q) r) similar to these for or instead of and (not (and p q)) == (or (not p)(not q)) (not (or p q)) == (and (not p)(not q)) (and p (or q r)) == (or (and p q)(and p r)) (or p (and q r)) == (and (or p q)(or p r)) (imp p q) == (or (not p) q) (eqv p q) == (or (and p q)(and (not p)(not q)))

Some terms Vocabulary for a logic formula: set of symbols containing all those that occur in the formula (and maybe some more) Interpretation for a logic formula: a mapping from a vocabulary for it, to truth-values T or F Model for a logic formula: an interpretation where the value of the formula is T Joint vocabulary for two (or more) logic formulas: something that is a vocabulary for each one of them (but there is of course a smallest joint vocabulary) p == q holds if they have the same value for all interpretations in all their joint vocabularies

Some terms Vocabulary for a logic formula: set of symbols containing all those that occur in the formula (and maybe some more) Interpretation for a logic formula: a mapping from a vocabulary for it, to truth-values T or F Model for a logic formula: an interpretation where the value of the formula is T Joint vocabulary for two (or more) logic formulas: something that is a vocabulary for each one of them (but there is of course a smallest joint vocabulary) p == q holds if they have the same value for all interpretations in all their joint vocabularies p |= q holds if, considering all interpretations in their joint vocabularies, if p is true in one then q is also true there. This is pronounced p entails q (Swedish: "innebär")

Some entailment rules (and p q) |= p General rule: if p == q then p |= q General rule: if p |= q and q |= p then p == q (not p) |= (imp p q) (and p (not p)) |= q Also, write p, q |= r if r is true in all interpretations where both p and q are true. (Allow two or more premises). Obtain: p, (imp p q) |= q /modus ponens/ p, q |= (and p q)

A very simple proof (imp a b) (imp b c) (imp (and c d) e) a d --------------------------- b c (and c d) e

Inference rule For organizing proofs, it is customary to identify a limited number of entailment rules and to require that only these shall be used each time a line is added to the proof The selected entailment rules are called the inference rules of this particular inference system Uses of the inference rules are expressed using |- instead of |= A proof with C1, C2, ... Cn as premises and D as conclusion is summarized as C1, C2, ... Cn |- D

Resolution Method for Theorem-Proving Transformations on Given Premises Rewrite all uses of imp and eqv using and, or, not Move all uses of not inwards, (not (and a b)) == (or (not a)(not b)) Replace (not (not p)) by p Move uses of or inside and, (or p (and q r)) == (and (or p q)(or p r)) Rewrite (or p (or q r)) as (or p q r), with arbitrary number of arguments, and similarly for and The result is an expression on conjunctive normal form Consider the arguments of and as separate formulas, obtaining a set of or-expressions with literals as their arguments Consider these or-expressions as having a set of literals as its argument, not a sequence of them. Results are called clauses. For convenience we shall write (not p) as -p

Resolution Method for Theorem-Proving Inference Rule Only one single inference rule is sufficient: {p}  A, {-p}  B |- A  B assuming p is not a member of A and -p is not a member of B Some examples: {a, b, c}, {-a, d, e} |- {b, c, d, e} {a, b, c}, {-a, b, e} |- {b, c, e} {a, b, c}, {-a, -b, d} |- {b, c, -b, d} This conclusion is correct, but useless for further proof! {a}, {b}, {c}, {-a, -b, -c, d} |- {d}

Resolution Method for Theorem-Proving Metatheorems If there exists a proof with C1, C2, ... Cn as premises and D as the consequence, then we write C1, C2, ... Cn |- D Metatheorems: Soundness: if C1, C2, ... Cn |- D then C1, C2, ... Cn |= D Completeness: if C1, C2, ... Cn |= D then C1, C2, ... Cn |- D Both of these hold - but completeness is more difficult to prove Soundness is automatic provided that each inference rule is in fact an entailment rule

Proof by Contradiction Given a set of propositions C1, C2, ... Cn and desired conclusion D We want to show that C1, C2, ... Cn |= D Transform the problem by considering C1, C2, ... Cn, (not D) and try to prove a contradiction e.g. (and p (not p)) for some p Write contradiction as  , or as {} in the case of resolution If this can be proved then we have a proof for what we want to show. In the case of resolution proofs: Add (not D) to the given propositions before the conversion to clause form Technically, we have a meta-theorem: if C1, C2, ... Cn, (not D) |=  then C1, C2, ... Cn |= D