Fixed point theorem of provability logic

https://arbital.com/p/fixed_point_theorem_provability_logic

by Jaime Sevilla Molina Jul 27 2016 updated Mar 2 2017

Deal with those pesky self-referential sentences!


[summary:

Let ϕ(p,q1,,qn) be a modal sentence modalized in p. Then there exists a sentence H(q1,..,qn) such that GL[pϕ(p,q1,,qn)][pH(q1,..,qn)].

This result can be used to give us insight about self-referent sentences of arithmetic. ]

The fixed point theorem of provability logic is a key result that gives a explicit procedure to find equivalences for sentences such as the ones produced by the Diagonal lemma.

In its most simple formulation, it states that:

Let ϕ(p) be a modal sentence modalized in p. Then there exists a letterless H such that GL[pϕ(p)][pH] %%note: Notation: A=AA%%.

This result can be generalized for cases in which letter sentences other than p appear in the original formula, and the case where multiple formulas are present.

Fixed points

The H that appears in the statement of the theorem is called a fixed point of ϕ(p) on p.

In general, a fixed point of a formula ψ(p,q1,qn) on p will be a modal formula H(q1,,qn) in which p does not appear such that GL[pψ(p,q1,,qn)][piH(q1,,qn)].

The fixed point theorem gives a sufficient condition for the existence of fixed points, namely that ψ is modalized in ψ. It is an open problem to determine a necessary condition for the existence of fixed points.

Fixed points satisfy some important properties:

If H is a fixed point of ϕ on p, then GLH(q1,,qn)ϕ(H(q1,,qn),q1,,qn). This coincides with our intuition of what a fixed point is, since this can be seen as an argument that when fed to ϕ it returns something equivalent to itself.

%%hidden(Proof): Since H is a fixed point, GL[pψ(p,q1,,qn)][piH(q1,,qn)]. Since GL is [ normal], it is closed under substitution. By substituing p for H, we find that GL[H(q1,,qn)ψ(H(q1,,qn),q1,,qn)][H(q1,,qn)H(q1,,qn)].

But trivially GL[H(q1,,qn)H(q1,,qn), so GL[H(q1,,qn)ψ(H(q1,,qn),q1,,qn)]. %%

H and I are fixed points of ϕ if and only if GLHI. This is knows as the uniqueness of fixed points.

%%hidden(Proof): Let H be a fixed point on p of ϕ(p); that is, GL(pϕ(p))(pH).

Suppose I is such that GLHI. Then by the first substitution theorem, GLF(I)F(H) for every formula F(q). If F(q)=(pq), then GL(pH)(pI), from which it follows that GL(pϕ(p))(pI).

Conversely, if H and I are fixed points, then GL(pH)(pI), so since GL is closed under substitution, GL(HH)(HI). Since GL(HH), it follows that GL(HI). %%

Special case fixed point theorem

The special case of the fixed point theorem is what we stated at the beginning of the page. Namely:

Let ϕ(p) be a modal sentence modalized in p. Then there exists a letterless H such that GL[pϕ(p)][pH].

There is a nice semantical procedure based on Kripke models that allows to compute H as a truth functional compound of sentences n %%note:nA=,,,n-timesA %%. (ie, H is in [ normal form]).

A-traces

Let A be a modal sentence modalized in p in which no other sentence letter appears (we call such a sentence a p-sentence). We want to calculate A's fixed point on p. This procedure bears a resemblance to the [ trace] method for evaluating letterless modal sentences.

We are going to introduce the notion of the A-trace of a p-sentence B, notated by [[B]]A. The A-trace maps modal sentences to sets of natural numbers, and is defined recursively as follows:

Lemma: Let M be a finite, transitive and irreflexive Kripke model in which (pA)isvalid,andBapsentence.ThenM,w\models Biff\rho(w)\in [[B]]_A$.

%%hidden(Proof): Coming soon [todo: proof] %%

Lemma: The A-trace of a p-sentence B is either finite or cofinite, and furthermore either it has less than n elements or lacks less than n elements, where n is the number of s in A.

%%hidden(Proof): Coming soon [todo: proof] %%

Those two lemmas allow us to express the truth value of A in terms of world ranks for models in which pA is valid. Then the fixed point H will be either the union or the negation of the union of a finite number of sentences n+1n %%note:Such a sentence is only true in worlds of rank n%%

In the following section we work through an example, and demonstrate how can we easily compute those fixed points using a [ Kripke chain].

Applications

For an example, we will compute the fixed point for the modal [ Gödel sentence] p¬p and analyze its significance.

We start by examining the truth value of ¬p in the 0th rank worlds. Since the only letter is p and it is modalized, this can be done without problem (remember that B is always true in the rank 0 worlds, no mater what B is). Now we apply to p the constraint of having the same truth value as ¬p.

We iterate the procedure for the next world ranks.

world= p(p)¬(p)012

Since there is only one in the formula, the chain is guaranteed to stabilize in the first world and there is no need for going further. We have shown the truth values in world 2 to show that this is indeed the case.

From the table we have just constructed it becomes evident that [[p]]¬p=N{0}. Thus H=0+10=¬.

