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It is not hard to see that ε-transitions do not add to the accepting power of the model. The underlying idea is that whenever an ID (q, σ v) directly computes another (p, v) via a path that includes some number of ε-transitions (before the σ-transition, after it or both), we can get the same effect by extending the transition relation to include a σ-transition directly from q to v. So, in the example we could add ‘a' edges from 0 to 1 (accounting for the path 0 3) and from 1 to 3 (accounting for the path 1 3) and ‘b' edges from 1 to 3 (accounting for the path 1 3), from 3 to 2 (accounting for the path 3 2), and from 1 to 2 (accounting for the path 1 2), Note that in each of these cases this corresponds to extending δ(q, σ) to include all states in ˆ δ(q, σ). The remaining effect of the ε-transition from 0 to 2 is the fact that the automaton accepts ‘ε'. This can be obtained, of course, by simply adding 0 to F. Formalizing this we get a lemma.
how to find whether the language is cfl or not?
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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