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