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In Alpern,Schneider 86 is described how to extract the automata that recognize safety and liveness properties from a Buchi automaton $m$. This shows that any property rapresented by a Buchi automaton is equivalent to the intersection of these two automata. In particular, the automata for the safety properties is represented by making all states of $m$ accepting. The question is: although it is a Buchi, can this automaton be determinized by using the subset construction?

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Yes. The automata in question are often called "Looping" automata (so you have a keyword to start from). A possible starting point is the following paper: https://faculty.idc.ac.il/udiboker/files/MullerAutomata.pdf

A looping automaton can be described as $\left<Q,\Sigma,\delta,Q_0\right>$ where the components are states, alphabet, transition function $\delta:Q\times \Sigma\to 2^Q$ and $Q_0$ are the initial states. Then, the acceptance condition is that there exists a run on the word. That is, $w$ is accepted iff the automaton has some run on it.

It is easy to see (using Kőnig's Lemma) that the acceptance condition can be equivalently described as follows: for a word $w\in \Sigma^\omega$, let $w_i$ be its prefix up to letter $i$, then $w$ is accepted iff $\delta^*(Q_0,w_i)\neq \emptyset$ for all $i\in \mathbb{N}$.

This shows that it's enough to track the subset construction in order to determine acceptance. What you end up is a deterministic automaton whose states are all accepting except for one, that corresponds to $\emptyset$.

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  • $\begingroup$ thanks for the detailed answer. Do you have some reference in which is discussed this particular case? $\endgroup$
    – kafka
    Feb 23 at 17:42
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    $\begingroup$ I added a reference in the answer. I'm not sure it has a proof of the claim, since the proof is pretty much identical to the case of NFA->DFA translation. $\endgroup$
    – Shaull
    Feb 23 at 18:13

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