Intelligent Systems

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Intelligent Systems

Lecture III – xx 2009Predicate Logic

Dieter Fensel and Dumitru Roman

www.sti-innsbruck.at 2

Where are we?

# Date Title

1 Introduction

2 Propositional Logic

3 Date Predicate Logic

4 Theorem Proving, Logic Programming, and Description Logics

5 Search methods

6 CommonKADS

7 Problem-Solving Methods

8 Planning

9 Agents

10 Rule learning

11 Inductive Logic Programming

12 Formal Concept Analysis

13 Neural Networks

14 Semantic Web and Exam Preparation


Outline

• Motivation• Technical Solution

– Formulas, Models, Tableaux– Deductive Systems– Resolution

• Illustration by Larger Examples• Extensions• Summary• References

Note: Most of the content of this lecture is based on Chapters 5-7 of the book Mathematical Logic for Computer Science (by Mordechai Ben-Ari)http://www.springer.com/computer/foundations/book/978-1-85233-319-5


MOTIVATION

4


Predicate logic

• Propositional logic is not expressive enough– Suppose we want to capture the knowledge that

Anyone standing in the rain will get wet.and then use this knowledge. For example, suppose we also learn that

Jan is standing in the rain.– We'd like to conclude that Jan will get wet. But each of these sentences would

just be a represented by some proposition, say P, Q and R. What relationship is there between these propositions? We can say

P /\ Q → RThen, given P /\ Q, we could indeed conclude R. But now, suppose we were told

Pat is standing in the rain.– We'd like to be able to conclude that Pat will get wet, but nothing we have stated

so far will help us do this– The problem is that we aren't able to represent any of the details of these

propositions. It's the internal structure of these propositions that make the reasoning valid. But in propositional calculus we don't have anything else to talk about besides propositions!

A more expressive logic is needed Predicate logic (occasionally referred to as First-order logic)


TECHNICAL SOLUTION

6


Predicate Logic – Overview

• All the logical machinery – formulas, interpretations, proofs, etc. – that was developed for the propositional logic (see previous lecture) can be applied to predicates– The presence of a domain upon which predicates are interpreted

considerably complicates the technical details but not the basic concepts

• Syntax– Predicate symbols are used to represent relations (functions from a

domain to truth values)– Quantifiers allow a purely syntactical expression of the statement that

the relation represented by the predicate is true for some or all elements of the domain

• Semantics– An interpretation consists of a domain and an assignment of relations to

the predicate letters– The semantics of the Boolean operators remains unchanged, but the

evaluation of the truth value of the formula must take the quantifiers into account


Predicate Logic – Overview (cont’)

• Semantic tableaux– The systematic search for a model is potentially infinite because

the domains like the integers are infinite– The construction of a tableau may not terminate, so there is no

decision procedure for satisfiability in the predicate logic, however, if a tableau happens to close, the formula is unsatisfiable, conversely, a systematic tableau for an unsatisfiable formula will close

• There are Gentzen and Hilbert deductive systems which are sound and complete– A valid formula is provable and we can construct a proof of the

formula using tableaux, but given an arbitrary formula we cannot decide if it is valid and hence provable


Predicate Logic – Overview (cont’)

• The syntax of predicate logic is extended with function symbols that are interpreted as functions on the domain– The predicate logic with function symbols is used in the

resolution procedure

• There are canonical interpretations called Herbrand interpretations– If a formula has a model, it has a model which is an Herbrand

interpretation, so to check satisfiablity, it is sufficient to check if there is a Herbrand model for a formula


Predicate Logic – Formulas, Models, Tableaux

• Relations and Predicates• Predicate Formulas• Interpretations• Equivalence and Substitution• Semantic Tableaux• Finite and Infinite Models• Undecidability of the Predicate Logic


Relations and Predicates

• The axioms and theorems of mathematics are defined on arbitrary sets such as the set of integers– We need to be able to write and manipulate logical formulas that

contain relations on values from arbitrary sets

• The predicate logic extends the propositional logic with predicate symbols that are interpreted as relations on a domain

• Let R be an n-ary relation on a domain D, i.e. R is a subset Dn. A relation can be represented by a Boolean-valued function R: Dn → {T,F}, by mapping an n-tuple to T iff it is included in the relation:

R(d1,…,dn) = T iff in R(d1,…,dn) in R


Predicate Formulas

• Predicate (relation) symbols • Constant symbols• Variables

• BNF Grammar for Predicate Formulas:


Predicate Formulas (cont’)

• Examples:

• Suppose A is a predicate formula. An occurrence of a variable x in A is a free variable of A if it is not within the scope of any quantifierA variable which is not free is said to be bound

• Examples:


Interpretations

• Let U be a set of formulas such that {p1, p2, . . . , pm} are all predicate symbols appearing in U and {a1, a2, . . . , ak} are all constant symbols appearing in U. An interpretation I of U is a tripe

I = (D, {R1,…,Rm}, {d1,…,dk})

where – D is a non-empty set (domain of I)

– Ri is an assignment of an ni-ary relation on D to the ni-ary predicate symbol pi

– di is an assignment of an element of D to the constant ai


Interpretations (cont’)

• Example: Consider the formulaSome of its possible interpretations are:

“For every natural number x, 0 ≤ x.”

