In Sec 2.1 of Principles of Model Checking, by Christel Baier and Joost-Pieter Katoen, the authors talk about what constitutes a deterministic transition system (T.S.), and mention two kinds - action-deterministic and state-deterministic. Regarding the latter, they say that a T.S. is state-deterministic if "for any state label $A \in 2^{AP}$ and any state there is at most one outgoing transition leading to a state with label A". This seems like too broad of a formulation of determinism to me. Intuitively, I would expect a non-deterministic system to be one in which for which the machine has to "guess" which alternative to take in order to generate a given trace (and there may be more than one way to do so). So by this criterion, if there were a T.S. in which one transition from a state $s$ lead to a state in which $p$ holds, and another transition from $s$ to a state in which $\{p,q\}$ holds, this system would be non-deterministic. However, it would qualify as a deterministic system according to the definition in the book. Is my understanding of non-determinism off?
On a slightly related note, would it be correct to say that any T.S. that accepted user input (from any non-empty alphabet such as $\{a,b,c...\}$) is deterministic provided there was at most one transition for each alphabet symbol from any given state? Although such a T.S. may appear from the graph to be non-deterministic because there may be multiple $\epsilon$ transitions from a given state leading to states $a$/$b$/$c$ etc, the system has deterministic execution on any specific user input because at most one of the states $a$/$b$/$c$ is possible.
