Undecidability Results Wolfgang Thomas Francqui Lecture, Mons, - - PowerPoint PPT Presentation

Undecidability Results

Wolfgang Thomas Francqui Lecture, Mons, April 2013

Fighting the Undecidable

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Overview

• 1. Undecidability?
• 2. The grid
• 3. Defining addition and multiplication
• 4. Undecidability in weak arithmetics
• 5. Conclusion

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Undecidability?

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Example: Hilbert’s 10th Problem (1900)

Given a Diophantine equation with any number of unknowns and with rational integral numerical coefficients: To devise a process (“Verfahren”) according to which it can be determined in a finite number of operations whether the equation is solvable in rational integers.

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Axel Thue (1863-1922)

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The “First Tree”

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Thue’s Problem (1910)

Given two terms s, t and a set of xioms in the form of equations u(x1, . . . , xn) = v(x1, . . . , xn) decide whether from s one can obtain t in finitely many steps by applications of axioms. Thue’s suspicion: Eine L¨

• sung dieser Aufgabe im allgemeinsten Falle d¨

urfte vielleicht mit un¨ uberwindlichen Schwierigkeiten verbunden sein. (A solution of this problem in the general case might perhaps be connected with insurmountable difficulties.)

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The Grid

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The Infinite Grid

The infinite grid is the structure

G2 = (N × N, (0, 0), S1, S2)

where S1(i, j) = (i + 1, j),

S2(i, j) = (i, j + 1)

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Undecidability of Monadic Grid-Theory

The monadic second-order theory of the infinite grid is undecidable. Proof by reduction of the halting problem for Turing machines: For any TM M construct a sentence ϕM of the monadic second-order language of G2 such that

M halts when started on the empty tape iff G2 |

= ϕM.

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Configurations of M

Assume that M works on a left-bounded tape. A halting computation of M can be coded by a finite sequence

• f configuration words

C0, C1, . . . , Cm.

We can arrange the configurations row by row in a right-infinite rectangular array:

q0 a0 a0 a0 a0 a0 a0 . . . a1 q1 a0 a0 a0 a0 a0 . . . q0 a1 a2 a0 a0 a0 a0 . . . a3 q2 a2 a0 a0 a0 a0 . . .

etc.

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Describing an M-Run

The sentence ϕM will express over G2 the existence of such an array of configurations.

a0, . . . , an are the tape symbols (a0 is the blank) q0, . . . , qk are the states of M, special halting state qs

We use set variables X0, . . . , Xn, Y0, . . . , Yk

Xi collects the grid positions where ai occurs, Yi collects the grid positions where state qi occurs. ϕM :

∃X0, . . . , Xn, Y0, . . . , Yk (Partition(X0, . . . , Yk) ∧ “the first row is the initial M-configuration” ∧ “a successor row is the successor configuration of the

preceding one”

∧ “at some position the halting state is reached”)

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A Hidden Grid

Consider the expansion of the tree T2 by the two first-letter-adding functions:

p0(w) = 0 · w, p1(w) = 1 · w

The MSO-theory of (T2, p0, p1) is undecidable. Proof: Define the grid on the domain 0∗1∗.

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Another Hidden Grid

Consider the binary tree with Equal-Level Predicate E

E(u, v) :⇔

|u| = |v|

Obtain (T2, E). The MSO-theory of (T2, E) is undecidable. Proof: Use E to define again the grid 0∗1∗.

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Path Logic over the Grid

In path logic we have first-order quantifiers and set quantifiers ranging only over paths. The finite-path theory of G2 is undecidable.

[W. Th. Path logics with synchronization, in K. Lodaya et al., Perspectives in Concurrency Theory, IARCS, Universities Press, India, 2009]

Idea: Transform 2-counter machine M into a finite-path sentence ϕM such that

M stops when started with counters (0, 0) iff G2 |

= ϕM

M-configuration:

(instruction label, value of counter 1, value of counter 2)

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A 2-Counter Machine M

• 1. if X2 = 0 goto 5
• 2. decr(X2)
• 3. incr(X1)
• 4. goto 1
• 5. stop

Configurations: (1, 3, 2), (2, 3, 2), (3, 3, 1), (4, 4, 1), . . . , , (5, 5, 0)

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Desribing an M-Configuration over G2

We use three paths Y, X1, X2

❄ ✲ ✲ ✲ ✲ ✲ ✲ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ✲ ✲ ✲ ❄

Sm Sℓ Sn

Coding configuration (ℓ, m, n) = (4, 2, 5).

