In this section we use the equivalence between standard and∀-symmetric acceptance games to show that the tree languages recognized byMSO-automata are closed under complementation.
Definition A.4. Letϕ∈For+(A)be a formula. Thedualϕδ∈FO+(A)ofϕis defined by induction as follows.
(a(x))δ ∶= a(x) (⊺)δ ∶= ()δ ∶= ⊺ (x≈y)δ ∶= x/≈y (x/≈y)δ ∶= x≈y (ϕ∧ψ)δ ∶= (ϕ)δ∨(ψ)δ (ϕ∨ψ)δ ∶= (ϕ)δ∧(ψ)δ ∀x.ψ ∶= ∃x.(ψ)δ ∃x.ψ ∶= ∀x.(ψ)δ
LetXbe a set andm∶A→ ℘(X)a marking. Thedual mδ∶A→ ℘(X)ofmis defined by putting
mδ(a) ∶= X∖m(a)
for eacha∈A. ⊲
The following property of the dual transformation is immediate by definition A.4.
Proposition A.5. Letϕ∈FO+(A)be a sentence, X a set and m∶A→ ℘(X)a marking. The following are equivalent.
1. (X,m) ⊧ϕ.
2. (X,mδ) /⊧ ϕδ.
We have now all the ingredients to prove the complementation lemma forMSO-automata.
Proof of proposition A.1 LetA= ⟨A,aI,∆,Ω⟩be anMSO-automaton. Define anMSO-automatonA= ⟨A,aI,∆δ,Ωδ⟩ by puttingΩδ(a) ∶=Ω(a)+1 and∆δ(a,c) ∶= (∆(a,c))δfor eacha∈Aandc∈C. LetTbe a tree. We want to show
that
AacceptsT iff Adoes not acceptT.
By proposition A.3 and determinacy of parity games, it suffices to prove that∃has a winning strategy inG∃=
A(A,T)if and only if∀has a winning strategy inG∀= A∀(A,T).
(⇒) Suppose that∃has a winning strategyf∃inG∃. We define a strategy f∀for∀inG∀, by induction on the construction of a matchπ∀ofG∀, while maintaining anf∃-conform shadow matchπ∃ofG∃. For each roundzthat is played inπ∀andπ∃, we want maintain the same basic position in the two matches.
At the initial round we initialize the two matches from position(aI,sI). Inductively, suppose that we are at
roundziand the same basic position(a,s) ∈A×T occurs inπ∀andπ∃. Letm∃∶A→ ℘(σR(s))be the suggestion of
f∃from position(a,s)inπ∃. We letm∀∶= (m∃)δbe the suggestion off∀from position(a,s)inπ∀. By proposition A.5, the markingm∀makes(∆(a,σC(s)))δ=∆δ(a,σC(s))false inσR(s), meaning that it is a legitimate move for ∀inπ∀. At this point we distinguish two cases.
1. Ifm∃made∀get stuck inπ∃, it means thatm∃(b) = ∅for allb∈A. Then, by definition,m∀(b) =σR(s)
for allb∈A. By definition ofG∀, no move is available for∃and she gets stuck in the matchπ∀, which is immediately won by∀.
2. Otherwise, let(b,t)be the next position picked by∃inπ∀. By definition ofG∀the nodetis not inm∀(b). By definition ofm∀, this means thattis inm∃(b). Therefore(b,t)is a legitimate move for∀inπ∃and we let(b,t)be the next basic position inπ∃. In this way the same basic positions are maintained in the two matches at roundzi+1.
By construction, either the two matchesπ∀andπ∃end in the same round, respectively with player∃and player
∀getting stuck, or they are both infinite. In the latter case, by the fact thatπ∃is f∃-conform and f∃is winning for∃, the minimum paritynoccurring infinitely often inπ∃in even. By definition ofΩδ, the minimum parity occurring infinitely often inπ∀isn+1, which is odd. Therefore∀winsπ∀.
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