Is $((p\rightarrow q) \land \neg p) \rightarrow \neg q$ a tautology? Is this proposition a tautology?

$((p\rightarrow q)  \land \neg p) \rightarrow \neg q$

Knowing that $\alpha \rightarrow \beta$ is equivalent to $\neg \alpha \lor \beta$, I have come up with 
$(\neg p \lor q) \land \neg p) \rightarrow \neg q$
$(\neg p \lor q) \land ((\neg \neg p \lor \neg q)$
$(\neg p \lor q) \land \neg (\neg p \lor q)$
$(\neg p \lor q) \land (p \lor \neg q)$
If this is correct, which I am not sure it is, then I think that this is not a tautology. Can someone confirm/refute my work here?
 A: Consider this counterexample:


*

*$p:\mathbf{F}$

*$q:\mathbf{T}$


Then
$$
[(p\to q)\land\neg p]\to\neg q\equiv[(\mathbf{F}\to\mathbf{T})\land\mathbf{T}]\to\mathbf{F}\equiv\mathbf{T}\to\mathbf{F}\equiv\mathbf{F}.
$$
How did I come up with this counterexample? Consider the following:
\begin{align}
[(p\to q)\land\neg p]\to\neg q&\equiv \neg[(\neg p\lor q)\land\neg p]\lor\neg q\\[1em]
&\equiv (p\land\neg q)\lor p\lor\neg q\\[1em]
&\equiv [p\land(\neg q\lor p)]\lor\neg q\\[1em]
&\equiv (p\lor\neg q)\land(\neg q\lor p)\\[1em]
&\equiv p\lor\neg q.
\end{align}
A: The proposition concerned: $$((p\rightarrow q) \land \neg p) \rightarrow \neg q$$
Applying that $\alpha \rightarrow \beta$ is equivalent to $\neg \alpha \lor \beta$, the proposition becomes: $$((\neg p \lor q) \land \neg p) \rightarrow \neg q$$
Applying again: $$\neg((\neg p \lor q) \land \neg p) \lor \neg q$$
Particularly, since we know that it is the fallacy of denying the antecedent, we know that it will be false when $p=0$ and $q=1$.

A review of your work:
Step 1: $(\neg p \lor q) \land \neg p) \rightarrow \neg q$
This is a correct usage of the rule.
Step 2: $(\neg p \lor q) \land ((\neg \neg p \lor \neg q)$
However, step 2 dealt with the parentheses incorrectly. This assumed step 1 to be: $$(\neg p \lor q) \land (\neg p \rightarrow \neg q)$$instead of: $$(\neg p \lor q) \land \neg p) \rightarrow \neg q$$which are different.
Step 3: $(\neg p \lor q) \land \neg (\neg p \lor q)$
This step forgot to change the $\lor$ to the $\land$ after applying De Morgan's laws.
Step 4: $(\neg p \lor q) \land (p \lor \neg q)$
The same goes with this step. However, the effect cancelled, making this equivalent with step 2. After all, two wrongs make a right, right?