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Update of Configuration

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An Intermediate Summary

MTh(T2) is decidable MTh(T2, E) is undecidable MTh(G2) is undecidable. PathTh(G2) is undecidable. We now show: ChainTh(T2, E) is decidable.

[W. Th., Infinite trees and automaton definable relations over ω-words, TCS 103 (1992)]

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Back to Tree with Equal-Level Predicate

We consider a “path logic” over T2, or even over any regular tree equipped with the equal-level predicate. We call chain logic the fragment of MSO logic where all set quantifications are restricted to subsets of paths (“chains”).

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Chain Logic over Regular Trees

The chain theory of a regular (binary) tree with equal level predicate is decidable. Idea: Reduction to the MSO-theory of (N, +1) Code a chain C in (T2, E, P) by a pair (αC, βC) of ω-words over {0, 1}:

αC is the sequence d0d1d2 . . . of “directions” βC(i) = 1 iff d0 . . . di−1 ∈ C

A third sequence γC signals membership of the reached vertices in P This result gives decidability of CTL∗-model-checking even when the “synchronization” via E is added.

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Defining Addition and Multiplication

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Quantification over Binary Relations

By the results of G¨

• del, Tarski, Turing we know:

The first-order theory of (N, +, ·, 0, 1) is undecidable. Already G¨

• del remarked in 1931:

In the second-order language (with quantifiers over elements and relations) one can define define + and · in (N, +1). Consequence: The second-order theory of (N + 1) is undecidable.

x + y = z

iff

∀R([R(0, x) ∧ ∀s, t(R(s, t) → R(s + 1, t + 1))] → R(y, z))

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Adding Double Function to (N, +1)

double(x) := 2x.

Robinson 1958: The (weak) MSO-theory of (N, +1, double) is undecidable. We follow a proof idea of Elgot and Rabin [JSL 31 (1966)]. Code a relation R = {(m1, n1), . . . , (mk, nk)} by a set MR = {m′

1 < n′ 1 < . . . < m′ k < n′ k}

For each n we need an infinite set of code numbers. Take as codes of n all numbers 2i · (double(n) + 1)

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Example

R = {(2, 1), (0, 2)}

A code set MR contains

1 · 5 < 2 · 3 < 8 · 1 < 2 · 5

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A Remark

There is an MSO-formula OddPos(X, x) that expresses

X(x)

in the <-listing of X-elements, x occurs on an odd position. Use ψ(X, z, z′) :

X(z) ∧ X(z′)

∧ there is precisely one y between z, z′ with X(y)

OddPos(X, x) : ψ∗(X, min(X), x) Next(X, x, y) says “in X, y is the next element after x

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Definability of Decoding

Let ϕ2(z, z′) :=

double(z) = z′

Then “s is a code of x”: ∃y(double(x) + 1 = y ∧ϕ∗

2(y, s))

Translation of ∃R(R(x, y) . . .):

∃X(∃s∃t(s is code of x ∧ t is code of y ∧OddPos(X, s) ∧ Next(X, s, t))

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A Sharper Result

Let f : N → N be strictly increasing,

f − idN be monotone and unbounded.

Then MTh(N, +1, 0, f) is undecidable.

[W. Th., A note on undecidable extensions of monadic second order arithmetic, Arch math. Logik 17 (1975)]

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Undecidability of Weak Arithmetics

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Successor Structure + Unary Predicate

Consider (N, +1, P)

χP is the characteristic function of P χP = 0 0 1 1 0 1 0 1 0 0 . . .

