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Trakhtenbrot's theorem

In logic, finite model theory, and computability theory, Trakhtenbrot's theorem (due to Boris Trakhtenbrot) states that the problem of validity in first-order logic on the class of all finite models is undecidable. In fact, the class of valid sentences over finite models is not recursively enumerable (though it is co-recursively enumerable).

Trakhtenbrot's theorem implies that Gödel's completeness theorem (that is fundamental to first-order logic) does not hold in the finite case. Also it seems counter-intuitive that being valid over all structures is 'easier' than over just the finite ones.

The theorem was first published in 1950: "The Impossibility of an Algorithm for the Decidability Problem on Finite Classes".[1]

Mathematical formulation edit

We follow the formulations as in Ebbinghaus and Flum[2]

Theorem edit

Satisfiability for finite structures is not decidable in first-order logic.
That is, the set {φ | φ is a sentence of first-order logic that is satisfied by all finite structures} is undecidable.

Corollary edit

Let σ be a relational vocabulary with one at least binary relation symbol.

The set of σ-sentences valid in all finite structures is not recursively enumerable.

Remarks

  1. This implies that Gödel's completeness theorem fails in the finite since completeness implies recursive enumerability.
  2. It follows that there is no recursive function f such that: if φ has a finite model, then it has a model of size at most f(φ). In other words, there is no effective analogue to the Löwenheim–Skolem theorem in the finite.

Intuitive proof edit

This proof is taken from Chapter 10, section 4, 5 of Mathematical Logic by H.-D. Ebbinghaus.

As in the most common proof of Gödel's First Incompleteness Theorem through using the undecidability of the halting problem, for each Turing machine   there is a corresponding arithmetical sentence  , effectively derivable from  , such that it is true if and only if   halts on the empty tape. Intuitively,   asserts "there exists a natural number that is the Gödel code for the computation record of   on the empty tape that ends with halting".

If the machine   does halt in finite steps, then the complete computation record is also finite, then there is a finite initial segment of the natural numbers such that the arithmetical sentence   is also true on this initial segment. Intuitively, this is because in this case, proving   requires the arithmetic properties of only finitely many numbers.

If the machine   does not halt in finite steps, then   is false in any finite model, since there's no finite computation record of   that ends with halting.

Thus, if   halts,   is true in some finite models. If   does not halt,   is false in all finite models. So,   does not halt if and only if   is true over all finite models.

The set of machines that does not halt is not recursively enumerable, so the set of valid sentences over finite models is not recursively enumerable.

Alternative proof edit

In this section we exhibit a more rigorous proof from Libkin.[3] Note in the above statement that the corollary also entails the theorem, and this is the direction we prove here.

Theorem

For every relational vocabulary τ with at least one binary relation symbol, it is undecidable whether a sentence φ of vocabulary τ is finitely satisfiable.

Proof

According to the previous lemma, we can in fact use finitely many binary relation symbols. The idea of the proof is similar to the proof of Fagin's theorem, and we encode Turing machines in first-order logic. What we want to prove is that for every Turing machine M we construct a sentence φM of vocabulary τ such that φM is finitely satisfiable if and only if M halts on the empty input, which is equivalent to the halting problem and therefore undecidable.

Let M= ⟨Q, Σ, δ, q0, Qa, Qr⟩ be a deterministic Turing machine with a single infinite tape.

  • Q is the set of states,
  • Σ is the input alphabet,
  • Δ is the tape alphabet,
  • δ is the transition function,
  • q0 is the initial state,
  • Qa and Qr are the sets of accepting and rejecting states.

Since we are dealing with the problem of halting on an empty input we may assume w.l.o.g. that Δ={0,1} and that 0 represents a blank, while 1 represents some tape symbol. We define τ so that we can represent computations:

τ := {<, min, T0 (⋅,⋅), T1 (⋅,⋅), (Hq(⋅,⋅))(q ∈ Q)}

Where:

  • < is a linear order and min is a constant symbol for the minimal element with respect to < (our finite domain will be associated with an initial segment of the natural numbers).
  • T0 and T1 are tape predicates. Ti(s,t) indicates that position s at time t contains i, where i ∈ {0,1}.
  • Hq's are head predicates. Hq(s,t) indicates that at time t the machine is in state q, and its head is in position s.

