Infinite models for propositional calculi
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Infinite Models for Propositional Calculi. Zachary Ernst University of Missouri-Columbia [email protected] The Gist. Finite matrix models are equivalent to finite state bottom-up tree automata. So perhaps, more powerful automata can play the role of infinite matrix models. The Problem.

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Infinite Models for Propositional Calculi

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Infinite models for propositional calculi

Infinite Models for Propositional Calculi

Zachary Ernst

University of Missouri-Columbia

[email protected]


The gist

The Gist

  • Finite matrix models are equivalent to finite state bottom-up tree automata.

  • So perhaps, more powerful automata can play the role of infinite matrix models.


The problem

The Problem

  • Finite matrix models are good for showing that formulae are not theorems of propositional logics.

  • But many systems require infinite models.

  • These are hard to enumerate, and there is no good, flexible framework for describing them.


Another example

Another Example

  • System due to J. Anderson:

    • Cxx

    • CCIxxy (where Ix=Cxx)

    • CCIxyxCCIIxyz

    • Modus Ponens and Universal Substitution

  • Theorems are all of the form:

    • CCIII…Ixxy, for any number of I’s.


A hyperfinite system

A “Hyperfinite” System

  • Anderson’s system is “hyperfinite”:

    • Any finite model that respects modus ponens and uniform substitution validates every formula.

  • This is easy to show, and the proof is informative about the limits of finite models.


The proof

The Proof

Consider the following infinite sequence of theorems:

Ix

IIx

IIIx

IIIIIIx

III…Ix


The proof1

The Proof

If M is some arbitrary finite matrix model…

Ix

IIx

IIIx

IIIIIIx

III…Ix

…then there must be some pair of formulae in the sequence that M “identifies”.


The proof2

The Proof

Ix

IIx

IIIx

IIIIIIx

III…Ix

Suppose M “thinks” that

IIx = IIIIIIx.


The proof3

The Proof

Ix

IIx

IIIx

IIIIIIx

III…Ix

Suppose M “thinks” that

IIx = IIIIIIx.

Then according to M:

CIIIIIIxIIx = Cxx, which is a theorem.


The proof4

The Proof

According to M:

CIIIIIIxIIx = Cxx, which is a theorem.

Now consider:

CCIIIIIIxIIxy=CCIIII(IIx)(IIx)y, which is of the form:

CCIII…IXXY, which is a theorem (where X=IIx).


The proof5

The Proof

According to M:

CIIIIIIxIIx = Cxx, which is a theorem.

Now consider:

CCIIIIIIxIIxy=CCIIII(IIx)(IIx)y, which is of the form:

CCIII…IXXY, which is a theorem (where X=IIx).

So one application of modus ponens yields y.

Therefore, the model must validate everything.


What happened

What Happened?

  • Finite matrix models must “identify” two elements of any sufficiently long list of formulae.

  • So it will incorrectly think that when those formulae are combined, the resulting formula will be equivalent to Cxx.

  • No finite matrix model validates exactly the instances of Cxx (Gödel).

  • If Cxx is a theorem, then the model will validate the formula.


How to use a matrix model

How to Use a Matrix Model

1

2

CpCqq

1

p

2

q

q


How to use a matrix model1

How to Use a Matrix Model

1

2

CpCqq

1

1

1

p

2

1

q

q

2

2


Finite matrices as finite automata

Finite Matrices as Finite Automata

CpCqq

  • Using a finite matrix model is like letting an automaton run over a tree.

  • “Designated Values” are like “Accept States”.

1

1

p

1

q

q

2

2


The disanalogy vocabulary

The Disanalogy -- Vocabulary

  • Finite tree automata have a finite input language.

  • Logics have an infinite language with countably many variables.

  • This matters for models, but not for countermodels.


Restricting the input vocabulary

Restricting the Input Vocabulary

  • Suppose {Cpq, r}s, by condensed detachment, and suppose s has fewer distinct variables than one of the premises.

  • Then there is a substitution  such that:

    • p= r; q=s, and

    • Cpq and r have no variables not appearing in s.

  • Therefore, if P is a set of premises, and there is a proof of C from P, then there is a proof of C from P containing only variables occurring in C.


Restricting the input vocabulary1

Restricting the Input Vocabulary

  • So we know in advance how many variables are necessary for a proof of C from P, if such a proof exists.

  • Thus, we do not need a countermodel containing infinitely many variables; if C has a single variable, then the countermodel is only required have an interpretation for only one variable.

  • So it does not matter that tree automata have a finite input language; they still might serve as countermodels.


A stronger automaton

A Stronger Automaton

  • Weighted Tree Automata use weights from a semiring:

    • Suppose semiring is

    • Every transition has a transition cost from

    • The costs for each successful run are multiplied using the semiring multiplication.

    • The total costs for all runs are added using the semiring addition.

    • The automaton accepts a tree if the cost associated with the tree is in some subset


Are weighted automata strong enough for infinite models

Are Weighted Automata Strong Enough for Infinite Models?

  • For some infinite sequence of formulae, a weighted automaton must be able to assign a different weight to each member of the sequence.

  • It is easy to construct an automaton that calculates the binary value of a tree. In other words, there is an automaton such that


Weighted automata and reflexivity

Weighted Automata and Reflexivity

  • Recall that Gödel showed that no finite model accepts exactly the instances of Cxx.

  • But if the binary value of then there is an automaton such that:

Terminology: We say that A “0-accepts” only the instances of Cxx.


Weighted automata and anderson s hyperfinite system

Weighted Automata and Anderson’s Hyperfinite System

  • Recall that the theorems of Anderson’s system are

  • We can construct an automaton such that:

  • So let


Infinite models for propositional calculi

YQE

  • Show that CCxyCCxzCyz does not imply CxCyCxy, with the rule modus ponens and uniform substitution.

  • Ted Ulrich has shown that if YQE does not imply CxCyCxy, then it will take an infinite model to show this.


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