“For every natural number x, 1|x.”

“For every string x over alphabet {0, 1}, empty string is a substring of x.”

“For every vertex x of G, (a, x) is an edge in G.”



• Suppose I is an interpretation for a predicate formula A. An assignment

is a function which assigns a value in the domain D to any variable appearing in the formula A

• Suppose A is formula, I an interpretation for A, and σI an assignment. We define vσI(A), the truth value of A under σI, inductively:(a) If A = p(c1, c2,…, cn) is an atomic formula, where each ci is either a variable xj or a constant symbol aj, then

(b) vσI(¬A) = ¬vσI (A)(c) vσI(A1 /\ A2) = vσI (A1) /\ vσI (A2)(d) vσI(A1 \/ A2) = vσI (A1) \/ vσI (A2)(e) (f)



• Theorem: If A is a closed formula, then vσI(A) does not

depend on σI. In that case, we write vI(A).

• Theorem: Let A' = A(x1, x2, . . . , xn) be a non-closed formula and let I be an interpretation. Then:

(a) vσI(A) = T for assignment σI iff

(b) vσI(A) = T for all assignments σI iff



• A closed formula A is true in I, or I is a model for A, if vI(A) = T

I |= A• A closed formula A is satisfiable if, for some

interpretation I,I |= A

• A is valid if, for all interpretations I,I |= A

• A is unsatisfiable if it is not satisfiable, and falsifiable if it is not valid



• Examples:


Equivalence and Substitution

• Suppose A1,A2 are two closed formulas. If, for all interpretations I

vI(A1) = vI(A2) we say that A1 and A2 are equivalent, and we write

A1 ≡ A2

• Suppose U is a set of closed formulas, and A a closed formula

U |= Ameans that, in all interpretations I in which all formulas from U are true, we also have

vI(A) = T• Examples:


Equivalence and Substitution (cont’)

• Theorem:

A ≡ B iff |= A ↔ B

U |= A iff |= A1 /\ A2 /\ ... /\ An → A

• Examples of valid formulas:


Semantic Tableaux

• Recall that a tableaux is systematic search for a counter example

• Example: We will try to show that

is a valid formula.

We consider its negation

and try to show that it is unsatisfiable.


Semantic Tableaux (cont’)

• Semantic tableaux for :



• Example: Now, consider the formula

which is satisfiable but not valid (so its negation should also be satisfiable). Semantic tableaux (we obtain a closed tableau for a satisfiable formula):



• Question: What went wrong in the previous example?– The same constant a was used twice to eliminate two distinct existential

quantifiers– We were forced to use the same constant since, once we eliminated the

universal quantifier in

we replaced it with a and were forced to work with that constant exclusively from that point on

• Solution: We will not delete universal quantifiers from nodes of the tableau; instead, we introduce some instance of that variable but keep writing the universal quantifier. E.g.



• Using these guidelines, if a tableau for the formula from the previous example is correctly constructed, one branch ends with the open leaf

In fact, this leaf gives a model for this satisfiable formula; the domain is

D = {a, b}and the unary relations are subsets

p = {a}, q = {b}(This is what we will be defined as an Herbrand model for this formula later on in this lecture)



• Example: Consider the formulas

Check whether A = A1 /\ A2 /\ A3 is a satisfiable formula and, if so, find one model for A.Solution: First construct a semantic tableau for the formula:



• The tableau in the previous example does not terminate; namely, every time an universal or an existential quantifier is dropped, a new constant symbol ai can be introduced, to get an infinite sequence of constants:

a1, a2,…, an,…The formula does have an obvious infinite model:

I = (N, {<})Furthermore, one can prove, using the formulas A2 and A3 (see the proof of Theorem 5.24 in the textbook) that every model of

A = A1 /\ A2 /\ A3

must be infinite. So, the tableau construction effectively produces a “generic” infinite model for A.

• One major difference in comparison with semantic tableaux for propositional logic is (as seen in the previous example) that a tableau of a predicate formula may not terminate

– The reason for this anomaly is that, in propositional logic, nodes of a tableau simplify in terms of the formula complexity. In predicate logic, this is not the case, since we can never eliminate universal quantifiers


Semantic Tableaux - Algorithm for Semantic Tableaux

• As in propositional case, we generalize the rules into two cases, γ-rules for universal formulas and δ-rules for existential formulas:

• A literal is a closed atomic formula p(a1, a2,…, an) or the negation of such a formula

• Algorithm for construction of a semantic tableau:Input: A predicate formula A Output: Semantic tableau T for A; all branches are either infinite, or finite with leaves marked × (closed) or o (open).A semantic tableau is a tree T each node of which will be labeled with a set of formulas. T is built as follows.


