Infinite Models for Propositional Calculi

1
Infinite Models for Propositional Calculi

• Zachary Ernst
• University of Missouri-Columbia
• ernstz_at_missouri.edu

2
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.

3
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.

4
Another Example

• System due to J. Anderson
• Cxx
• CCIxxy (where IxCxx)
• CCIxyxCCIIxyz
• Modus Ponens and Universal Substitution
• Theorems are all of the form
• CCIIIIxxy, for any number of Is.

5
A Hyperfinite System

• Andersons 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.

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The Proof
Consider the following infinite sequence of
theorems
Ix IIx IIIx IIIIIIx IIIIx
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The Proof
If M is some arbitrary finite matrix model
Ix IIx IIIx IIIIIIx IIIIx
then there must be some pair of formulae in the
sequence that M identifies.
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The Proof
Ix IIx IIIx IIIIIIx IIIIx
Suppose M thinks that IIx IIIIIIx.
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The Proof
Ix IIx IIIx IIIIIIx IIIIx
Suppose M thinks that IIx IIIIIIx.
Then according to M CIIIIIIxIIx Cxx, which is
a theorem.
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The Proof
According to M CIIIIIIxIIx Cxx, which is a
theorem.
Now consider CCIIIIIIxIIxyCCIIII(IIx)(IIx)y,
which is of the form CCIIIIXXY, which is a
theorem (where XIIx).
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The Proof
According to M CIIIIIIxIIx Cxx, which is a
theorem.
Now consider CCIIIIIIxIIxyCCIIII(IIx)(IIx)y,
which is of the form CCIIIIXXY, which is a
theorem (where XIIx).
So one application of modus ponens yields
y. Therefore, the model must validate everything.
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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.

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How to Use a Matrix Model
1
2
CpCqq
1 2
1 1
1
2
p
q
q
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How to Use a Matrix Model
1
2
CpCqq
1
1 2
1 1
1
1
2
p
1
q
q
2
2
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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
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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.

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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 ?qs, 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.

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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.

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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

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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

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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.
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Weighted Automata and Andersons Hyperfinite
System

• Recall that the theorems of Andersons system are
  
• We can construct an automaton such that
• So let

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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.