In this section we formalize two essential intuitions about non-deterministic Büchi automata:
1. checking whether a non-deterministic Büchi automatonBaccepts a treeTreduces to check a condition on
2. checking whether the intersection of the languages of two non-deterministic Büchi automata is not empty reduces to the construction of a finite sequence of well-founded trees with certain properties (proposition 4.9).
The idea is that a run of a non-deterministic Büchi automaton on a treeTcan be split into several tasks
concerning well-founded subtrees (and prefixes, which are just a particular kind of well-founded subtrees) ofT, and
there is never the need to considerTas a whole. We informally refer to this as theBounded Information Property
of non-deterministic Büchi automata. Intuitively, this is the key property allowing to describe winning strategies for non-deterministic Büchi automata by means ofWFMSO-formulae - under certain conditions that we will see in the sequel.
Definition 4.6. LetB= ⟨B,bI,∆,F⟩be a non-deterministic Büchi automaton andTa tree. Let f be a surviving
strategy for∃inA(B,T)@(bI,sI). Letγ≤ωbe an ordinal. A γ-accepting sequence for f overBandTis a
sequence(Ei)i<γsuch that, for alli<γ: 1. Eiis a prefix ofT;
2. Ft(Ei) <Ft(Ei+1);
3. for allsin the frontier ofEi, there is a uniquea∈Asuch that(a,s) ∈Dom(f); in addition,ais inF. ⊲
Intuitively, fork<ω, ak-accepting sequence for a surviving strategy f witnesses the fact that f ‘behaves as’ a winning strategy for∃in the prefixEkofT. For each prefixEiin the sequence, the condition that eachs∈Ft(Ei)is
associated with auniqueaccepting state is motivated by the fact that f can be assumed to be full and functional,B being non-deterministic.
Proposition 4.7. LetB= ⟨B,bI,∆,F⟩be a non-deterministic Büchi automaton andTa tree. The following are
equivalent.
• Player∃has a winning strategy inA(B,T)@(bI,sI).
• Player∃has a surviving strategy f inA(B,T)@(bI,sI)and there is anω-accepting sequence for f overB
andT.
Proof (⇒) Let f be a winning strategy for∃inA(B,T)@(bI,sI). By proposition 2.36 and remark 4.3, we can
assume f to be full and functional. LetTf the tree representation of f.
The idea of the proof is that the sequence(Ei)i<ωis easily definable onTf: since a branch ofTf corresponds to
the sequence of basic positions visited along an f-conform matchπ, there are infinitely many positions occurring in πof the form(b,t)withb∈F. We intersectπwith each frontier in one of these positions. Then it is sufficient to project this construction on the treeTto obtain an analogous construction as in the statement.
We provide the formal details following these intuitions. Given a branch S of T, since f is full there
is an f-conform match π that is played along S. Since f is winning, there are infinitely many positions
(a0,s0),(a1,s1),(a2,s2), . . .occurring alongπsuch that eachaiis inF. We can order these positions in a sequence
(ai,si)i<ωaccording to the round ofπin which they occur, such that(ai,si)is visited inπbefore(ai+1,si+1). An infinite sequence(si)i<ωof nodes is induced, such thatsiR+si+1and all nodes in the sequence are in the branchS. For eachi<ω, we letEibe the prefix ofTinduced by defining its frontierFt(Ei)as follows:
⋆ for each branch S ofT, let(si)i<ωbe the associated sequence of nodes of S, given as above. We put
Ft(Ei)∩S= {si}.
By construction of(si)i<ωit is straightforward to check that(Ei)i<ωis anω-accepting sequence forf overBandT.
(⇐) Let f be a surviving strategy for∃inGB= A(B,T)@(bI,sI)and(Ei)i<ωanω-accepting sequence for f overBandT. We claim that f is in fact a winning strategy for∃inA(B,T)@(bI,sI).
For this purpose, letπbe an infinite f-conform match ofGB. LetSbe the branch ofTon whichπis played.
The branchSintersects the frontierFt(Ei)of each well-founded subtreeEifrom the sequence(Ei)i<ω, in a node
si∈S∩Ft(Ei). This induces a sequence(si)i<ωof nodes inS. By the fact thatFt(Ei) <Ft(Ei+1)for eachi<ω, all nodes in(si)i<ωare distinct, and furthermore for eachi<ωthere is a uniqueb∈Bsuch that(b,si) ∈Dom(f)and
b∈F. This means that the matchπplayed onSvisits infinitely many basic positions associated with accepting states ofB. Therefore∃winsπ. This suffices to show that f is a winning strategy for∃inGB. ◻
For non-deterministic Büchi automataB1andB2and a treeT∈L(B1)∩L(B2), let(G1i)i<ωand(G2i)i<ωbe ω-accepting sequences respectively forB1andB2onT. We introduce the notion ofk-trapforB1andB2. The idea is that ak-trap is a finite sequence(Ei)i≤kwitnessing the interleaving of sequences(G1i)i<ωand(G2i)i<ωup tok.
