08-decidability

# 08-decidability - Proofs using Deciders Let L1 and L2 be...

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Proofs using Deciders Proposition: Let L 1 and L 2 be decidable languages, then the concatenation L = L 1 L 2 is also decidable. Proof: We show decidability of L by constructing a decider for it. Let M 1 and M 2 be deciders for L 1 and L 2 , respectively, then we can construct a decider M for L as follows: M = "On input w , 1. For each way to split w into two parts, w = w 1 w 2 , do: 2. Run M 1 on w 1 . 3. Run M 2 on w 2 . 4. If for any combination M 1 and M 2 accept, accept ; otherwise, reject ." – p. 1/1

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Decidability Let us study some "standard" machines that are deciders. It turns out that these standard machines help us construct more complicated proofs. – p. 2/1
Decidability Theorem: The language A DFA = {( B,w )| B is a DFA that accepts string w } is decidable. Proof: We construct a decider M DFA for A DFA . M DFA = "On input ( B,w ) , where B is a DFA and w is a string: 1. Simulate B on input w . 2. If the simulation ends in an accept state of the DFA, accept ; otherwise, reject ." – p. 3/1

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Decidability Theorem: The language A NFA = {( B,w )| B is an NFA that accepts string w } is decidable. Proof: We construct a decider M NFA for A NFA . M NFA = "On input ( B,w ) , where B is an NFA and w is a string: 1. Convert NFA B into an equivalent DFA B (this is algorithmic, so a TM can do it). 2. Run M DFA on input ( B , w ) . 3. If M DFA accepts, accept ; otherwise, reject ." – p. 4/1
Decidability Theorem: The language A REX = {( R,w )| R is a regular expression that generates string w } is decidable. Proof: We construct a decider M REX for A REX . M REX = "On input ( R,w ) , where R is a regular expression and w is a string: 1. Convert regular expression R into an equivalent DFA R (this is algorithmic, so a TM can do it). 2. Run M DFA on input ( R ,w ) . 3. If M DFA accepts, accept ; otherwise, reject ." – p. 5/1

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Decidability Theorem: The language E DFA = {( A )| A is a DFA and L ( A ) = ∅} is decidable.
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