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### Finite Model TheoryLecture 10

Second Order Logic

Second Order Logic

- Add second order quantifiers:9 X.f or 8 X.f
- All 2nd order quantifiers can be done before the 1st order quantifiers [ why ?]
- Hence: Q1 X1. … Qm Xm. Q1 x1 … Qn xn. f, where f is quantifier free

Fragments

- MSO = X1, … Xm are all unary relations
- 9 SO = Q1, …, Qm are all existential quantifiers
- 9 MSO = [ what is that ? ]
- 9 MSO is also called monadic NP

Games for MSO

The MSO game is the following. Spoiler may choose between point move and set move:

- Point move Spoiler chooses a structure A or B and places a pebble on one of them. Duplicator has to reply in the other structure.
- Set move Spoiler chooses a structure A or B and a subset of that structure. Duplicator has to reply in the other structure.

Games for MSO

Theorem The duplicator has a winning strategy for k moves if A and B are indistinguishable in MSO[k]

[ What is MSO[k] ? ]

Both statement and proof are almost identical to the first order case.

EVEN Ï MSO

Proposition EVEN is not expressible in MSO

Proof:

- Will show that if s = ; and |A|, |B| ¸ 2k then duplicator has a winning strategy in k moves.
- We only need to show how the duplicator replies to set moves by the spoiler [why ?]

EVEN Ï MSO

- So let spoiler choose U µ A.
- |U| · 2k-1. Pick any V µ B s.t. |V| = |U|
- |A-U| · 2k-1. Pick any V µ B s.t. |V| = |U|
- |U| > 2k-1 and |A-U| > 2k-1. We pick a V s.t. |V| > 2k-1 and |A-V| > 2k-1.
- By induction duplicator has two winning strategies:
- on U, V
- on A-U, A-V
- Combine the strategy to get a winning strategy on A, B. [ how ? ]

EVEN 2 MSO + <

- Why ?

MSO Games

- Very hard to prove winning strategies for duplicator
- I don’t know of any other application of bare-bones MSO games !

9MSO

Two problems:

- Connectivity: given a graph G, is it fully connected ?
- Reachability: given a graph G and two constants s, t, is there a path from s to t ?
- Both are expressible in 8MSO [ how ??? ]
- But are they expressible in 9MSO ?

9 MSO

Reachability:

- Try this:F = 9 X. f
- Where f says:
- s, t 2 X
- Every x 2 X has one incoming edge (except t)
- Every x 2 X has one outgoing edge (except s)

9 MSO

- For an undirected graph G:s, t are connected , G ²F
- Hence Undirected-Reachability29 MSO

9 MSO

- For an undirected graph G:s, t are connected , G ²F
- But this fails for directed graphs:
- Which direction fails ?

s

t

9 MSO

Theorem Reachability on directed graphs is not expressible in 9 MSO

- What if G is a DAG ?
- What if G has degree · k ?

Games for 9MSO

The l,k-Fagin game on two structures A, B:

- Spoiler selects l subsets U1, …, Ul of A
- Duplicator replies with L subsets V1, …, Vl of B
- Then they play an Ehrenfeucht-Fraisse game on (A, U1, …, Ul) and (B, Vl, …, Vl)

Games for 9MSO

Theorem If duplicator has a winning strategy for the l,k-Fagin game, then A, B are indistinguishable in MSO[l, k]

- MSO[l,k] = has l second order 9 quantifiers, followed by f2 FO[k]
- Problem: the game is still hard to play

Games for 9MSO

- The l, k – Ajtai-Fagin game on a property P
- Duplicator selects A 2 P
- Spoiler selects U1, …, Ulµ A
- Duplicator selects B Ï P,then selects V1, …, Vlµ B
- Next they play EF on (A, U1, …, Ul) and (B, V1, …, Vl)

Games for 9MSO

Theorem If spoiler has winning strategy, then P cannot be expressed by a formula in MSO[l, k]

Application: prove that reachability is not in 9MSO [ in class ? ]

MSO and Regular Languages

- Let S = {a, b} and s = (<, Pa, Pb)
- Then S*\' STRUCT[s]
- What can we express in FO over strings ?
- What can we express in MSO over strings ?

MSO on Strings

Theorem [Buchi] On strings: MSO = regular languages.

- Proof [in class; next time ?]

Corollary. On strings: MSO = 9MSO = 8MSO

MSO and TrCl

TheoremOn strings, MSO = TrCl1

However, TrCl2 can express an.bn [ how ? ]

Question: what is the relationship between these languages:

- MSO on arbitrary graphs and TrCl1
- MSO on arbitrary graphs and TrCl

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