Definition 4.8. LetB1= ⟨B1,bI1,∆1,F1⟩andB2= ⟨B2,b2I,∆2,F2⟩beNDBautomata and letTbe a tree. Given some
fixedk<ω, let(Ei)i≤kbe a sequence of prefixes ofTsuch thatE0= {sI}andEi⊊Ei+1for alli≤k. We say thatTand(Ei)i≤kconstitute ak-trap forB1andB2if there exist
1. a strategy f1for∃inA(B1,T)@(b1I,sI)which is surviving inEk,
2. a strategy f2for∃inA(B2,T)@(b2I,sI)which is surviving inEk,
3. ak-accepting sequence(G1i)i≤kfor f1overB1andT,
4. ak-accepting sequence(G2i)i≤kfor f2overB2andT,
such that, for alli<k, the following conditions hold: • Ft(Ei) ≤Ft(Gi1) <Ft(Ei+1);
• Ft(Ei) ≤Ft(Gi2) <Ft(Ei+1).
We say that the strategies f1and f2witnessthe k-trap forB1andB2.
Figure 4.1: initial part of ak-trap
Proposition 4.9([28]). LetB1andB2be NDB automata and let m be the product of the cardinalities of their carriers. If there exists an m-trap forB1andB2then L(B1)∩L(B2) ≠ ∅.
Proof sketch A detailed proof of this result can be found in [28], proof of theorem 27. In the sequel we confine ourselves to a sketch giving the idea of the argument.
LetB1= ⟨B1,bI1,∆1,F1⟩andB2= ⟨B2,b2I,∆2,F2⟩beNDBautomata. Our initial assumption is that anm-trap for
B1andB2exists. LetTbe the tree associated with them-trap as in definition 4.8. The idea is that, for each pair of
states(b1,b2)inB1×B2, them-trap gives to∃:
• a subtreeTb1,b2 ofTwith rootsIand a prefixEb1,b2 ofTb1,b2; • the information on how to playG1= A(B1,Tb1,b2)@(b1,sI)andG
2= A(
B2,Tb1,b2)@(b2,sI), in such a way that along each match ofGnwhich is played inEb1,b2 she never gets stuck and a basic position with an accepting stateb∈Fnoccurs, for eachn∈ {1,2}.
Putting together infinitely many copies of those subtrees we can construct a tree Tω, whose membership in
L(B1)∩L(B2)is witnessed by patching together the strategies suggested by them-trap.
For this purpose, we define a sequence of relations(Hi)i<ω, where eachHiis a subset ofB1×B2. For the base case, we setH0=B1×B2. For the inductive step, we put(b1,b2) ∈Hi+1if and only if the following conditions hold.
2. There is a tree Tb1,b2 with root sI, prefixesEb1,b2, Gb1 and Gb2 ofTb1,b2, strategies fb1 and fb2 for ∃ respectively inA(B1,Tb1,b2)@(b1,sI)andA(B2,Tb1,b2)@(b2,sI), such that the following holds.
a) The strategy fb1 is surviving for∃inEb1,b2. b) The strategy fb2 is surviving for∃inEb1,b2. c) Ft(Gb1) <Ft(Eb1,b2).
d) Ft(Gb2) <Ft(Eb1,b2).
e) For alls∈Ft(Gb1), there is a unique position(b
′
1,s) ∈B1×Tb1,b2 such that(b
′
1,s)is a node of the tree representationTfb1 of fb1; in addition,b
′
1is inF1. f) For alls∈Ft(Gb2), there is a unique position(b
′
2,s) ∈B2×Tb1,b2 such that(b
′
2,s)is a node of the tree representationTfb2 of fb2; in additionb
′
2is inF2. g) For alls∈Ft(Eb1,b2), there is a unique pair(b
′
1,b′2)such that(b′1,s)and(b′2,s)are nodes respectively ofTfb1 andTfb2; in addition,(b
′
1,b′2)is inHi.
Provided this construction, the proof of the main statement is reduced to the proof of the following two facts. 1. If there exists anm-trap forB1andB2then(b1I,b2I)is inHm.
