Download Monadic second order Ehrenfeucht-Fraïssé game and the finiteness

Monadic second order Ehrenfeucht-Fraïssé game and
the finiteness of the universe
A small project for the course by Jouko Väänänen: Second order logic - a philosophical and mathematical appraisal,
spring 2011
Mikko Männikkö
Department of Mathematics and Statistics
University of Helsinki
Spring 2012
0. Introduction
The goal of this paper is to prove that in monadic second order (MSO) logic there is no sentence
that can characterize the finiteness of the universe.
In second order (SO) logic the sentence
∀f(∀x∀y (x ≠ y → f(x) ≠ f(y)) → ∀y∃x (y = f(x)))
is true in a structure iff every injective function f is a bijection i.e. the universe of the discourse is a
finite set. This proofs that in SO logic we can characterize the finiteness of the universe.
In first order (FO) logic we typically use the Ehrenfeucht-Fraïssé (EF) game to prove that FO logic
is not strong enough to characterize finiteness. In very much the same style I show that in MSO
logic we have the MSO Ehrenfeuch-Fraïssé (MSOEF) game that leads to the same result.
I assume that reader is familiar with such concepts as FO logic, SO logic and mathematical games.
1. FO case
Definition 1.1 (5.13)[2]. Suppose L is a vocabulary and M, M’ are L-structures. A partial mapping
π : M → M’ is a partial isomorphism M → M’ if there is an isomorphism π* : [dom(π)]M →
[rng(π)]M’ extending π. We use Part(M, M’) to denote the set of partial isomorphism M → M’.
In the EFn(A, B) game there are two players called I and II. The EFn(A, B) game is played on
two models, denoted here by A and B. The goal of I is to show that the two models are different.
The goal of II is to show that the models are the same.
The EFn(A, B) game lasts n rounds. Each round consists of the following steps
(1) I starts by choosing an element from A or from B.
(2) II responds by choosing an element from the other model.
We denote by ai the element chosen from Aon the ith round, by bi we denote the element chosen
from the B on the ith round.
After the n rounds II wins if f = {ai → bi} ∈ Part(A, B). Otherwise I wins.
Definition 1.2 (6.2)[2]. The quantifier rank of a formula φ, denoted QR(φ), is defined as follows:
QR(t = t´) = QR(Rt1…tn) = 0,
QR(¬φ) = QR(φ),
QR((φ ∧ ψ)) = QR((φ ∨ ψ)) = max{QR(φ), QR(ψ)},
QR(∃x φ) = QR(∀x φ) = QR(φ) + 1.
Let
A ≡n B
denote the condition that A and B satisfy the same sentences of FO logic of QR ≤ n.
Theorem 1.3 (Corollary 2.2.9)[1]. For structures A, B and n ≥ 0 the following are equivalent:
(1) II wins EFn(A, B).
(2) A ≡n B.
Theorem 1.4. FO logic is not strong enough to characterize finiteness.
Proof:
For any FO sentence φ with QR(φ) ≤ n, we can choose two models A and B with no structure such
that |A| = n and |B| = ω. II wins the game EFn(A, B) by using the following strategy:
case 1: If in round n I chooses an element an (or bn) that has already been chosen in an
earlier round m, then II chooses bn = bm (or an = am).
case 2: If in round n I chooses an element an (or bn) that has not yet been chosen, then II
chooses an element bn (or an) that has not yet been chosen.
In n rounds II never runs out of elements to choose. By theorem 1.3 this means that φ is true in A
if and only if it is true in B. Since φ is any FO sentence this means that we can always find a finite
model A and an infinite model B such that φ cannot tell the difference between them i.e. FO logic
is not strong enough to characterize finiteness. á
2. MSO case
Monadic second order logic is an extension of the first order logic that allows quantification over
monadic predicate. This means that we can quantify over not only elements of the universe but also
over subsets of the universe. We use uppercase letters like X, Y, Z for second order relation
variables.
Semantics of MSO logic are as in FO logic with the addition of a rule for quantification over
monadic predicate:
A ⊨ ∃X φ(X) ↔ there is some R ⊆ A such that A, R ⊨ φ(X).
In the monadic second order logic Ehrenfeucht-Fraïsse game the FO EF-game is modified to allow
a new kind of move that takes care of the second-order quantifications.
Definition 2.1. In the MSOEFn(A, B) game there are two players called I and II. The MSOEFn(A,
B) game is played on two models, denoted here by A and B. The goal of I is to show that the two
models are different. The goal of II is to show that the models are the same.
The MSOEFn(A, B) game lasts n rounds. Each round consists of the following steps
(1) I starts by choosing an element or a subset from A or from B.
(2) II responds by choosing an element (if I has chosen an element) or a subset (if I has
chosen a subset) from the other model.
We denote by ai the ith element chosen from A, by bi we denote the ith element chosen from the B.
We denote by Ai the ith subset chosen from A, by Bi we denote the ith subset chosen from the B.
After the n rounds II wins if f = {ai → bi} ∈ Part((A, A1, …), ( B, B1, ...)). Otherwise I wins.