Semantic Tableaux - Algorithm for Semantic Tableaux (cont’)

(1) Initially, T is a single node, labeled {A}(2) We build the tableau inductively by choosing an unmarked leaf l,

labeled U(l), and applying one of the following rules:– If U(l) is a set of literals and γ -formulas containing a pair of

complementary literals {p(a1, a2,…, an), ¬p(a1, a2,…, an)}, mark it as closed (×)

– If U(l) is not a set of literals, choose a formula A in U(l) which is not a literal:

• α- and β-rules are applied just as in propositional logic• If A is a γ-formula, add a new node l', a child of l, and label it

U(l') = U(l) U {γ(a)}where a is a constant appearing in U(l). If U(l) consists of literals and γ -formulas only, mark it × or o, depending on whether there is a set of complementary literals.

• If A is a δ-formula, create a new node l' as a child of l and label itU(l') = (U(l) − {A}) U {δ(a)}

where a is some constant that does not appear in U(l).


Semantic Tableaux - Algorithm for Semantic Tableaux (cont’)

• A branch in T is closed if it terminates in a leaf marked ×. Otherwise, it is open.

• Theorem (Soundness): Suppose A is a predicate formula and T its semantic tableau. If T closes, then A is unsatisfiable.

• Theorem (Completeness): Suppose A is a valid formula. Then, the systematic semantic tableau for A terminates and is closed.(Note: see section 5.5 in the textbook for a detailed description of systematic semantic tableau)


Finite and Infinite Models

• Theorem (Löwenheim): If a formula is satisfiable, then it is satisfiable in a countable model

• Theorem (Löwenheim - Skolem): If a countable set of predicate formulas is satisfiable, then it is satisfiable in a countable model

• Theorem (Compactness Theorem): Let U be a countable set of formulas. If all finite subsets of U are satisfiable, then so is U


Undecidability of the Predicate Logic

• Church, building on the work of Turing, proved that there is no decision procedure for validity in predicate logic

• Turing machines can be viewed as devices which compute functions on natural numbers; i.e. given a Turing machine T, we can associate to it a function

fT : N → Nso that fT(n) = m if T halts with the tape consisting of m 1’s when started on the tape with the input of n consecutive 1’s. If T never halts on the input of n consecutive 1’s, then fT(n) is undefined

• Theorem (Church): It is undecidable whether a Turing machine, started on a blank tape, will halt– In other words, it is undecidable, given a Turing machine T, whether fT

(0) is defined


Undecidability of the Predicate Logic (cont’)

• Two-register machine (or, a Minsky machine) M consists of a pair of registers (x, y) which can store natural numbers, and a program P = {L0, L1,…, Ln}, which is a sequential list of instructions. Ln is always the command “halt”, and for 0 ≤ i < n, Li has one of the two forms– r := r + 1, for r in {x, y}– if r = 0 then go to Lj else r := r − 1, for r in {x, y}, 0 ≤ j ≤ n

• Execution of M: sequence of states sk = (Li , x, y) where Li is the current instruction during the execution, and x,y are current contents of the two registers.– Initial state: s0 = (L0,m, 0), for some m– If sk = (Ln, x, y), for some k then M halts and y = f (m) is computed by M

• Theorem: For every Turing machine T that computes f : N → N, a two-register machine M can be constructed which computes the same function.– Corollary: It is undecidable whether, given a two-register machine M,

whether fM(0) exists or not



• Theorem (Church): Validity in predicate calculus is undecidableSketch of the Proof:To each two-register machine M, we associate a predicate formula SM such that M halts when started at (L0, 0, 0) |= SM

We use the language:– Binary relations: pi (x, y) (i = 0, 1,…, n)– Unary function: s(x)– Constant symbol: a

Intended interpretation:– pi (x, y): M is at the state (Li, x, y)– s(x): successor function s(x) = x + 1– a: a = 0



Finally, define

SM says the following: if a machine with the program P = {L0, L1, . . . , Ln} is started at the initial state (L0, 0, 0), then the computation will halt with the values at the registers being (z1, z2), for some natural numbers z1, z2

Si is defined by cases of the instruction Li :

Since the Halting Problem for two-register machines is undecidable, it is impossible to verify algorithmically whether |= SM or not.


Predicate Calculus: Deductive Systems

• Gentzen System G• Hilbert System H


Gentzen System G

• As in propositional logic, Gentzen proof system is based on the reversal of a semantic tableau for a formula

• Example: Prove that

Solution: start by constructing a tableau for the negation

:


Deductive System G

• Axioms: Any set of formulas U containing a complementary pair of literals

• Rules: α- and β-rules are the same as in propositional logic, plus

where

and a is an arbitrary constant.• Theorem (Soundness and Completeness) Let U be a set

of formulas. There is a Gentzen proof for U if and only if there is a closed semantic tableau for U.


