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This close relationship between the SL2 languages and the recognizable languages lets us use some of what we know about SL2 to discover properties of the recognizable languages. Because they are SL2 languages, the runs of an automaton A (and, equivalently, the strings of pairs licensed by G2A) will satisfy the 2-suffix substitution closure property. This means that every recognizable language L is a homomorphic image of some language L′ (over an alphabet Σ′ , say) for which
u′1σ′v′1 ∈ L′ and u′2 σ′v′2 ∈ L′⇒ u′1σ′v′2( and u′2σ′v′1) ∈ L′.
Moreover, u′1σ′v′1 ∈ L′ and u′1σ′v′2 ∈ L′⇒ u′2σ′v′2 ∈ L′
The hypothetical u′1σ′ and u′2σ′ are indistinguishable by the language. Any continuation that extends one to a string in L′ will also extend the other to a string in L′ ; any continuation that extends one to a string not in L′ will extend the other to a string not in L′.
For the SL2 language L′ the strings that are indistinguishable in this way are marked by their ?nal symbol. Things are not as clear for the recognizable language L because the homomorphism may map many symbols of Σ′ to the same symbol of Σ. So it will not generally be the case that we can easily identify the sets of strings that are indistinguishable in this way. But they will, nevertheless, exist. There will be pairs of strings u1 and u2 - namely the homomorphic images of the pairs u′1σ′ and u′2σ′-for which any continuation v, it will be the case that u1v ∈ L iff u2v ∈ L.
This equivalence between strings (in the sense of being indistinguishable by the language in this way) is the key to characterizing the recognizable languages purely in terms of the strings they contain in a way analogous to the way suffix substitution closure characterizes the SL2.
Differentiate between DFA and NFA. Convert the following Regular Expression into DFA. (0+1)*(01*+10*)*(0+1)*. Also write a regular grammar for this DFA.
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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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