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CS621: Artificial Intelligence. Pushpak Bhattacharyya CSE Dept., IIT Bombay Lecture–10: Soundness of Propositional Calculus 12 th August, 2010. Soundness, Completeness & Consistency. Soundness. Semantic World ---------- Valuation, Tautology. Syntactic World ---------- Theorems,

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cs621 artificial intelligence

CS621: Artificial Intelligence

Pushpak BhattacharyyaCSE Dept., IIT Bombay

Lecture–10: Soundness of Propositional Calculus

12th August, 2010

soundness completeness consistency
Soundness, Completeness &Consistency

Soundness

Semantic

World

----------

Valuation,

Tautology

Syntactic

World

----------

Theorems,

Proofs

Completeness

*

*

slide3
Soundness
      • Provability Truth
  • Completeness
      • Truth Provability
slide4
Soundness:Correctness of the System
      • Proved entities are indeed true/valid
  • Completeness:Power of the System
      • True things are indeed provable
slide5

TRUE

Expressions

Outside

Knowledge

System

Validation

consistency
Consistency

The System should not be able to

prove both P and ~P, i.e., should not be

able to derive

F

slide7
Examine the relation between

Soundness

&

Consistency

Soundness Consistency

slide8
If a System is inconsistent, i.e., can derive

F , it can prove any expression to be a

theorem. Because

F P is a theorem

inconsistency unsoundness
InconsistencyUnsoundness

To show that

FP is a theorem

Observe that

F, PF ⊢ F By D.T.

F ⊢ (PF)F; A3

⊢ P

i.e. ⊢ FP

Thus, inconsistency implies unsoundness

unsoundness inconsistency
UnsoundnessInconsistency
  • Suppose we make the Hilbert System of propositional calculus unsound by introducing (A /\ B) as an axiom
  • Now AND can be written as
    • (A(BF ))F
  • If we assign F to A, we have
    • (F (BF )) F
    • But (F (BF )) is an axiom (A1)
    • Hence F is derived
slide11
Inconsistency is a Serious issue.

Informal Statement of Godel Theorem:

If a sufficiently powerful system is complete it is inconsistent.

Sufficiently powerful: Can capture at least Peano Arithmetic

introduce semantics in propositional logic
Introduce Semantics in Propositional logic

Valuation Function V

Definition of V

V(F ) = F

Where F is called ‘false’ and is one of the two symbols (T, F)

Syntactic ‘false

Semantic ‘false’

slide13
V(F) = F

V(AB) is defined through what is called the truth table

V(A) V(B) V(AB)

T F F

T T T

F F T

F T T

tautology
Tautology

An expression ‘E’ is a tautology if

V(E) = T

for all valuations of constituent propositions

Each ‘valuation’ is called a ‘model’.

slide15
To see that

(FP) is a tautology

two models

V(P) = T

V(P) = F

V(FP) = T for both

slide16
FP is a theorem

FP is a tautology

Soundness

Completeness

slide17
If a system is Sound & Complete, it does not

matter how you “Prove” or “show the validity”

Take the Syntactic Path or the Semantic Path

syntax vs semantics issue
Syntax vs. Semantics issue

Refers to

FORM VS. CONTENT

Tea

(Content)

Form

form content
Form & Content

Godel, Escher, Bach

By D. Hofstadter

painter

musician

logician

problem
Problem

(P Q)(P Q)

Semantic Proof

A B

P Q P Q P Q AB

T F F T T

T T T T T

F F F F T

F T F T T

slide21
To show syntactically

(P Q) (P Q)

i.e.

[(P (Q F )) F ]

[(P F ) Q]

slide22
If we can establish

(P (Q F )) F ,

(P F ), Q F ⊢ F

This is shown as

Q F hypothesis

(Q F ) (P (Q F)) A1

slide23
QF; hypothesis

(QF)(P(QF)); A1

P(QF); MP

F; MP

Thus we have a proof of the line we started with

soundness proof
Soundness Proof

Hilbert Formalization of Propositional

Calculus is sound.

“Whatever is provable is valid”

slide25
Statement

Given

A1, A2, … ,An|- B

V(B) is ‘T’ for all Vs for which V(Ai) = T

slide26
Proof

Case 1 B is an axiom

V(B) = T by actual observation

Statement is correct

slide27
Case 2 B is one of Ais

if V(Ai) = T, so is V(B)

statement is correct

slide28
Case 3 B is the result of MP on Ei & Ej

Ejis Ei B

Suppose V(B) = F

Then either V(Ei) = F or V(Ej) = F

.

.

.

Ei

.

.

.

Ej

.

.

.

B

slide29
i.e. Ei/Ej is result of MP of two expressions coming before them

Thus we progressively deal with shorter and shorter proof body.

Ultimately we hit an axiom/hypothesis.

Hence V(B) = T

Soundness proved

tourist in a country of truth sayers and liers
Tourist in a country of truth-sayers and liers
  • Facts and Rules: In a certain country, people either always speak the truth or always lie. A tourist T comes to a junction in the country and finds an inhabitant S of the country standing there. One of the roads at the junction leads to the capital of the country and the other does not. S can be asked only yes/no questions.
  • Question: What single yes/no question can T ask of S, so that the direction of the capital is revealed?
diagrammatic representation
Diagrammatic representation

Capital

S (either always says the truth

Or always lies)

T (tourist)

deciding the propositions a very difficult step needs human intelligence
Deciding the Propositions: a very difficult step- needs human intelligence
  • P: Left road leads to capital
  • Q: S always speaks the truth
meta question what question should the tourist ask
Meta Question: What question should the tourist ask
  • The form of the question
  • Very difficult: needs human intelligence
  • The tourist should ask
    • Is R true?
    • The answer is “yes” if and only if the left road leads to the capital
    • The structure of R to be found as a function of P and Q
get form of r quite mechanical
Get form of R: quite mechanical
  • From the truth table
    • R is of the form (P x-nor Q) or (P ≡ Q)
get r in english hindi hebrew
Get R in English/Hindi/Hebrew…
  • Natural Language Generation: non-trivial
  • The question the tourist will ask is
    • Is it true that the left road leads to the capital if and only if you speak the truth?
  • Exercise: A more well known form of this question asked by the tourist uses the X-OR operator instead of the X-Nor. What changes do you have to incorporate to the solution, to get that answer?