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The upper string r ∈ Q+ is the sequence of states visited by the automaton as it scans the lower string w ∈ Σ*. We will refer to this string over Q as the run of A on w. The automaton A accepts w iff the run of A on w ends in an accepting state. (If A is non-deterministic there will potentially be many runs with the automaton accepting if any one of them ends in an accepting state.) Note that the set of runs of an automaton is an SL2 language, recognized by the SL2 automaton (over Q) one gets by projecting away the third component of the triples of GA. Thus there is some kind of close relationship between the strictly local languages and the recognizable languages.
To get at this we will start by working in the other direction, extending our tiles to hold four symbols. The idea is to include, for each tile (q, p, σ) ∈ GA, a tile extended with σ′ for each σ′ ∈ Σ. (We don't actually need tiles for all such σ′ , only for those that occur on tiles (x, q, σ′) which might precede this one in a tiling, but including all of them will be harmless-the ones that do not occur on such tiles will just be useless.)
Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second
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s->0A0|1B1|BB A->C B->S|A C->S|null find useless symbol?
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Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con
S-->AAA|B A-->aA|B B-->epsilon
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