1. Second Order Monadic Theory of One Successor Presented By: Tamar Aizikowitz Spring 2007 Automata Seminar

2. 2 Second Order Monadic Logic Variables: Variables over natural numbers: x, y, z… Variables over sets (functions  → {0,1} ): σ, τ, δ… Constant: The natural number 0 Successor function: S(x) = x + 1 Binary Predicates: σ(x) = 0 σ(x) = 1

3. 3 Examples of MSO Formulae σ is a subset of τ : F(σ,τ) =  x (σ(x) → τ(x)) σ is singleton: F(σ) =  x (σ(x)   y (σ(y) → x=y)) x < y: F(x,y) =  (x=y)   σ [σ(y)   z  z’ (σ(z)  S(z’)=z → σ(z’)) → σ(x)]

4. 4 Theorem 1 Let F(σ 1,…,σ n ) be an MSO formula, then the following infinitry language over the alphabet {0,1} n is ω -regular: L(F) = {σ 1 (0)  σ n (0)  σ 1 (k)  σ n (k)  | F(σ 1,…,σ n )} Proof: (1) Prove that F can be transformed to normal form (2) Prove that a Büchi automata can be built s.t. it accepts L(F), for all normal form F.

5. 5 Part 1 – Normal Form (1) Lemma 1: Every formula F(σ 1,…,σ n,x 1,…,x m ) is equivalent to an MSO formula of the form Q 1  Q i Q i+1  Q j G where: (1) G is a formula with no quantifiers (2) Q 1  Q i are function quantifiers (3) Q i+1  Q j are numerical quantifiers

6. 6 Part 1 – Normal Form (2) Proof of Lemma 1: Assume F is in prennix normal form Q 1  Q k F’ where F’ contains no quantifiers. Q i is out-of-order if it is a number quantifier with a function quantifier after it. Let Q i be the rightmost out-of-order quantifier. The weight of Q i is the number of function quantifiers that appear after it. We prove the claim by induction on the number of out-of-order quantifiers and the weight of the rightmost one.

7. 7 Part 1 – Normal Form (3) Proof of Lemma 1 continued… Assume x quantifier is rightmost out-of-order:  x  σ Q 1  Q k H   σ  x Q 1  Q k H  x  σ Q 1  Q k H   σ  x Q 1  Q k H  x  σ Q 1  Q k H   δ  σ Q 1  Q k  x  y(δ(x)=1  (δ(y)=1→ H))  x  σ Q 1  Q k H   δ  σ Q 1  Q k  x  y(δ(x)=0  (δ(y)=1  H))

8. 8 Part 1 – Normal Form (4) Simple structure:  x  i=1,…,k  j=0,…,n σ i (x+j) = ε ij ; ε ij = 0,1 i.e.  x (σ 1 (x) = ε 10    σ 1 (x+n) = ε 1n   ) Lemma 2: Every formula has an equivalent of the form Q 1  Q k G where Q i are function quantifiers and G is a prepositional combination of simple structures and atomic formulae.

9. 9 Part 2 – Büchi Automata (1) Lemma 3: A is atomic  L(A) is ω -regular. Proof of Lemma 3: A is of the form σ(x) = 0/1 “Count” until x Verify that the value is 0/1 accordingly Go to (non-)accepting sink state Lemma 4: B is a basic structure  L(B) is ω -regular. Proof of Lemma 4: Skip x-1 letters from {1,0} k (“guess x ” non-deterministically) Verify next n+1 letters match ε ij values Go to (non-)accepting sink state

10. 10 Part 2 – Büchi Automata (2) Proof of Theorem 1: Assume F(σ 1,…,σ n ) is in normal form (i.e. Q 1  Q k G ). We prove by induction on the number of Boolean connectives in G that L(G) is ω -regular: Base: G is an atomic formula or a basic structure  the claim follows from Lemmas 3 and 4. Closure: L(G 1  G 2 ) = L(G 1 )  L(G 2 ) L(G 1  G 2 ) = L(G 1 )  L(G 2 ) L(  G) = L(G) C  the claim follows from the closure properties of ω -regular languages.

11. 11 Part 2 – Büchi Automata (3) Proof of Theorem 1 continued… Now we prove the claim for F by induction on the number of quantifiers Q i : Base:no quantifiers  already proven Closure: L(  σ i H(σ 1,…,σ m )) is the language h(L(H)) where h: {0,1} m →({0,1} m-1 ) * is a homomorphism s.t. h(ε 1  ε i-1 ε i ε i+1  ε m ) = ε 1  ε i-1 ε i+1  ε m L(  σH) = L(  σ  H) C

12. 12 Decidability of MSO Corollary 1: An algorithm exists which determines for a given closed formula F whether F is valid. Proof of Corollary 1: Assume F is of the form Qσ G(σ). Therefore: If Q =  then F is valid iff L(G)   If Q =  then F is valid iff L(G) = {0,1} ω, which is equivalent to L(G) C =  The claim follows from the fact that emptiness is decidable for Büchi Automata.