Deductive System G

• Example: The proof in G for

is


Hilbert System H

• Axioms: The three axioms for the Hilbert system in propositional logic, plus– Axiom 4. – Axiom 5.

assuming x is not free in A

• Rules of Inference: Modus Ponens, plus

(Generalization:)


Hilbert System H (cont’)

• There is a problem with the Generalization Rule if it is not being used judiciously; consider the following derivation in the set N with the unary predicate even(x):

1. even(2) ├ even(2) Assumption2. even(2) ├ even(x) Gen. Rule 1

We derived a wrong conclusion that every natural number is even, starting from the true assumption that 2 is even. What went wrong?Answer: We should not be able to generalize based on a constant included in the assumptions. Namely, assumptions may contain very specific facts and not simply general logical truths.The Generalization Rule must be modified as follows (and with the modification the deduction rule can be further defined)


Hilbert System H (cont’)

• Generalization Rule:

provided a does not appear in U.• Deduction Rule:

• Theorem (Soundness and Completeness): Hilbert proof system H for predicate logic is sound and complete.

• Axiom 4 and MP justify the following derived rule (Specification Rule):

for any constant a.


Some relevant theorems and proofs in H

• Theorem: Proof:

• Theorem: Proof:


Some relevant theorems and proofs in H (cont’)

• Theorem: Proof:

• This proves a more general version of the Generalization Rule:


Some relevant theorems and proofs in H (cont’)

• Theorem: Proof:

• Theorem: Proof:


Resolution

• Functions and Terms• Clausal Form• Herbrand Models• Herbrand’s Theorem• Ground Resolution• Substitutions• Unification• General Resolution


Functions and Terms

• We will now allow the language for predicate logic to contain function symbols; they will be interpreted as functions (of appropriate arity) in the domain of an interpretation

• Example: Consider the formula p(x, y) → p(x, f (y)) where p is a binary predicate symbol, and f is a unary function symbol; two possible interpretations of this formula– I1 = (N, {<}, {succ(x)}), where succ(x) = x + 1

– I2 = ({0, 1}*, {substr}, {f }), where the relation substr(w1,w2) means that w1 is a substring of w2, and f(w) = w0, is the function appending 0 to the right end of word w


Functions and Terms (cont’)

• Terms: Suppose

then

• Suppose a, b are constant symbols, p a binary predicate symbol, f is a binary function symbol, and g a unary function symbol– Examples of terms: a, g(a), f (x, y), f (x, g(a)), f (f (a, x), b), f (f (x, y), f (b,

g(a))),…– Examples of atomic formulas: p(a, a), p(f (x, y), g(y)), p(g(b), f (a, g(a))),

…


Functions and Terms (cont’)

• A term or atom is ground if it contains no variables; a formula is ground if it has no quantifiers and no variables

• A' is a ground instance of a quantifier-free formula A if it can be obtained from A by substituting ground terms for free variables

• Examples:– Examples of ground terms: f(a, a), g(b), f(f (a, b), g(a)),…– Examples of ground formulas: ¬p(a, a), p(f (a, b), b) → p(a, a),…– The formula ¬p(f(a, b), b) \/ p(a, f(a, a)) is a ground instance of

the formula ¬p(f(x, b), y) \/ p(x, f(x, x))


Clausal Form

• A formula A is in prenex conjunctive normal form (PCNF), iff it is of the form

Q1x1Q2x2…Qnxn M(x1, x2,..., xn)where Qi (i = 1, 2,…, n) are quantifiers and M(x1, x2,..., xn) is a quantifier-free formula in CNF, called the matrix of A.

• A closed formula is in clausal form if it is in PCNF and all quantifiers Qi are universal.

• A literal is an atomic formula or its negation; e.g. p(x, f(a, y)),¬p(g(x), f(x, g(x)))

A literal is said to be ground if it contains no variables.A clause is a disjunction of literals; e.g.

¬p(x, y) \/ p(x, f (x))


Clausal Form (cont’)

• C' is a ground clause if it is a ground instance of a clause C; e.g. ifC = ¬p(x, y) \/ p(x, f (x)),

the substitution x ← f(a), y ← a produces the ground clauseC' = ¬p(f(a), a) /\ p(f(a), f(f(a)))

• Example of formula in clausal form:

This clausal form can be written in the following way:{{p(x, f(y)), q(z)}, {¬q(f(x)),¬p(f(y), z)}}

An even more simplified version of this clausal form is:{pxfyqz,¬qfx¬pfyz}



• The notation A ≈ B is used to denote the fact that a formula A is satisfiable iff B is satisfiable

• Theorem (Skolem): Suppose A is a closed formula. Then, there exists a formula A' in clausal form such that

