Proving Logical Equivalence with Equality I found a logical equivalence in a book, 
(a ∧ b) = (a ∧ c)  ≡  ¬a ∨ (b = c)

but without a proof so I'm trying to prove it without a truth table.
I'm not sure if my thought process is correct. See below
My thought process:
(a ∧ b) = (c ∧ a) (Commutative)

a ∧ (b = c) ∧ a   (Associative ??)

a ∧ a ∧ (b = c)   (Commutative)

a ∧ (b = c)       (AND simplification)

¬a ∨ ¬(b = c)     (De Morgan's)

¬a ∨ (b = c)      (Double Negation of (b = c))

I'm new to predicate logic so I'm not sure if this proof is sound. In particular, is the second step justified? If my proof is not correct, please offer me suggestions on how to make it more sound. 
 A: Your proof attempt is incorrect both for the reason you state, and because you're missing a negation on the penultimate line which I assume is a typo but correcting it requires more than just adding in a $\neg$.
Associative laws typically only apply to the same connective. You can't just arbitrarily re-parenthesize things. I assume what you are writing as = is what is more commonly referred to as $\leftrightarrow$ in this context. You either have some rules for explicitly manipulating = or you have a definition of it. Probably the latter, e.g. $P\leftrightarrow Q :\equiv (P\to Q)\land(Q\to P)$ or $P\leftrightarrow Q:\equiv (P\land Q)\lor(\neg P\land \neg Q)$.
You can substitute the latter definition into your formulas and probably handle it from there.
A: Your proof is not correct.
The second line is not generally true. 
Also note that you applied negation using De Morgan's Law on line 5, which is basically proving that $(a \land b) = (a \land c) \equiv \neg (\neg a \lor (b=c))$.
To prove this equivalence, note that equal sign can be interpreted as the following forms:
$$ 
\begin{align}
x=y 
&\equiv (x\land y) \lor (\neg x \land \neg y) && (1)\\
&\equiv (\neg x \lor y) \land (x \lor \neg y) && (2)
\end{align}
$$
Pick one of the forms you like to rewrite the original statement purely in conjunctions ($ \land $), disjunctions ($ \lor $), and negations ($ \neg $). Then start your derivation from there.
Below is my proof. (I picked the second form, you can try the first one yourself)
$$
\begin{align*}
& \quad (a \land b) = (a \land c)\\
& \equiv (\neg (a \land b)\lor (a \land c)) \land ((a \land b)\lor \neg (a \land c)) 
&& \text{from (2)} \\
&\equiv (\neg a \lor \neg b \lor (a \land c)) \land ((a \land b) \lor \neg a \lor \neg c)
&& \text{(De Morgan's Law)}\\
& \equiv \neg a \lor [(\neg b \lor (a \land c)) \land (\neg c \lor (a \land b))]
&& \text{(Reverse Distributive)}\\
&\equiv \neg a \lor [(\neg b\lor a)\land (\neg b \lor c) \land (\neg c \lor a) \land (\neg c \lor b)]
&& \text{(Distributive Law)}\\
&\equiv \neg a \lor [(a \lor (\neg b \land \neg c)) \land (\neg b \lor c) \land (b \lor \neg c)]
&& \text{(Reverse Distributive)}\\
&\equiv [\neg a \lor a \lor (\neg b \land \neg c)] \land [\neg a \lor ((\neg b \lor c) \land (b \lor \neg c))]
&& \text{(Distributive Law)}\\
&\equiv [\mathbf{t} \lor (\neg b \land \neg c)] \land[\neg a \lor (b = c)]
&& \text{(Negation Law)}\\
&\equiv \mathbf{t} \land [\neg a \lor (b = c)]
&& \text{(Universal Bound)} \\
&\equiv \neg a \lor (b = c)
&& \text{(Identity Law)}
\end{align*}
$$
