Is the reasoning $\neg p\vee\neg q;\;p\therefore\neg q$ valid? 
Is it valid to say that: $$\begin{array}{l}\neg p\vee\neg q\\p\\\hline\therefore\neg q\end{array}$$ knowing that?: $$\begin{array}{l}p\vee q\\\neg p\\\hline\therefore q\end{array}$$ (the last reasoning is called "Disjunctive syllogism").


I think yes, because:
$$\begin{array}{lll}
(1)&\neg p\vee\neg q&\text{Premise}\\
(2)&p&\text{Premise}\\
(3)&p\to\neg q&\text{Conditional equivalence in (1)}\\
(4)&\neg q&\text{Modus Ponens (2)-(3)}\\
\end{array}$$
Therefore the first reasoning is valid.
Is my deduction correct?
 A: Yes, this is valid. As you mention, it's called the Disjunctive Syllogism (or Modus Tollendo Ponens) and is one of the fundamental building blocks (inference rules) of propositional calculus.
The "standard" proof goes something like this, using conjunction introduction and De Morgan:


*

*Given $\neg p \vee \neg q$

*Given $p$

*Hypothetically suppose $q$:


*

*Conjunction introduction gives $p \wedge q$

*De Morgan gives $\neg (\neg p \vee \neg q)$

*Let $X$ stand for $\neg p \vee \neg q$

*We now have $X \wedge \neg X$: a contradiction


*Therefore $q \rightarrow \bot$

*Therefore, by reductio ad absurdum, $\neg q$
You can also get this without De Morgan if you have to:


*

*Given $\neg p \vee \neg q$

*Given $p$

*Hypothetically suppose $\neg p$:


*

*Now we have $p \wedge \neg p$, which is a contradiction


*So $\neg p \rightarrow \bot$, and $\bot \rightarrow \neg q$ (principle of explosion)

*Therefore $\neg p \rightarrow \neg q$

*$\neg q \rightarrow \neg q$, by conditional introduction

*So $\neg q$ by disjunction elimination

