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$$p\rightarrow q$$ $$\neg p$$ $$\overline{\therefore \neg q}$$

We were given a task to prove the validity of this argument (assuming the statement above is the same as: $(p \rightarrow q) \wedge \neg p \Rightarrow \neg q$).

I concluded with the argument to be valid. If $q$ is $0$, then the conclusion is $1$. For $p \rightarrow q$ to be true, then $p$ has to be $0$. Then $\neg p$ is $1$. This makes LHS and RHS to be $1$. Did I understand this correctly, or am I mistaken?

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    I was not aware that you had asked this exact question before. Next time please clarify on the original question. However, I can see that the answer you got on your duplicate question didn't actually address your error in reasoning, so I'm glad I could clear it up. Feel free to ask below my answer (not a new question) if you need to clarify further. – 2017-01-18
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    The answer of the previous questions was however mis-approved. He/she used another statement (switching out $\neg p$ with $q$), which is my reason to re-post the question. Glad I did so, you made it quite clear to me the idea of proving an argument. Thanks a lot! – 2017-01-18
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    The previous answer ended with : "For example, take $q=1,p=0$" and this is enough to prove invalidity. Thus, you haven't read it, (mis-)approved it and re-asked... – 2017-01-18
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    @MauroALLEGRANZA: I agree with you that the previous answer is enough to show invalidity, but I think more should have been said about why the reasoning was wrong, hence I believe my answer complements it. – 2017-01-18

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You're mistaken. The argument is only correct if every possible assignment of truth values that satisfies the premises also satisfies the conclusion. You cannot work backwards from the truth of the conclusion (you haven't proven it!) to determine what should or should not be true.

In your specific case, there are 3 ways that $(p \to q)$ can be true, and in 2 of them $\neg p$ is also true, and 1 of those does not satisfy $\neg q$, so there is 1 counter-example scenario to the argument. Hence it is invalid.