The sentence φM states that (i) <, min, Ti's and Hq's are interpreted as above and (ii) that the machine eventually halts. The halting condition is equivalent to saying that Hq∗(s, t) holds for some s, t and q∗ ∈ Qa ∪ Qr and after that state, the configuration of the machine does not change. Configurations of a halting machine (the nonhalting is not finite) can be represented as a τ (finite) sentence (more precisely, a finite τ-structure which satisfies the sentence). The sentence φM is: φ ≡ α ∧ β ∧ γ ∧ η ∧ ζ ∧ θ.

We break it down by components:

  • α states that < is a linear order and that min is its minimal element
  • γ defines the initial configuration of M: it is in state q0, the head is in the first position and the tape contains only zeros: γ ≡ Hq0(min,min) ∧ ∀s T0 (s, min)
  • η states that in every configuration of M, each tape cell contains exactly one element of Δ: ∀s∀t(T0(s, t) ↔ ¬ T1(s, t))
  • β imposes a basic consistency condition on the predicates Hq's: at any time the machine is in exactly one state:
 
  • ζ states that at some point M is in a halting state:
 
  • θ consists of a conjunction of sentences stating that Ti's and Hq's are well behaved with respect to the transitions of M. As an example, let δ(q,0)=(q',1, left) meaning that if M is in state q reading 0, then it writes 1, moves the head one position to the left and goes into the state q'. We represent this condition by the disjunction of θ0 and θ1:
 

Where θ2 is:

 

And:

 

Where θ3 is:

 

s-1 and t+1 are first-order definable abbreviations for the predecessor and successor according to the ordering <. The sentence θ0 assures that the tape content in position s changes from 0 to 1, the state changes from q to q', the rest of the tape remains the same and that the head moves to s-1 (i. e. one position to the left), assuming s is not the first position in the tape. If it is, then all is handled by θ1: everything is the same, except the head does not move to the left but stays put.

If φM has a finite model, then such a model that represents a computation of M (that starts with the empty tape (i.e. tape containing all zeros) and ends in a halting state). If M halts on the empty input, then the set of all configurations of the halting computations of M (coded with <, Ti's and Hq's) is a model of φM, which is finite, since the set of all configurations of halting computations is finite. It follows that M halts on the empty input iff φM has a finite model. Since halting on the empty input is undecidable, so is the question of whether φM has a finite model   (equivalently, whether φM is finitely satisfiable) is also undecidable (recursively enumerable, but not recursive). This concludes the proof.

Corollary

The set of finitely satisfiable sentences is recursively enumerable.

Proof

Enumerate all pairs   where   is finite and  .

Corollary

For any vocabulary containing at least one binary relation symbol, the set of all finitely valid sentences is not recursively enumerable.

Proof

From the previous lemma, the set of finitely satisfiable sentences is recursively enumerable. Assume that the set of all finitely valid sentences is recursively enumerable. Since ¬φ is finitely valid iff φ is not finitely satisfiable, we conclude that the set of sentences which are not finitely satisfiable is recursively enumerable. If both a set A and its complement are recursively enumerable, then A is recursive. It follows that the set of finitely satisfiable sentences is recursive, which contradicts Trakhtenbrot's theorem.

References edit

  1. ^ Trakhtenbrot, Boris (1950). "The Impossibility of an Algorithm for the Decidability Problem on Finite Classes". Proceedings of the USSR Academy of Sciences (in Russian). 70 (4): 569–572.
  2. ^ Ebbinghaus, Heinz-Dieter; Flum, Jörg (1995). Finite Model Theory. Springer Science+Business Media. ISBN 978-3-540-60149-4.
  3. ^ Libkin, Leonid (2010). Elements of Finite Model Theory. Texts in Theoretical Computer Science. ISBN 978-3-642-05948-3.
  • Boolos, Burgess, Jeffrey. Computability and Logic, Cambridge University Press, 2002.
  • Simpson, S. "Theorems of Church and Trakhtenbrot". 2001.[1]