A ≈ A‘Proof (Algorithm for converting A to A'):We start with a particular formula

and we will illustrate each step of the algorithm by showing how it applies to this particular exampleInput: Closed formula AOutput: formula A' in clausal form, such that A ≈ A‘The following steps are performed (see next slide)



(1) Rename bound variables (if necessary) so that no variable appears in the scope of two different quantifiers:

(2) Eliminate all propositional connectives, except for negation, conjunction, and disjunction

(3) Push all negations inward and eliminate double negations:

(4) Extract quantifiers from the matrix:



(5) Transform the matrix of the formula into CNF:

(6) Eliminate existential quantifiers:– If we have replace x with a new constant symbol a– If we have , replace y with f(x1, x2, . . .

, xn), where f is a new n-ary function symbol

• The method of converting a closed formula A into a clausal form A' so that A ≈ A‘ is also called Skolemization of A



• Example: Transform the formula

into a clause form A' ≈ A

Solution:


Herbrand Models

• Once we allow function symbols in the language of predicate logic, the models can become rather complicated since, given any function symbol, there are a large number of functions on the domain that can interpret it

• We will show that, if a set of clauses has a model (i.e. is satisfiable), it has a particular canonical (or, generic) model


Herbrand Models (cont’)

• Let S be a set of clauses and– : set of constant symbols appearing in S– : set of function symbols appearing in S

• We define HS, the Herbrand universe HS for S is defined inductively, as follows:– a HS, for every a– f(t1, t2,…, tn), for every n-ary function symbol f ,

and (t1, t2,…, tn) HS

If there are no constant symbols in S ( = Ø), we initialize the inductive definition by putting some arbitrary new constant symbol a in HS.

• Remark: The Herbrand universe HS for S will be infinite as soon as there is a function symbol in S



• Examples:– If S1 = {pxy¬qa, qa¬pbx}, then

HS = {a, b}

– If S2 = {¬pxf (y), pwg(w)}, then

HS = {a, f(a), g(a), f(f(a)), f(g(a)), g(f(a)), g(g(a)),…}

– If S3 = {¬paf (x, y), pbf (x, y)}, then

HS = {a, b, f(a, a), f(a, b), f(b, a), f(b, b), f(a, f(a, a)),

f(f(a, a), a),…}



• Suppose HS is the Herbrand universe for a set of clauses. The Herbrand base BS is the set of ground atomic formulas formed using the predicate symbols in S and terms from HS.

• Example:– For S3 = {¬paf (x, y), pbf (x, y)}, the Herbrand base is

BS = {p(a, f(a, a)), p(a, f(a, b)), p(a, f(b, a)), p(a, f(b, b)), …, p(a, f(f(a, a), a)), . . . , p(b, f(a, a)), p(b, f(a, b)), p(b, f(b, a), p(b, f(b, b))

• An Herbrand model for a set of clauses S consists of the Herbrand universe HS as the domain, the function symbols and constants are interpreted by themselves, and the relations are interpreted in some way which would make all the clauses in S true; in other words, the true relations form some subset of BS which makes all clauses true



• Example: For S3 = {¬paf(x, y), pbf(x, y)}, the Herbrand universe is

HS = {a, b, f(a, a), f(a, b), f(b, a), f(b, b), f(a, f(a, a)), f(f(a, a), a),…}

The relations in Herbrand model are given as

v(p(a, f(a, a))) = F, v(p(a, f(b, a))) = F, v(p(a, f(a, b))) = F

v(p(a, f(b, b))) = F,

v(p(b, f(a, a))) = T, v(p(b, f (b, a))) = T,

v(p(b, f(a, b))) = T, v(p(b, f(b, b))) = T, …

• Theorem: Let S be a set of clauses. S has a model if and only if it has an Herbrand model.



• Theorem (Herbrand’s Theorem - Semantic Form): A set of clauses S of predicate logic is unsatisfiable if and only if a finite set of ground instances of clauses from S is unsatisfiable

• Example: Consider the following unsatisfiable formula

Its clausal form is {¬p(x) \/ q(x), p(y),¬q(z)}One set of ground instances for this set of clauses is:

{¬p(a) \/ q(a), p(a),¬q(a)}obtained by substitution x ← a, y ← a, z ← aNow, we can use any method for proving unsatisfiability from propositional logic (semantic tableaux, resolution, etc.):



• Herbrand’s theorem gives us the handle needed to define an efficient semi-decision procedure for validity of formulas in the predicate calculus:

1. Negate the formula2. Transform the formula into a clausal form3. Generate a finite set of ground instances of clauses4. Check if the set of ground clauses from (3) is unsatisfiable

Step 3 is highly problematic; namely, there are infinitely many ground terms (if there is at least one function symbol) so it may be difficult to find a correct substitution for resolution. In fact, if the set of clauses is satisfiable, there is no such substitution, so our search for it may run forever. This is not surprising, however, considering that we already know that the validity in predicate logic is undecidable.


