How to solve Distributivity of $\lor$ over $\land$ The problem I need to prove is $p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)$
I am trying to get the RHS equivalent to the LHS
So I change 
$(p \lor q) \land (p \lor r)$
(using the Golden Rule)
(Golden Rule is- $p \land q \equiv p \equiv q \equiv p \lor q$) 
$p \lor q  \equiv p \lor r \equiv (p \lor q) \lor (p \lor r)$
?
?
and I know it ends like
$p \lor (q \equiv r \equiv (q \lor r))$
(the I reintroduce Golden Rule again)
$p \lor (q \land r)$
I am missing a few steps in the middle? but I'm not quite sure what
 A: I'm using George Tourlakis, Mathematical Logic (2008); see page 42-43 for the rules and page 74 : 2.4.23 Theorem. (Distributivity: $\lor$ over $\land$ and $\land$ over $\lor$)

$(p \lor q ) \land (p \lor r)$

$p \lor q \lor p \lor r \equiv p \lor q \equiv p \lor r$
apply the Golden Rule (using Equanimity and Leibniz Merged, taht is a "derived rule"; see Tourlakis, page 57, 2.1.16 Theorem. (Eqn + Leib Merged) :  $C[p := A], A \equiv B \vdash  C[p := B]$; we call it "E+L")

NOTE. Due to the use of $p$ and $q$ in the formulas, I will describe the substitution in the “C-part” as “... sub-formula”

$p \lor p \lor q \lor r \equiv p \lor q \equiv p \lor r$
apply E+L, where the ‘C-part’ is $... \equiv p \lor q \equiv p \lor r$, using the “derived axiom” : $(((p \lor q) \lor p) \lor r) \equiv (((p \lor p) \lor q) \lor r)$ (proved with Associativity and Symmetry of $\lor$)
$p \lor q \lor r \equiv p \lor q \equiv p \lor r$
apply E+L where the ‘C-part’ is $...\lor q \lor r \equiv p \lor q \equiv p \lor r$, using the axiom Idempotency of $\lor$, i.e. $p \lor p \equiv p$
$p \lor q \lor r \equiv p \lor (q \equiv r)$
apply E+L where the ‘C-part’ is $p \lor q \lor r \equiv …$, using the axiom Distributivity of $\lor$ over $\equiv$, i.e. $p \lor (q \equiv r) \equiv p \lor q \equiv p \lor r$;
Now, read $p \lor q \lor r \equiv p \lor (q \equiv r)$ as $p \lor (q \lor r) \equiv p \lor (q \equiv r)$ and apply Distributivity of $\lor$ over $\equiv$, i.e. $p \lor (A \equiv B) \equiv p \lor A \equiv p \lor B$, obtaining :
$p \lor (q \lor r \equiv q \equiv r)$
Finally, apply E+L where the ‘C-part’ is $p \lor …$, using the Golden Rule, i.e. $(q \lor r \equiv q \equiv r) \equiv q \land r$:

$p \lor (q \land r)$

