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# Infinite Models for Propositional Calculi PowerPoint PPT Presentation

Infinite Models for Propositional Calculi. Zachary Ernst University of Missouri-Columbia ernstz@missouri.edu. 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.

Infinite Models for Propositional Calculi

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## Infinite Models for Propositional Calculi

Zachary Ernst

University of Missouri-Columbia

ernstz@missouri.edu

### 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

• 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

• 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

• 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

Consider the following infinite sequence of theorems:

Ix

IIx

IIIx

IIIIIIx

III…Ix

### 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 Proof

Ix

IIx

IIIx

IIIIIIx

III…Ix

Suppose M “thinks” that

IIx = IIIIIIx.

### 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 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 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?

• 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.

1

2

CpCqq

1

p

2

q

q

1

2

CpCqq

1

1

1

p

2

1

q

q

2

2

### 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

• 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

• 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 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

• 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?

• 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

• 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

• Recall that the theorems of Anderson’s system are

• We can construct an automaton such that:

• So let

### 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.