I see that Σ* is claimed to be decidable in many documents, but I have never seen an example or easy demostration that it is decidable.
What is the proof that Σ* is decidable?
Asked By : Charles
Answered By : Andrej Bauer
Theorem: The set $\Sigma^{*}$ of all words is decidable.
Proof. According to the definition of decidability, we must provide a computable function $d$ which takes a word $w$ and outputs $1$ if $w \in \Sigma^{*}$, and outputs $0$ if $w \not\in \Sigma^{*}$. Such a function is very easily constructed, it is $$d(w) = 1,$$ That is, because every word is in $\Sigma^{*}$, the decision function always says "yes". QED.
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Question Source : http://cs.stackexchange.com/questions/32726
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