Therefore, GL[p¬p][p¬]. Thus, the fixed point for the modal Gödel sentence is the Consistency of arithmetic!

By employing the [ arithmetical soundness of GL], we can translate this result to PA and show that PAPA[G¬PAG]PA[G¬PA] for every sentence G of arithmetic.

Since in PA we can construct G by the Diagonal lemma such that PAG¬PAG, by necessitation we have that for such a G then PAPA[G¬PAG]. By the theorem we just proved using the fixed point, then PAPA[G¬PA]. SInce [ everything PA proves is true] then PAG¬PA.

Surprisingly enough, the Gödel sentence is equivalent to the consistency of arithmetic! This makes more evident that G is not provable [godelsecondincompletenesstheorm unless PA is inconsistent], and that it is not disprovable unless it is [omegainconsistency ω-inconsistent].

Exercise: Find the fixed point for the [ Henkin sentence] HH.

%%hidden(Show solution): world= p(p)01 Thus the fixed point is simply . %%

General case

The first generalization we make to the theorem is allowing the appearance of sentence letters other than the one we are fixing. The concrete statement is as follows:

Let ϕ(p,q1,,qn) be a modal sentence modalized in p. Then ϕ has a fixed point on p.%%.

There are several constructive procedures for finding the fixed point in the general case.

One particularly simple procedure is based on k-decomposition.

K-decomposition

Let ϕ be as in the hypothesis of the fixed point theorem. Then we can express ϕ as B(D1(p),,Dk(p)), since every p occurs within the scope of a (The qis are omitted for simplicity, but they may appear scattered between B and the Dis). This is called a k-decomposition of ϕ.

If ϕ is 0-decomposable, then it is already a fixed point, since p does not appear.

Otherwise, consider Bi=B(D1(p),,Di1(p),,Di+1(p),,Dk(p)), which is k1-decomposable.

Assuming that the procedure works for k1 decomposable formulas, we can use it to compute a fixed point Hi for each Bi. Now, H=B(D1(H1),,Dk(Hk)) is the desired fixed point for ϕ.

%%hidden(Proof): [todo: proof] There is a nice semantic proof in Computability and Logic, by Boolos et al. %%

This procedure constructs fixed points with structural similarity to the original sentence.

Example

Let's compute the fixed point of p¬(qp).

We can 1-decompose the formula in B(d)=¬d, D1(p)=qp.

Then B1(p)=¬=, which is its own fixed point. Thus the desired fixed point is H=B(D1())=¬¬q.

Exercise: Compute the fixed point of p[(pq)(pr)].

%%hidden(Show solution): One possible decomposition of the the formula at hand is B(a)=a, D1(p)=(pq)(pr).

Now we compute the fixed point of B(), which is simply .

Therefore the fixed point of the whole expression is B(D1(p=))=[(q)(r)]=[(q)(r)] %%

Generalized fixed point theorem

Suppose that Ai(p1,,pn) are n modal sentences such that Ai is modalized in pn (possibly containing sentence letters other than pjs).

Then there exists H1,,Hn in which no pj appears such that GLin{(piAi(p1,,pn)}in{(piHi)}.

%%hidden(Proof): We will prove it by induction.

For the base step, we know by the fixed point theorem that there is H such that GL(p1Ai(p1,,pn))(p1H(p2,,pn))

Now suppose that for j we have H1,,Hj such that GLij{(piAi(p1,,pn)}ij{(piHi(pj+1,,pn))}.

By the [ second substitution theorem], GL(AB)[F(A)F(B)]. Therefore we have that GL(piHi(pj+1,,pn)[(pj+1Aj+1(p1,,pn))(pj+1Aj+1(p1,,pi1,Hi(pj+1,,pn),pi+1,,pn))].

If we iterate the replacements, we finally end up with GLij{(piAi(p1,,pn)}(pj+1Aj+1(H1,,Hj,pj+1,,pn)).

Again by the fixed point theorem, there is Hj+1 such that GL(pj+1Aj+1(H1,,Hj,pj+1,,pn))[pj+1Hj+1(pj+2,,pn)].

But as before, by the second substitution theorem, GL[pj+1Hj+1(pj+2,,pn)][(piHi(pj+1,,pn))(piHi(Hj+1,,pn)).

Let Hi stand for Hi(Hj+1,,pn), and by combining the previous lines we find that GLij+1{(piAi(p1,,pn)}ij+1{(piHi(pj+2,,pn))}.

By [ Goldfarb's lemma], we do not need to check the other direction, so GLij+1{(piAi(p1,,pn)}ij+1{(piHi(pj+2,,pn))} and the proof is finished


One remark: the proof is wholly constructive. You can iterate the construction of fixed point following the procedure implied by the construction of the Hi to compute fixed points.

%%

An immediate consequence of the theorem is that for those fixed points Hi and every Ai, GLHiAi(H1,,Hn).

Indeed, since GL is closed under substitution, we can make the change pi for Hi in the theorem to get that GLin{(HiAi(H1,,Hn)}in{(HiHi)}.

Since the righthand side is trivially a theorem of GL, we get the desired result.