Using logical equivalences to show $(A \Delta B) \cap C = (A \cap C) \Delta (B \cap C)$ I'm trying to show $(A \Delta B) \cap C = (A \cap C) \Delta (B \cap C)$ using logical equivalences.
(NOTE: "$\Delta$" denotes exclusive or, i.e. $A \Delta B = (A \cup B) \setminus (A \cap B) = (A \setminus B) \cup (B \setminus A)$.)
My workings so far:
\begin{align*}
x ∈ (A △ B) ∩ C 
  & \iff x ∈ (A △ B) ∧ x ∈ C  &&\text{(definition of ∩)} 
\\& \iff [(x ∈ A \lor x ∈ B)  ∧ ¬(x ∈ A  ∧ x ∈ B)] ∧ x ∈ C  &&\text{(definition of △)}
\\ \text{Let $P = x ∈ A$, $Q = x ∈ B$, $R = x ∈ C$} 
\\& \iff [(P \lor Q)  ∧ ¬(P  ∧ Q)] ∧ R
\\& \iff [(P \lor Q) ∧ R] ∧ ¬(P ∧ Q)  &&\text{(associativity + commutativity)}
\end{align*}
This is close to the result, I just need to turn $P \land Q$ into $P \land R \land Q \land R$, but I'm not sure how to do this. I tried using the tautology law $P ∧ ⊤ ⟺ P$ to "create" an $R$ term. However, this also introduces an $¬R$ term, which I cannot get rid of:
\begin{align*} 
[(P \lor Q) ∧ R] ∧ ¬(P ∧ Q) 
& \iff [(P \lor Q) ∧ R] ∧ ¬[(P ∧ Q) ∧ (R \lor ¬R)]  &&\text{(P ∧ ⊤ ⟺ P)}
\\& \iff [(P \lor Q) ∧ R] ∧ ¬[(P ∧ (R \lor ¬R)) ∧ (Q ∧ (R \lor ¬R))]  &&\text{(distributivity)}
\end{align*}
Could someone please give me a hint as to what I could have used in place of $P ∧ ⊤ ⟺ P$?
I also tried $(P ∧ Q) ∧ (R \lor T)$, but this collapses back into $(P ∧ Q)$ via the absorption law.

Suppose that x is in (A XOR B) $\cap$ C. Two cases, WLOG assume that x is in A, not in B and in C.  But then x is certainly in A $\cap$ C, and since it is not in B, it cannot be in B $\cap$ C. But this suffices.
For the other direction, again assume WLOG that x is in $A \cap C$ and not in $B \cap C$. It follows that x is in A, x is in C. Suppose that x were in B. Then it would be in B $\cap C$ since it is also in C. Hence, x is not in B. But this suffices.