trakhtenbrot, theorem, logic, finite, model, theory, computability, theory, boris, trakhtenbrot, states, that, problem, validity, first, order, logic, class, finite, models, undecidable, fact, class, valid, sentences, over, finite, models, recursively, enumera. In logic finite model theory and computability theory Trakhtenbrot s theorem due to Boris Trakhtenbrot states that the problem of validity in first order logic on the class of all finite models is undecidable In fact the class of valid sentences over finite models is not recursively enumerable though it is co recursively enumerable Trakhtenbrot s theorem implies that Godel s completeness theorem that is fundamental to first order logic does not hold in the finite case Also it seems counter intuitive that being valid over all structures is easier than over just the finite ones The theorem was first published in 1950 The Impossibility of an Algorithm for the Decidability Problem on Finite Classes 1 Contents 1 Mathematical formulation 1 1 Theorem 1 2 Corollary 2 Intuitive proof 3 Alternative proof 4 ReferencesMathematical formulation editWe follow the formulations as in Ebbinghaus and Flum 2 Theorem edit Satisfiability for finite structures is not decidable in first order logic That is the set f f is a sentence of first order logic that is satisfied by all finite structures is undecidable Corollary edit Let s be a relational vocabulary with one at least binary relation symbol The set of s sentences valid in all finite structures is not recursively enumerable Remarks This implies that Godel s completeness theorem fails in the finite since completeness implies recursive enumerability It follows that there is no recursive function f such that if f has a finite model then it has a model of size at most f f In other words there is no effective analogue to the Lowenheim Skolem theorem in the finite Intuitive proof editThis proof is taken from Chapter 10 section 4 5 of Mathematical Logic by H D Ebbinghaus As in the most common proof of Godel s First Incompleteness Theorem through using the undecidability of the halting problem for each Turing machine M displaystyle M nbsp there is a corresponding arithmetical sentence ϕM displaystyle phi M nbsp effectively derivable from M displaystyle M nbsp such that it is true if and only if M displaystyle M nbsp halts on the empty tape Intuitively ϕM displaystyle phi M nbsp asserts there exists a natural number that is the Godel code for the computation record of M displaystyle M nbsp on the empty tape that ends with halting If the machine M displaystyle M nbsp does halt in finite steps then the complete computation record is also finite then there is a finite initial segment of the natural numbers such that the arithmetical sentence ϕM displaystyle phi M nbsp is also true on this initial segment Intuitively this is because in this case proving ϕM displaystyle phi M nbsp requires the arithmetic properties of only finitely many numbers If the machine M displaystyle M nbsp does not halt in finite steps then ϕM displaystyle phi M nbsp is false in any finite model since there s no finite computation record of M displaystyle M nbsp that ends with halting Thus if M displaystyle M nbsp halts ϕM displaystyle phi M nbsp is true in some finite models If M displaystyle M nbsp does not halt ϕM displaystyle phi M nbsp is false in all finite models So M displaystyle M nbsp does not halt if and only if ϕM displaystyle neg phi M nbsp is true over all finite models The set of machines that does not halt is not recursively enumerable so the set of valid sentences over finite models is not recursively enumerable Alternative proof editIn this section we exhibit a more rigorous proof from Libkin 3 Note in the above statement that the corollary also entails the theorem and this is the direction we prove here Theorem For every relational vocabulary t with at least one binary relation symbol it is undecidable whether a sentence f of vocabulary t is finitely satisfiable ProofAccording to the previous lemma we can in fact use finitely many binary relation symbols The idea of the proof is similar to the proof of Fagin s theorem and we encode Turing machines in first order logic What we want to prove is that for every Turing machine M we construct a sentence fM of vocabulary t such that fM is finitely satisfiable if and only if M halts on the empty input which is equivalent to the halting problem and therefore undecidable Let M Q S d q0 Qa Qr be a deterministic Turing machine with a single infinite tape Q is the set of states S is the input alphabet D is the tape alphabet d is the transition function q0 is the initial state Qa and Qr are the sets of accepting and rejecting states Since we are dealing with the problem of halting on an empty input we may assume w l o g that D 0 1 and that 0 represents a blank while 1 represents some tape symbol We define t so that we can represent computations t lt min T0 T1 Hq q Q Where lt is a linear order and min is a constant symbol for the minimal element with respect to lt our finite domain will be associated with an initial segment of the natural numbers T0 and T1 are tape predicates Ti s t indicates that position s at time t