Consequence of B¨ uchi’s analysis of MTh(N, +1): For each monadic formula ϕ(X) one can construct a B¨ uchi (or Muller) automaton Aϕ such that

(N, +1) | = ϕ[P]

iff Aϕ accepts χP. Acceptance Problem Acc(P): Given a B¨ uchi autoamaton A, does A accept χP? Then MTh(N, +1, P) is decidable iff Acc(P) is decidable.

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The Prime Predicate P

Can we decide for any B¨ uchi automaton A whether

A accepts χP = 0 0 1 1 0 1 0 1 . . .?

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Prime Numbers

Decidability of MTh(N, +1, P) (and even of FOTh(N, +1, <, P)) is open. Twin prime hypothesis TPH:

∀x∃y

• x < y ∧ P(y) ∧ P(y + 1 + 1)
• Dirchlet’s Theorem:

Let Am,n := {m + i · n | i ≥ 0} If m, n are relatively prime, then |Am,n ∩ P| = ∞ For fixed m, n, this claim is expressible in MTh(N, +1, P)

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More on Arithmetical Progressions

An arithmetic progression of length k in P is a sequence

m, m + d, . . . , m + (k − 1) · d

• f successive prime numbers
• B. Green, T. Tao (2006):

For each k there are infinitely many arithmetical progressions

• f length k in P.

Illustration (Frind, Underwood, Jobling (2004)):

m = 56211383760397, d = 44546738095860, k = 22

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Undecidability: An Example

There is a recursive set P ⊆ N such that FOTh(N, +1, P) is undecidable.

• Proof. Let M be an enumerable but undecidable set with

enumeration m0, m1, m2, . . .. Consider the ω-word

10m010m110m2 · · ·

Let P be the associated set. It is recursive. Given m let

ϕm : ∃x

• Px ∧ ¬P(x + 1) ∧ ¬P(x + 2) ∧ . . . ∧ P(x + m + 1)
• Then

m ∈ M ⇔ (N, +1, P) |

= ϕm

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Classifying Undecidability

We identify sentences with natural numbers. A theory is then coded by a set of natural numbers. The undecidable sets are classified in the arithmetical hierarchy: A set A belongs to the class Σ0

n iff

for some decidable relation R:

x ∈ A ⇔ ∃y1∀y2 . . . ∃/∀ynR(x, y1, . . . , yn) Π0

n contains the complements of the Σ0 n-sets.

The Σ0

1-sets are the recursively enumerable ones.

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Complexity of MTh(N, +1, P)

If P is recursive, then MTh(N, +1, P) is on level Σ0

3 ∩ Π0 3 of

the arithmetical hierarchy. Consider Muller automaton A = (Q, {0, 1}, q0, δ, F)

A accepts χP ⇔

• F∈F

(

q∈F

∃ωi δ(q0, χP[0, i]) = q ∧

q∈F

∃<ωi δ(q0, χP[0, i]) = q)

This is a Boolean combination of Σ2-conditions. So {A | A accepts χP} ∈ Σ3 ∩ Π3 Consequence: If P is recursive, then in MTh(N, +1, P)

+ and · are not definable.

(So MTh(N, +1) is a “weak arithmetic”.)

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Expanding T2 by a Predicate

For recursive P ⊆ {0, 1}∗, the theory MTh(S2, P) belongs to the class ∆1

2, and there is a recursive P ⊆ {0, 1}∗ such that

MT(S2, P) is Π1

1-hard.

One constructs a recursive P such that a known Π1

1-complete

set is reducible to MT(S2, P). As Π1

1-complete set use a coding of finite-path trees. [W. Th., On monadic theories of monadic predicates, LNCS 6300 (2010)]

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(Q, <) and (R, <)

An exercise: MTh(Q, <) is decidable. For a hint see Rabin’s landmark paper of 1969

M.O. Rabin, Decidability of second-order theories and automata on infinite trees, Trans. AMS 141 (1969)

Much more than an exercise: MTh(R, <) is undecidable. For a condensed hint see the last 10 pages of Shelah’s landmark paper of 1975

• S. Shelah, The monadic theory of order, Ann. Math. 102 (1975)

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