Ground Resolution

• Suppose C1,C2 are ground clauses containing a pair of clashing literals l, lc , say, l in C1, lc in C2. The resolvent of C1 and C2 is the clause

C1 and C2 are called the parent clauses in resolution

• Theorem: The resolvent C is satisfiable if and only if C1 and C2 are simultaneously satisfiable


Substitutions

• Suppose x1, x2,..., xn are distinct variables and t1, t2,…, tn terms such that ti is not equal to xi (i = 1, 2,…, n). The substitution

{x1 ← t1, x2 ← t2,…, xn ← tn}

is the mapping assigning the term ti to the variable xi.– Generally, we will use Greek letters Θ, σ, ζ, μ,… to denote substitutions

• An expression is any term, literal, a clause, or a set of clauses. If E is an expression and

σ = {x1 ← t1, x2 ← t2,…, xn ← tn}is a substitution, the instance Eσ is obtained by simultaneously applying the substitution σ to all occurrences of xi’s in E

• Example: Suppose the expression E is the clause p(x, y) \/ ¬q(f(x)) and σ is the substitution {x ← a, y ← f(x)}. Then the instance Eσ of E is:

Eσ = p(a, f(x)) \/ ¬q(f(a))


Substitutions (cont’)

• SupposeΘ = {x1 ← t1, x2 ← t2,…, xn ← tn}σ = {y1 ← s1, y2 ← t2,…, yk ← sk}

are two substitutions with X = {x1, x2,..., xn }, Y = {y1, y2,..., yk}.The composition of Θ and σ is the substitution

Θσ = {xi ← tiσ : xi ≠ tiσ} U {yj ← sj : yj in Y, yj not in X}In plain English: first apply the substitution σ to the terms ti that appear in Θ. If some substitutions become xi ← xi, delete them. Finally, append all yj ← sj to the list, for which yj is not one of the variables xi in Θ.

• Lemma: If E is an expression and Θ and σ are two substitutions, E(Θσ) = (EΘ)σ

• Lemma: The composition of substitutions is associative: if Θ, σ, and ζ are substitutions,

Θ(σζ) = (Θσ)ζ


Unification

• Consider the pair of literalsp(f (x), g(y)), ¬p(f(f(a)), g(z)).

This pair cannot be resolved using the method of ground resolution. However, the substitution

{x ← f(a), y ← a, z ← a}will turn the pair into

p(f(f(a)), g(a)), ¬p(f(f(a)), g(a))to which ground resolution can be applied.In fact, a simpler substitution

{x ← f(a), y ← z}will also accomplish this, even though we will not end up with a pair of ground literals, but

p(f(f(a)), g(z)), ¬p(f(f(a)), g(z)),which are still clashing.


Unification (cont’)

• Given a set of atoms, a unifier is a substitution which makes the atoms identical. A most general unifier, or m.g.u, for short, is a unifier μ such that, if Θ is any unifier for the set of atoms

Θ = μ ζfor some substitution ζ.In other words, every unifier for that set of atoms can be obtained from an m.g.u. through further substitution.

• Example: The unifierμ = {x ← f(a), y ← z}

is an m.g.u. for the set of atoms p(f(x), g(y)) and p(f(f(a)), g(z)). In fact,

{x ← f(a), y ← a, z ← a} = μ{z ← a}



• Question: When is it impossible to unify two atomic formulas?1. if they start with different predicate symbols (obvious)

2. A trickier obstacle: consider the pair of atoms

p(a, x), p(a, f(x))

No matter what substitution we attempt to use, the problem is that we will never be able to make the terms x and f(x) identical.

The reason for this is that the two terms we are attempting to unify contain the same variable.



• If we want to unify two atomsp(t1, t2,…, tn), p(t’1, t’2,…, t’n)

we are trying to unify pairs of terms t1 and t’1 , t2 and t’2,…, tn and t’n.We can view this problem as attempting to solve a system of n term equations:

t1 = t’1t2 = t’2

...tn = t’n

• Example: The problem of trying to unify the pair of atomsp(x, f(y)), p(f(f(y)), g(a))

can be viewed as solving the system of two term equations:x = f(f(y))f(y) = g(a)

Based on the above remark, this system cannot be solved since the terms in the second equation start with different function symbols



• A set of term equations is in solved form if:– All equations are of the form

x = t, x − variable , t − term not containing x– Every variable x which appears in the left-hand side of one of the

equations does not appear in any other equation in the system.

If the solved form of the system is

x1 = t1, x2 = t2,…, xn = tn

the solving substitution is

σ = {x1 ← t1, x2 ← t2,…, xn ← tn}


Unification Algorithm

Input: A set of term equationsOutput: A set of term equations in solved form or the answer “not

unifiable”(1) Transform every equation of the form t = x (x-variable, t-term which

is not a variable) into x = t;(2) Delete all trivial equations of the form x = x (x-variable)(3) Suppose t' = t'' is an equation where neither term t', t'' is a variable.

