Using the laws of logic prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology Could someone please tell me if I am correct and if I am not, tell me where I went wrong? 
Using the laws of logic prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology.
First I used the Implication law $(p \rightarrow q) \equiv (\neg p \vee q)$ to show that $$[\neg q \land (p \rightarrow q)] \equiv [\neg q \land (\neg p \vee q)]$$
Then I "factored" (?) the $\neg$ out and had 
$$
\neg [q \vee (p \land \neg q)]
$$
And since $(p \land \neg q)$ denotes to "$p$ but not $q4" then I assumed I could leave $q$ out, leaving me with
$$
¬[q∨(p)]
$$
Which is
$$
\equiv (\neg q \land \neg p)
$$
Which says "not $q$ and not $p$".
Since it is "not $p$", does that mean that
$$
\equiv (\neg q \land \neg p) \rightarrow \neg p
$$
and prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology?
 A: As Henning notes, you really do need to specify what laws are at your disposal. With that in mind, here is one approach (assuming you are able to use what I use):
\begin{align}
[\neg q\land(p\to q)]\to\neg p&\equiv\neg[\neg q\land(\neg p\lor q)]\lor\neg p\tag{material implication}\\[1em]&\equiv [q\lor(p\land\neg q)]\lor\neg p\tag{De Morgan}\\[1em]
&= (q\lor\neg p)\lor(p\land\neg q)\tag{associativity}\\[1em]
&\equiv \neg(p\land\neg q)\lor(p\land\neg q)\tag{De Morgan}\\[1em]
&\equiv \neg M\lor M\tag{let $M\equiv p\land\neg q$}\\[1em]
&\equiv \mathbf{T}\tag{negation}.
\end{align}
A: $(-q \wedge (p \rightarrow q) ) \rightarrow -p$
$= -(-q\wedge(p\rightarrow q)) \vee -p$
$= (--q\vee-(-p\vee q))\vee -p$
$= (--q \vee (--p \wedge -q)) \vee -p$
$=(q \vee (p \wedge -q)) \vee -p$
$=((q \vee p) \wedge (q \vee -q)) \vee -p$
$=((q\vee p)\wedge 1) \vee (-p)$
$=(q \vee p) \vee (-p) = q \vee (p \vee -p) = q \vee 1 = 1$
Hence it is a tautology.
I used the rules $--a = a$ and $a \wedge 1= a$, $a \vee 1 = 1$
A: $[\neg q \land (p \rightarrow q)] \rightarrow \neg p$ 
$\equiv [\neg q \land (\neg p \vee q)] \rightarrow \neg p$ 
$\equiv [q \vee \neg(\neg p \vee q)] \vee \neg p$ 
$\equiv q \vee (p \land \neg q) \vee \neg p$ 
$\equiv [(q \vee p) \land (q \vee \neg q)] \vee \neg p$ 
$\equiv q \vee 1$
$\equiv 1$
