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Transcript
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

p, q |= (and p q)
/modus ponens/
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