How can logical equivalence be derived from this.. (p ∨∼q) ∧ (∼p ∨∼q) 
≡ (∼q ∨ p) ∧ (∼q ∨∼p)                
by (a) 
≡∼q ∨ (p ∧∼p)                  
by (b)
≡∼q ∨ c             
by (c) 
≡∼q                 
by (d) 
Therefore, (p ∨∼q) ∧ (∼p ∨∼q) ≡∼q


*

*c being a contradiction


What are the laws being used in each of the four steps?
Commutative, associative, distributive, identity, negation, double negative, idempotent, universal bound, de morgans, absorption or the negations of t and c.
Please help I have no idea were to start.
 A: For $(a)$ you have first used the fact that $\lor $ is commutative. In both brackets the expressions have first been changed to $( \lnot q \lor p)$ and $(\lnot q \lor  \lnot p)$. 
Then for $(b)$ the author has used the distribution law of the conjunction over the disjunction. That is, $ ( \lnot q \lor p) \land (\lnot q \lor  \lnot p) \equiv ( \lnot q \lor (p \land \lnot p  )) $
For $(c)$ the derivation should be obvious since $ (p \land \lnot p  ) $ is a contradiction. All he has done is a substitution. 
For $(d)$ I'm not sure how the book defines it. Sometimes this is used as one of the idempotent laws. But either way, you can write a small natural deduction proof to say that $ (p \lor c) \equiv p $. For the implication assume $ \lnot p$ and then arrive at a contradiction. For the reverse implication just use the introduction of a disjunction.  
A: (a) commutative
(b) distributive
(c) Law of (non-)contradiction (negated on both sides)
(d) identity
although some of these can be named differently depending on the author.