2. If(b1I,b2I)is inHmthenL(B1)∩L(B2) ≠ ∅.
We refer to [28] for a proof of the first fact, showing how the components witnessing (b1I,b2I) ∈Hm are
already provided by ourm-trap. Instead we focus on the second fact, which is proved by constructing a tree
Tω∈L(B1)∩L(B2), from the assumption that(b1I,b2I) ∈Hm. The key observation is that by definition the sequence (Hn)n<ωstabilizes atm. This means thatHm=Hm+k for allk<ω, as can be shown by a simple combinatorial
argument, using the fact thatm is the cardinality ofB1×B2. Roughly, the argument goes as follows: given
(b1,b2) ∈Hm, there is a sequence(bi1,bi2)i≤mof pairs, with(bi1,bi2) ∈Hifor eachi≤m, which is determined by the
property of point 2.gof the definition of(Hi)i<ω. Since(bi1,bi2)i≤mhasm+1 elements, there is at least one pair
which is repeated, that is,(bl1,bl2) = (b1j,b2j), for somel,jwithl<j<m+1. This means that we can suitably ‘plug’
(bn1,bn2)n<jin place of(bn1,bn2)n<l, to expand the sequence(bi1,bi2)i≤mto lengthm+1+(j−l). This witnesses that (b1,b2)was in fact a member ofHm+(j−l), and then we can repeat the argument.
Now we proceed with the construction of the treeTω∈L(B1)∩L(B2). The idea is to provideTωas the limit of a sequence(Ti)i<ωof well-founded trees, withTia prefix ofTi+1for eachi<ω. Givenn∈ {1,2}, we also construct the graph of a strategy fωnfor∃inA(Bn,Tω)as the limit of a sequence(fin)i<ω, with the graph of finstrictly
contained in the graph of fin+1, for eachi<ω. In the sequel we sketch the inductive construction of(Ti)i<ωand
(fin)i<ω.
• For the first elementT0in the sequence, we use the assumption that (b1I,bI2)is inHm=Hm+1to get a treeTb1
I,b
2
I
with rootsI, prefixesEb1
I,b 2 I ,Gb1 I andGb2 I ofTb1 I,b 2 I , strategies fb1 I and fb2 I for∃respectively in A(B1,Tb1 I,b2I)@(b 1 I,sI)andA(B2,Tb1 I,b2I)@(b 2
I,sI)with the properties given by definition ofHm+1. Then we putT0∶=Eb1
I,b2I.
Forn∈ {1,2}, the strategy f1nis defined to be the restriction of fbn
I to basic positions fromBn× (Eb1I,b2I∖
Ft(Eb1
I,b2I)).
• The treeTi+1will be given as a well-founded extension ofTi. By inductive hypothesis eacht∈Ft(Ti)is
associated with a pair(b1t,bt2) ∈Hm. This also means that(b1t,b2t)is inHm+1, and we can repeat the same argument that we used for the base case to get a treeTb1
t,b2t with roott, prefixesEb1t,b2t,Gb1t andGb2t of
Tb1
t,bt2, strategies fb1t and fb2t for∃respectively inA(B1,Tbt1,b2t)@(b
1
t,t)andA(B2,Tb1
t,bt2)@(b
2
t,t)with the
properties given by definition ofHm+1. We defineTi+1by putting Ti+1 = (Ti∖Ft(Ti))∪ ⋃
t∈Ft(Ti)
Eb1
t,b2t.
By constructionTi+1yields a tree structureTi+1, which is induced byTiandTbt1,bt2.
For eachn∈ {1,2}, andt∈Ft(Ti), let fbEn
t denote the restriction of fb n
t to basic positions fromBn×Eb1t,bt2. the
strategy fin+1is defined by putting
fin+1 = fin∪ ⋃
t∈Ft(Ti)
fbEn t,
Figure 4.2: construction ofTi+1.
where∪is the union of graphs of functions. In order to check that fi+1is indeed a function, observe that by inductive hypothesis finis a function with domainBn×(Ti∖Ft(Ti))and each fbEn
t has domainBn×Eb1t,b2t. All
these strategies have disjoint domains by construction. It follows that also fi+1is uniquely defined on each basic position in its domain, which is justBn×Ti+1by definition ofTi+1.
Letrbe a node inFt(Ti+1). By construction we have thatris inFt(Eb1
t,b2t)for somet∈Ft(Ti). Sincet
is associated with the pair(b1t,b2t) ∈Hm+1, then by definition ofHm+1the noderis associated with a pair
(b1r,b2r) ∈Hm. This suffices to maintain the inductive hypothesis in the next stagei+2.
The proof is concluded by checking that for each n∈ {1,2} the strategy fωn is winning for ∃ in Gω∶=
A(Bn,Tω)@(bnI,e). For this purpose, the key observation is that an fωn-conform matchπωofGωcan be seen as an infinite sequenceπ1,π2, . . . ,πnof partial matches ofGω, where eachπiis an fin-conform match played along the
well-founded subtreeTiofT. By definition of fin, alongπia basic position of the form(b,s) ∈Bn×Tiis visited
withb∈Fn. This means that someb∈Fnis visited infinitely often inπω, which is then won by∃. We refer to [28]
for further details. ◻