If starting function symbols of t' and t'' are not the same, output “not unifiable”; otherwise, if

t' = f(t'1,…, t'k), t'' = f (t''1,…, t''k),replace the equation with k new equations

t'1 = t''1,…, t'k = t''k(4) If x = t is an equation such that x appears elsewhere in the system:

– if x appears in t, output “not unifiable”– if x appears in other equations, replace all occurrences of x in those

equation with t


Unification Algorithm – Example

• Consider the system of term equations:g(y) = x, f(x, h(x), y) = f(g(z),w, z)

• Start by using Rule 3 to replace the second equation with three equations:g(y) = x, x = g(z), h(x) = w, y = z

• Next, use Rule 1 to rewrite the first and the third equation:x = g(y), x = g(z), w = h(x), y = z

• Using the first equation, eliminate x from the right-hand sides of the remaining equations:

x = g(y), g(y) = g(z), w = h(g(y)), y = z• Next, using Rule 3, rewrite the second equation:

x = g(y), y = z, w = h(g(y)), y = z• The last equation is redundant (it is identical to the second equation) and we

can delete it:x = g(y), y = z, w = h(g(y))

• Finally, use the second equation to eliminate y from the right-hand sides of other equations:

x = g(z), y = z, w = h(g(z))• This system is in solved form, so m.g.u is:

μ = {x ← g(z), y ← z,w ← h(g(z))}


General Resolution

• Suppose L = {l1, l2,…, lm} is a set of literals. Then, we define its complement as Lc = {lc1, lc2,…, lcm}

• General Resolution Rule: Suppose C1 and C2 are two clauses with no variables in common. Let

so that L1 and Lc2 can be unified using an m.g.u. σ. Then,

C1 and C2 are said to be clashing clauses and their resolvent is

Res(C1, C2) = (C1σ − L1σ) U (C2σ − L2σ)


General Resolution (cont’)

• Example: Consider the two clausesC1 = p(f(x), g(y)) \/ q(x, y)

C2 = ¬p(f(f(a), g(z)) \/ q(f(a), g(z))These two clauses contain the following literals:

L1 = {p(f(x), g(y))}L2 = {¬p(f(f (a), g(z))}

so that Lc2 = {p(f(f(a)), g(z))}.

It is easy to check that the literals in L1 and Lc2 are unifiable with an

m.g.u.σ = {x ← f (a), y ← z}

so the clauses C1 and C2 are clashing, and their resolvent is:Res(C1, C2) = (C1σ − L1σ) U (C2σ − L2σ)

= {q(f(a), z)} U {q(f(a), g(z))}so that we get the clause q(f(a), z) \/ q(f(a), g(z))


General Resolution (cont’)

• Most often, the two clauses we are trying to resolve will have variables in common, so we cannot use the Resolution Rule as stated before

• Before we attempt to resolve such pairs of clauses, we have to standardize apart; i.e. we have to rename the variables so that the clauses no longer have any variables in common

• Example: Consider the two literalsp(f(x), y), ¬p(x, a)

The two atoms contain the same variable x, so we have to standardize them apart before attempting to resolve; e.g. we can replace x in the second atom with z:

p(f(x), y), ¬p(z, a)It is easy to see that an m.g.u. is σ = {z ← f (x), y ← a}. The m.g.u. transforms the pair of literals into:

p(f(x), a), ¬p(f (x), a)Now, we can apply resolution to this pair of literals to obtain the empty clause.


General Resolution Procedure

• Input: A set of clauses S• Output: “satisfiable” or “not satisfiable”, in the case the algorithm

terminates• Set S0 := S• Suppose Si has been constructed• Choose a pair of clashing clauses C1, C2 in Si and find

C = Res(C1, C2)(we may need to standardize apart certain clauses during this step)

• If C = □ (i.e. empty clause), terminate the procedure and output “not• satisfiable”; otherwise, put

Si+1 := Si U {C}• • If Si+1 = Si, for all possible pairs of clashing clauses in Si, terminate

the procedure and output “satisfiable”


General Resolution Procedure – Example

• Example: Show, using the general resolution procedure, that the set of clauses 1-7 is unsatisfiable.

This is shown in lines 8-15 which are a refutation by the resolution procedure (each line contains the resolvent, the mgu and the number of the parent classes)


General Resolution Procedure – Soundness and Completeness

• Theorem (Soundness of Resolution) If the empty clause is derived by general resolution, then S is unsatisfiable.

• Theorem (Completeness of Resolution) If a set of clauses S is unsatisfiable, then the empty clause can be derived from S using general resolution.


