# Non-Deterministic Turing Machine Transition Function

Say a Non-Deterministic Turing Machine is in a state $q_1$ and it reads symbol $a$ from its tape. Is it possible to have two different transitions:

$$(q_1,a) \rightarrow (q_{accept}, <space>, L)$$

and

$$(q_1,a) \rightarrow (q_{reject}, <space>, L)$$

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Bluntly, yes. Though (depending on which precise definition of a TM you're using) the second will never get used, as a nondeterministic machine will always follow an accepting computation path, if it exists.

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Of course, if $(q_1,a)$ is reachable, then you will end up with a machine that for some inputs is able to accept as well as to reject. Whether that is a problem or not depends on how you're planning to interpret the machine's result.
The most common way to interpret non-determinism (such as when defining NP and like complexity classes) is that we're only concerned with whether it is possible for the nondeterministic machine to accept or not. In this view, there is no relevant difference between rejecting and diverging, and so it doesn't really matter whether the machines moves to $q_{\rm reject}$ or to an explicit infinite loop.