Formulas of MSO logic can have both first and second order variables free, which we indicate as in
φ(X, x). The free second order variables are among those listed as X just as the free first order
variables are among those listed as x. MSO quantifier rank is defined to represent the nesting depth
of first and second order quantifications (counted as one).
Definition 2.2. The MSO quantifier rank, denoted by QRMSO(ϕ), of a formula ϕ is defined as
follows:
QRMSO(t = t´) = QRMSO(Rt1…tn) = 0,
QRMSO(¬ϕ) = QRMSO(ϕ),
QRMSO((ϕ ∧ ψ)) = QRMSO((ϕ ∨ ψ)) = max{QRMSO(ϕ), QRMSO(ψ)},
QRMSO(∃xϕ) = QRMSO(∀xϕ) = QRMSO(∃Xϕ) = QRMSO(∀Xϕ) = QRMSO(ϕ) + 1
Let
A ≡nMSO B
denote the condition that A and B satisfy the same sentences of MSO logic of QRMSO ≤ n.
Theorem 2.3 (Theorem 3.3.1)[1]. For structures A, B and n ≥ 0 the following are equivalent:
(1) II wins MSOEFn(A, B).
(2) A ≡nMSO B.
Lemma 2.4. I uses subset moves to partition the models. After the first subset round A is divided
into two partitions {x ∈ A | x ∈ A1} and {x ∈ A | x ∉ A1} (and B is divided into two partitions {y
∈ B | y ∈ B1} and {y ∈ B | y ∉ B1}). After the first two subset rounds A is divided into four
partitions {x ∈ A | x ∈ A1, x ∈ A2}, {x ∈ A | x ∈ A1, x ∉ A2}, {x ∈ A | x ∉ A1, x ∈ A2} and {x ∈
A | x ∉ A1, x ∉ A2}. After the first n subset rounds models are divided into 2n partitions. We call
the partitions in A P1, P2, … and partitions in B Q1, Q2, …
We say that
Pi = {x ∈ A | x ∈ A(m), x ∉ A(n)}
corresponds to
Qi = {y ∈ B | y ∈ B(m’), x ∉ B(n’)}
iff m = m’ and n = n’.
Theorem 2.5. MSO logic is not strong enough to characterize finiteness.
Proof:
For any MSO sentence ϕ with QRMSO(ϕ) ≤ n, we can choose two models A and B with no structure
such that |A| = 2n and |B| = ω. If II wins the game MSOEFn(A, B) then by theorem 2.3 this means
that φ is true in A if and only if it is true in B. Since φ is any MSO sentence this means that we can
always find a finite model A and an infinite model B such that φ cannot tell the difference between
them i.e. MSO logic is not strong enough to characterize finiteness.
In the game MSOEFn(A, B) I can play subset moves and element moves arbitrarily. This makes
the formulating of winning strategy for II a messy business. We make the following observations
about the strategy of I:
The only way for I to win in this case is to prove that models A and B differ in size. Since models
are too big for this to show in just n element moves, I needs to partition the models by using subset
moves. In order to win I needs to find corresponding partitions Pi and Qi after k subset moves such
that |Pi| ≠ |Qi| and min{|Pi|, |Qi|} < n–k (rounds left in the game). For I to play an element move
before the right kind of corresponding partitions are found is in strategy wise just the same as
playing a subset move with a subset of size one. Therefore the game MSOEFn(A, B) collapses into
a game MSOEFn*(A, B) where I needs to play all the subset moves before the element moves.
This observation makes formulating a winning strategy for II much easier.
II wins the game MSOEFn*(A, B) by using a following strategy:
When I plays a subset move she splits every partition Pi in A (or B) in two partitions Pi’ and Pi’’.
Notice that Pi’ or Pi’’ can be empty sets. Before any subset moves are played the whole universe is
considered as a one big partition.
Case 1. |Pi| = |Qi| and I splits Pi (or Qi). II can simply split Qi in a same way as I and there in
no danger of losing there. Only problem for II is the case where some partitions Pi and Qi
differ in size (as is the case in the beginning of the game).
Case 2. |Pi| ≠ |Qi| and I splits Pi (or Qi) in two parts Pi’ and Pi’’. If |Pi’| < 2n-k II chooses Qi’ ⊆
Qi such that |Qi’| = |Pi’| and Qi’’ = Qi – Qi’ where |Qi’’| ≥ 2n-k. If |Pi’’| < 2n-k II chooses Qi’’ ⊆ Qi
such that |Qi’’| = |Pi’’| and Qi’ = Qi – Qi’’ where |Qi’| ≥ 2n-k.
Since in the beginning of the game |A| = 2n and |B| = ω the best that I can do in n subset moves,
when II is following this strategy, is to reach a position where there are corresponding partitions Pi
and Qi such that |Pi| = 1 and |Qi| = ω. With no moves left I can’t use the element moves to show the
difference in size. á
References:
1. Ebbinghaus, H. –D. and Flum, J. Finite Model Theory. Springer.
2. Väänänen, J. 2011. Models and Games. Cambridge University Press.