ILLUSTRATION BY LARGER EXAMPLES

80


Example 1

• Show thatis not valid using the semantic tableaux methodSolution: It suffices to show that the negation of the formula

is satisfiable. In fact, we will also find the smallest model for this negation, i.e. the smallest structure in which the original formula fails to be true.– From , we get the descendent

– As the next level of the tableau, we get:

– After introducing two new constant symbols a and b:


Example 1 (cont’)

– We can now generate the following instances of the universal formula:

– After this, it can be shown that, after branching off different possibilities for p(a) \/ q(a) and p(b) \/ q(b), we will never be able to terminate all the branches and mark them as closed.

• In fact, by examining what happens after the tree starts branching, one can show by following one of the branches that the smallest model for the negation has the universe {a, b} and that the relations are defined as follows:

p(a), ¬q(a), ¬p(b), q(b) which will falsify the original formula.


Example 2

• Prove, in the Hilbert proof system for predicate logic, that

Solution:


Example 2

• Transform each of the following formulas to the clausal form:

Solution:


Example 2 (cont’)


Example 3

• Prove the validity of

by resolution refutation of its negation.

Solution:

First, consider the negation of the formula:

Convert the above formula into a PCNF:


Example 3 (cont’)

Finally, try to refute the set of clauses{¬A(x) \/ B(x),A(f(x)),¬B(z)}

First, unify ¬A(x) \/ B(x) and A(f(y)) using{x ← f(y)}

to get as the resolventB(f(y))

Next, unify B(f(y)) and B(z) using the unifier{z ← f(y)}

and apply resolution to get the empty clause. So,

is unsatisfiable, which means that its complement is a valid formula of predicate logic


Example 4

• Find out if the formula

is a logical consequence of the set of formulas

Solution: We use resolution to determine whether

is a logical consequence of the formulas

Note that C is a logical consequence of A and B iff the formula A /\ B → C is valid. This, in turn, is equivalent to checking whether ¬(A /\ B → C) A /\ B /\ ¬C is unsatisfiable.


Example 4 (cont’)

Skolemize A, B, and ¬C to get the following set of universal formulas:

where f, g, h, i are new unary function symbols, and a, b are new constant symbolsSo, we need to check if the following set of clauses is unsatisfiable:

Each clause in S contains a negative literal. In general, if both premises of a resolution rule contain a negative literal, so does the conclusion. Thus, we can only derive clauses with negative literals from S (by resolution), but not the empty clause (a contradiction). Therefore, S is satisfiable and C cannot be a logical consequence of A and B.


EXTENSIONS

90


Extensions

• The characteristic feature of first-order logic is that individuals can be quantified, but not predicates; second-order logic extends first-order logic by adding the latter type of quantification

• Other higher-order logics allow quantification over even higher types than second-order logic permits– These higher types include relations between relations, functions from

relations to relations between relations, and other higher-type objects

• Ordinary first-order interpretations have a single domain of discourse over which all quantifiers range; many-sorted first-order logic allows variables to have different sorts, which have different domains


Extensions (cont’)

• Intuitionistic first-order logic uses intuitionistic rather than classical propositional calculus; for example, ¬¬φ need not be equivalent to φ; similarly, first-order fuzzy logics are first-order extensions of propositional fuzzy logics rather than classical logic

• Infinitary logic allows infinitely long sentences; for example, one may allow a conjunction or disjunction of infinitely many formulas, or quantification over infinitely many variables

• First-order modal logic has extra modal operators with meanings which can be characterised informally as, for example "it is necessary that φ" and "it is possible that φ“


SUMMARY

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Summary

• Predicate logic differentiates from propositional logic by its use of quantifiers– Each interpretation of predicate logic includes a domain of

discourse over which the quantifiers range.

• There are many deductive systems for predicate logic that are sound (only deriving correct results) and complete (able to derive any logically valid implication)

• The logical consequence relation in predicate logic is only semidecidable

• This lecture focused on three core aspects of the propositional logic: Syntax, Semantics, and Reasoning


REFERENCES

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References

• Mathematical Logic for Computer Science (2nd edition) by Mordechai Ben-Ari– http://www.springer.com/computer/foundations/book/978-1-8523

3-319-5

• Propositional Logic at Stanford Encyclopedia of Philosophy– http://plato.stanford.edu/entries/logic-classical/

• First-order logic on Wikipedia– http://en.wikipedia.org/wiki/First-order_logic

• Many online resources– http://www.google.com/search?q=predicate+logic – http://www.google.com/search?hl=en&q=First-order+logic


Next Lecture

# Date Title

1 Introduction

2 Propositional Logic

3 Predicate Logic

4 Theorem Proving, Logic Programming, and Description Logics

5 Search methods

6 CommonKADS

7 Problem-Solving Methods

8 Planning

9 Agents

10 Rule learning

11 Inductive Logic Programming

12 Formal Concept Analysis

13 Neural Networks

14 Semantic Web and Exam Preparation


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