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Infinite Models for Propositional CalculiPowerPoint Presentation

Infinite Models for Propositional Calculi

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Presentation Transcript

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

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.

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.

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.

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