NKU CSC 685 Kirby

First-Order Logic and Undecidability

A decision ("yes/no") problem is decidable if there is an algorithm that takes any problem instance and returns the correct true/false answer.

A decision problem is semi-decidable if there is an algorithm that takes any problem instance and returns true if and only if the correct answer is "true".

von Neumann Machine

Multitape TM

NondeterministicTM

Random AccessMachine

procedural pseudocode

your favorite algorithmic formalism

Church-Turing thesis

Turing Machine easy to prove stuff here

easy to do stuff here

Finite Automata

Context-Free Grammar

This sentence is false.

Some self-referential sentences make statementsthat are perfectly acceptable, like this one.

Do we "reject" self-reference?

Deep Computer Science Idea:Programs are strings.Programs can process strings.So: Programs can process programs.

>> program= "print sqrt(81)">> strlen( program ) 14>> execute( program )9>> execute( "for (;;)" )^C interrupted by user

Self-reference is inescapable in computer science.

bool halts( string pgm, string w ){ // pgm is the source code of a void function that takes one string parameter. // Return true if pgm halts on input w, false otherwise}

bool halts( string pgm, string w ){ // pgm is the source code of a void function that takes one string parameter. // Return true if pgm halts on input w, false otherwise}

Proof that the Halting Problem is Undecidable - 1

Example: void f( string w ){ if ( w[0]=='b' ) for (;;) ;}

void f( string w ){ if ( w[0]=='b' ) for (;;) ;}

p1

halts( p1, "hi" ) == true halts( p1, "bye") == false halts( p1, p1 ) == true

Suppose it were decidable. Then we'd have the following function:

Proof that the Halting Problem is Undecidable - 2

Of course, programs whose functions call functions are nbd...

Example: void g(){ for ( ;; ) ;}

void f( string w ){ if ( w[0]=='b' ) g() ;}

void g(){ for ( ;; ) ;}

void f( string w ){ if ( w[0]=='b' ) g() ;}

p2

halts( p2, "hi" ) == truehalts( p2, "bye") == falsehalts( p2, p2 ) == true

Proof that the Halting Problem is Undecidable - 3

What happens here?

Example:

p3

halts( p3, p3 ) == ?!!! true false true … contradiction

bool halts( string pgm, string w ){ // ...}void f( string w ){

if ( halts( w, w ) )for( ;; ) ;

}

bool halts( string pgm, string w ){ // ...}void f( string w ){

if ( halts( w, w ) )for( ;; ) ;

}

Conclusion: HP is unsolvable.

Halting Problem Unsolvable⇒Post Correspondence Problem Unsolvable⇒First-Order Logic Validity Problem Unsolvable

First-Order Logic Validity Problem Solvable⇒Post Correspondence Problem Solvable⇒Halting Problem Solvable⇒

Usually we work by proof by contradiction

Post Correspondence ProblemCan copies of a set of dominos be sequenced so that the string on top is the same as the string on the bottom?

1111

1111

1011110

1011110

100

100

Example problem instance:1

111

1111

1111

1111

1011110

1011110

1011110

1011110

100

100

100

100

Example problem instance:10

101

10101

01111

01111

101011

101011

Answer: No

10101

1010110

101

10101

01111

01111

01111

01111

101011

101011101011

101011

Answer: Yes 1011110

1011110

1111

1111

1111

1111

100

100

PCPCan copies of a set of dominos be sequenced so that the string on top is the same as the string on the bottom?

Example problem instance:

Answer: Yes

0010

0010

01011

01011

01101

01101

10001

10001

(but 66 dominos needed!)

PCP:Given {(r1,s1), ... (rn,sn)} where ri,si {0,1}*, does thereexist a sequence {i1...ik} such that ri1ri2...rik = si1si2...sik ?

0010

0010

01011

01011

01101

01101

10001

10001

HP solver

Post Correspondence Problem Solvable⇒Halting Problem Solvable

M,w instance converter

{(r1,s1), ... (rn,sn)}

(hypothetical)PCP solver

yes/no

Ensure:M halts on wiff {(r1,s1), ... (rn,sn)}has a solution

PROBLEM REDUCTION

PCP solver

PCP Solvable⇒First-Order Logic Validity Problem Solvable

instance converter

{ (r1,s1), ...(rn,sn) }

(hypothetical)Validity Detector

yes/no

Ensure:{(r1,s1), ... (rn,sn)}has a solutioniff is a valid wff.

PROBLEM REDUCTION

PCP reduction

PCP instance: 10101

10101

01111

01111

101011

101011

10101

10101gf

gfg

gfgfg

01111

01111fgggg

fgggg

101011

101011gfgfgg

gfgfgg

= ( P(f(g(e)), g(f(g(e)))) P(g(g(f(e))), g(g(e))) P(g(f(g(e))),g(g(f(e)))) vw { Pvw [ P(f(g(v)), g(f(g(w)))) P (g(g(f(v))), g(g(w))) P(g(f(g(v))),g(g(f(w)))) ] } )z Pzz

Wffify!

One constant: eTwo one-place function symbols: f,gOne two-argument predicate: P

Expressing unreachability

u v v is unreachable from u

R

In SECOND-ORDER logic:

unreachable( R,u,v):=Pxyz( Pxx (PxyPyz Pxz) Puv (Rxy Pxy))

Can we express this in first order logic?