contains i where i 0 1 Hq s are head predicates Hq s t indicates that at time t the machine is in state q and its head is in position s The sentence fM states that i lt min Ti s and Hq s are interpreted as above and ii that the machine eventually halts The halting condition is equivalent to saying that Hq s t holds for some s t and q Qa Qr and after that state the configuration of the machine does not change Configurations of a halting machine the nonhalting is not finite can be represented as a t finite sentence more precisely a finite t structure which satisfies the sentence The sentence fM is f a b g h z 8 We break it down by components a states that lt is a linear order and that min is its minimal element g defines the initial configuration of M it is in state q0 the head is in the first position and the tape contains only zeros g Hq0 min min s T0 s min h states that in every configuration of M each tape cell contains exactly one element of D s t T0 s t T1 s t b imposes a basic consistency condition on the predicates Hq s at any time the machine is in exactly one state t s q QHq s t s t q q Q q qHq s t Hq s t displaystyle forall t exists s bigvee q in Q H q s t land neg exists s exists t bigvee q q in Q q neq q H q s t land H q s t nbsp z states that at some point M is in a halting state s t q Qa QrHq s t displaystyle exists s exists t bigvee q in Q a cup Q r H q s t nbsp 8 consists of a conjunction of sentences stating that Ti s and Hq s are well behaved with respect to the transitions of M As an example let d q 0 q 1 left meaning that if M is in state q reading 0 then it writes 1 moves the head one position to the left and goes into the state q We represent this condition by the disjunction of 80 and 81 80 s t s min T0 s t Hq s t 82 displaystyle theta 0 equiv forall s forall t s neq underline min land T 0 s t land H q s t to theta 2 nbsp Where 82 is T1 s t 1 Hq s 1 t 1 s s s i 0 1Ti s t 1 Ti s t displaystyle T 1 s t 1 land H q s 1 t 1 land forall s s neq s to bigwedge i 0 1 T i s t 1 leftrightarrow T i s t nbsp And 81 s t s min T0 s t Hq s t 83 displaystyle theta 1 equiv forall s forall t s underline min land T 0 s t land H q s t to theta 3 nbsp Where 83 is T1 s t 1 Hq s t 1 s s s i 0 1Ti s t 1 Ti s t displaystyle T 1 s t 1 land H q s t 1 land forall s s neq s to bigwedge i 0 1 T i s t 1 leftrightarrow T i s t nbsp s 1 and t 1 are first order definable abbreviations for the predecessor and successor according to the ordering lt The sentence 80 assures that the tape content in position s changes from 0 to 1 the state changes from q to q the rest of the tape remains the same and that the head moves to s 1 i e one position to the left assuming s is not the first position in the tape If it is then all is handled by 81 everything is the same except the head does not move to the left but stays put If fM has a finite model then such a model that represents a computation of M that starts with the empty tape i e tape containing all zeros and ends in a halting state If M halts on the empty input then the set of all configurations of the halting computations of M coded with lt Ti s and Hq s is a model of fM which is finite since the set of all configurations of halting computations is finite It follows that M halts on the empty input iff fM has a finite model Since halting on the empty input is undecidable so is the question of whether fM has a finite model A displaystyle mathcal A nbsp equivalently whether fM is finitely satisfiable is also undecidable recursively enumerable but not recursive This concludes the proof Corollary The set of finitely satisfiable sentences is recursively enumerable ProofEnumerate all pairs A ϕ displaystyle mathcal A phi nbsp where A displaystyle mathcal A nbsp is finite and A ϕ displaystyle mathcal A models phi nbsp Corollary For any vocabulary containing at least one binary relation symbol the set of all finitely valid sentences is not recursively enumerable ProofFrom the previous lemma the set of finitely satisfiable sentences is recursively enumerable Assume that the set of all finitely valid sentences is recursively enumerable Since f is finitely valid iff f is not finitely satisfiable we conclude that the set of sentences which are not finitely satisfiable is recursively enumerable If both a set A and its complement are recursively enumerable then A is recursive It follows that the set of finitely satisfiable sentences is recursive which contradicts Trakhtenbrot s theorem References edit Trakhtenbrot Boris 1950 The Impossibility of an Algorithm for the Decidability Problem on Finite Classes Proceedings of the USSR Academy of Sciences in Russian 70 4 569 572 Ebbinghaus Heinz Dieter Flum Jorg 1995 Finite Model Theory Springer Science Business Media ISBN 978 3 540 60149 4 Libkin Leonid 2010 Elements of Finite Model Theory Texts in Theoretical Computer Science ISBN 978 3 642 05948 3 Boolos Burgess Jeffrey Computability and Logic Cambridge University Press 2002 Simpson S Theorems of Church and Trakhtenbrot 2001 1 Retrieved from https en wikipedia org w index php title Trakhtenbrot 27s theorem amp oldid 1202711952, wikipedia, wiki, book, books, library,

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