[summary:
Let ϕ(p,q1,…,qn) be a modal sentence modalized in p. Then there exists a sentence H(q1,..,qn) such that GL⊢⊡[p↔ϕ(p,q1,…,qn)]↔⊡[p↔H(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⊢⊡[p↔ϕ(p)]↔⊡[p↔H] %%note: Notation: ⊡A=A∧◻A%%.
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⊢⊡[p↔ψ(p,q1,…,qn)]↔⊡[pi↔H(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 GL⊢H(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⊢⊡[p↔ψ(p,q1,…,qn)]↔⊡[pi↔H(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 GL⊢H↔I. 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⊢⊡(p↔ϕ(p))↔(p↔H).
Suppose I is such that GL⊢H↔I. Then by the first substitution theorem, GL⊢F(I)↔F(H) for every formula F(q). If F(q)=⊡(p↔q), then GL⊢⊡(p↔H)↔⊡(p↔I), from which it follows that GL⊢⊡(p↔ϕ(p))↔(p↔I).
Conversely, if H and I are fixed points, then GL⊢⊡(p↔H)↔⊡(p↔I), so since GL is closed under substitution, GL⊢⊡(H↔H)↔⊡(H↔I). Since GL⊢⊡(H↔H), it follows that GL⊢(H↔I). %%
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⊢⊡[p↔ϕ(p)]↔⊡[p↔H].
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:
- [[⊥]]A=∅
- [[B→C]]A=(N∖[[B]]A)∪[[C]]A
- [[◻D]]A={m:∀i<mi∈[[D]]A}
- [[p]]A=[[A]]A
Lemma: Let M be a finite, transitive and irreflexive Kripke model in which (p↔A)isvalid,andBap−sentence.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 p↔A 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+1⊥∧◻n⊥ %%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)0⊥⊤⊥1⊤⊥⊤2⊤⊥⊤
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+1⊥∧◻0⊥=¬◻⊥.
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 PA⊢◻PA[G↔¬◻PAG]↔◻PA[G↔¬◻PA⊥] for every sentence G of arithmetic.
Since in PA we can construct G by the Diagonal lemma such that PA⊢G↔¬◻PAG, by necessitation we have that for such a G then PA⊢◻PA[G↔¬◻PAG]. By the theorem we just proved using the fixed point, then PA⊢◻PA[G↔¬◻PA⊥]. SInce [ everything PA proves is true] then PA⊢G↔¬◻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] H↔◻H.
%%hidden(Show solution): world= p◻(p)0⊤⊤1⊤⊤ 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),…,◻Di−1(p),⊤,◻Di+1(p),…,◻Dk(p)), which is k−1-decomposable.
Assuming that the procedure works for k−1 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↔¬◻(q→p).
We can 1-decompose the formula in B(d)=¬d, D1(p)=q→p.
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↔◻[◻(p∧q)∧◻(p∧r)].
%%hidden(Show solution): One possible decomposition of the the formula at hand is B(a)=a, D1(p)=◻(p∧q)∧◻(p∧r).
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 GL⊢∧i≤n{⊡(pi↔Ai(p1,…,pn)}↔∧i≤n{⊡(pi↔Hi)}.
%%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⊢⊡(p1↔Ai(p1,…,pn))↔⊡(p1↔H(p2,…,pn))
Now suppose that for j we have H1,…,Hj such that GL⊢∧i≤j{⊡(pi↔Ai(p1,…,pn)}↔∧i≤j{⊡(pi↔Hi(pj+1,…,pn))}.
By the [ second substitution theorem], GL⊢⊡(A↔B)→[F(A)↔F(B)]. Therefore we have that GL⊢⊡(pi↔Hi(pj+1,…,pn)→[⊡(pj+1↔Aj+1(p1,…,pn))↔⊡(pj+1↔Aj+1(p1,…,pi−1,Hi(pj+1,…,pn),pi+1,…,pn))].
If we iterate the replacements, we finally end up with GL⊢∧i≤j{⊡(pi↔Ai(p1,…,pn)}→⊡(pj+1↔Aj+1(H1,…,Hj,pj+1,…,pn)).
Again by the fixed point theorem, there is H′j+1 such that GL⊢⊡(pj+1↔Aj+1(H1,…,Hj,pj+1,…,pn))↔⊡[pj+1↔H′j+1(pj+2,…,pn)].
But as before, by the second substitution theorem, GL⊢⊡[pj+1↔H′j+1(pj+2,…,pn)]→[⊡(pi↔Hi(pj+1,…,pn))↔⊡(pi↔Hi(H′j+1,…,pn)).
Let H′i stand for Hi(H′j+1,…,pn), and by combining the previous lines we find that GL⊢∧i≤j+1{⊡(pi↔Ai(p1,…,pn)}→∧i≤j+1{⊡(pi↔H′i(pj+2,…,pn))}.
By [ Goldfarb's lemma], we do not need to check the other direction, so GL⊢∧i≤j+1{⊡(pi↔Ai(p1,…,pn)}↔∧i≤j+1{⊡(pi↔H′i(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 H′i to compute fixed points.
%%
An immediate consequence of the theorem is that for those fixed points Hi and every Ai, GL⊢Hi↔Ai(H1,…,Hn).
Indeed, since GL is closed under substitution, we can make the change pi for Hi in the theorem to get that GL⊢∧i≤n{⊡(Hi↔Ai(H1,…,Hn)}↔∧i≤n{⊡(Hi↔Hi)}.
Since the righthand side is trivially a theorem of GL, we get the desired result.