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When an FSA is deterministic the set of triples encoding its edges represents a relation that is functional in its ?rst and third components: for every q and σ there is exactly one state p such that hq, p, σi ∈ T. This function is called the transition function of the automaton and is usually denoted δ:
For any state q and symbol σ, then, δ(q, σ) is the state reached from q by following a single edge labeled σ. This can be extended to the path function, a function taking a state q and any string w ∈ Σ∗ which returns the statereached from q by following a path labeled w:
Note that ˆ δ is total (has some value for all q and w) and functional (that value is unique) as a consequence of the fact that δ is, which, in turn, is a consequence of the fact that the automaton is deterministic. In terms of the transition graph, this means that for any string w and any node q, there will always be exactly one path labeled w from q (which leads to δ(q,w)) and this is a consequence of the fact that there is always exactly one edge labeled σ from each node q of the graph and every σ ∈ Σ (which leads to δ(q, σ)).
Intuitively, closure of SL 2 under intersection is reasonably easy to see, particularly if one considers the Myhill graphs of the automata. Any path through both graphs will be a
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In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
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