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The path function δ : Q × Σ* → P(Q) is the extension of δ to strings:
This just says that the path labeled ε from any given state q goes only to q itself (or rather never leaves q) and that to ?nd the set of states reached by paths labeled wσ from q one ?rst ?nds all the states q′ reached by paths labeled w from q and then takes the set of all the states reached by an edge labeled σ from any of those q′.
We will still accept a string w i? there is a path labeled w leading from the initial state to a ?nal state, but now there may be many paths labeled w from the initial state, some of which reach ?nal states and some of which do not. When thinking in terms of the path function, we need to modify the de?nition of the language accepted by A so it includes every string for which at least one path ends at a ?nal state.
A common approach in solving problems is to transform them to different problems, solve the new ones, and derive the solutions for the original problems from those for the new ones
Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
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So we have that every language that can be constructed from SL languages using Boolean operations and concatenation (that is, every language in LTO) is recognizable but there are r
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