With :


  $(¬p∧¬r)∨(¬p∧q)∨(q∧¬r)∨(q∧q)$


you first apply $(q \land q) \equiv q$ to simplify it :


  $(¬p∧¬r)∨(¬p∧q)∨(q∧¬r)∨q$.


Consider now : $(q∧¬r)∨q$; applying Absorption Laws : $p∨(p∧q) \equiv p$, we have that it is equivalent to $q$.

Thus :


  $[(¬p∧¬r)∨(¬p∧q)]∨[(q∧¬r)∨q] \equiv (¬p∧¬r)∨(¬p∧q)∨q$.


Apply it again to : $(¬p∧q)∨q$, which is again equivalent to $q$, to get :


  
    $(¬p∧¬r)∨[(¬p∧q)∨q] \equiv (¬p∧¬r)∨q$.
  




Regarding absorption :


  
    $p∨(p∧q) \equiv p$
  


you can prove it with Natural Deduction or verify it via truth table :


if $p$ is TRUE, then $p \lor (p \land q)$ is TRUE, by truth table for $\lor$;
if $p$ is FALSE, then $(p \land q)$ is FALSE, by truth table for $\land$; so both $p$ and $(p \land q)$ are FALSE, and thus $p∨(p∧q)$ is FALSE, by truth table for $\lor$.


Don't really understand the absorption law
Let's say that I have a sampled signal x[n], it is being, in this exact order, up sampled by 2, down sampled by4, up sampled by 4 and down sampled by 2 to produce y[n].

It seems to me that it should be pretty self evident that since we up sampled the signal by 2 and down sampled it by 2, then up sampled it by 4 and down sampled by 4, I should just get the original x[n] back. 

Am I right?

So the real question is, can the various up/down sampling pieces be readily swapped?
