# Simplifying quantum circuits

I am very new to quantum circuits and am unsure how to simplify them. Say I'm given a very simple circuit:

and I want to simplify it using basic quantum gates. I'm not looking for an answer, but more just how to go about it. I started by computing the outputs with $$|0\rangle$$, $$|1\rangle$$, $$|+\rangle$$ and $$|-\rangle$$ states and thought that a simple NOT gate would suffice because $$|0\rangle \rightarrow |1\rangle$$, $$|1\rangle \rightarrow |0\rangle$$, $$|+\rangle \rightarrow |+\rangle$$. However, when it comes to the $$|-\rangle$$ state, the output is $$-|-\rangle$$, and I don't know how to account for that when doing the simplification.

Very basic question but any help is appreciated.

Computing the effects of the sequence of gates on the basis states and comparing the results to those of the known gates is indeed a common technique for circuit simplification. If you're careful about not discarding any global phases acquired when the gates are applied, you don't even need to compute the effects on $$|+\rangle$$ and $$|-\rangle$$ states, since the effects on a linear combination of basis states can be computed as a linear combination of effects on basis states, so if the results on $$|0\rangle$$ and $$|1\rangle$$ match, so will the results on any superposition states.
(Not sure why you call out getting $$-|-\rangle$$ state when applying the sequence of gates to $$|-\rangle$$, that's exactly the effect the NOT gate has on the $$|-\rangle$$ state: $$NOT(|0\rangle - |1\rangle) = |1\rangle - |0\rangle = -(|0\rangle - |1\rangle)$$.)