Equivalance proof of ((p → q) ∧ ¬q) → ¬p and ((p ∨ q) ∧ ¬p) → q I am trying to prove if the above propositional formulas is a tautology using the logic laws. However, I am really stuck, I don't know where else to go. 
Here is what I've done for the former formula.
((p → q) ∧ ¬q) → ¬p 
((¬p ∨ q) ∧ ¬q) → ¬p (implication law)
((q ∨ ¬p) ∧ ¬q) → ¬p (commutative law)

Here's what I've done for the latter formula:
((p ∨ q) ∧ ¬p) → q
((q ∧ p) ∧ ¬p) → q (commutative law)
(q ∧ (p ∧ q)) → q (associative law)
((q ∧ p) v (q ∧ q)) → q (distributive law)
((p v q) v (q ∧ q)) → q (commutative law)
((p v q) v q) → q  (idempotent law)
((p v (q v q)) → q (associative law)
I am completely lost and I don't know what to do after the last steps. 
Thanks 
 A: The two propositional formulas are equivalent because each one is a tautology. I'll do the first one (I've taken commutativity and associativity as given to keep the proof short):
\begin{align*}
((p \to q) \land \neg q) \to \neg p
&\equiv \neg ((\neg p \lor q) \land \neg q) \lor \neg p & \textsf{Implication Law} \\
&\equiv \neg (\neg p \lor q) \lor \neg\neg q \lor \neg p & \textsf{DeMorgan's Law} \\
&\equiv (\neg \neg p \land \neg  q) \lor \neg\neg q \lor \neg p & \textsf{DeMorgan's Law} \\
&\equiv (p \land \neg  q) \lor q \lor \neg p & \textsf{Double Negation Law} \\
&\equiv (p \land \neg  q) \lor (\top \land q) \lor \neg p & \textsf{Identity Law} \\
&\equiv (p \land \neg  q) \lor ((p \lor \neg p) \land q) \lor \neg p & \textsf{Complement Law} \\
&\equiv (p \land \neg  q) \lor (p \land q) \lor (\neg p \land q) \lor \neg p & \textsf{Distributive Law} \\
&\equiv (p \land (\neg  q \lor q)) \lor (\neg p \land q) \lor \neg p & \textsf{Distributive Law} \\
&\equiv (p \land \top) \lor (\neg p \land q) \lor \neg p & \textsf{Complement Law} \\
&\equiv p \lor (\neg p \land q) \lor \neg p & \textsf{Identity Law} \\
&\equiv (p \lor\neg p) \lor (\neg p \land q) & \textsf{Comm/Assoc Law} \\
&\equiv \top \lor (\neg p \land q) & \textsf{Complement Law} \\
&\equiv \top & \textsf{Domination Law} \\
\end{align*}
The proof for the other formula is nearly identical.
A: HINT
One really useful thing to do in these kinds of cases is to rewrite any implications as disjunctions, using:
$$P \rightarrow Q \Leftrightarrow \neg P \lor Q$$
This is useful because we have far more equivalence principles involving the basic Boolean operators $\land$, $\lor$, and $\neg$ than we have dealing with implications. In fact, depending on what equivalence principles you have, sometimes you simply cannot show the equivalence while only relying on equivalence principles involving the $\rightarrow$. Indeed, notice how you keep having $\rightarrow q$ at the end of all of your statements!
