Let $A$ and $B$ be sets. Define the symmetric difference of $A$ and $B$, written $A+B$, by $A+B=(A \cup B) \backslash (A \cap B)$.
Prove the following statement
$A + (B+C) = B+(A+C) = C+ (A+B)$
This is a mixture of associative law and commutative law.
Proof using associative law which states that for all statements $P,Q$ and $R$.
$(P \land Q) \land R \leftrightarrow P \land (Q \land R)$ and $(P \lor Q) \lor R \leftrightarrow P \lor (Q \lor R)$
$A + (B+C) $
$A + (B \cup C) \backslash (B \cap C)$
$$
\begin{array}{c}
A+(B+C)\\
A+[(B \cup C) \setminus (B \cap C)]\\
(A \cup [(B \cup C) \setminus (B \cap C)]) \setminus (A \cap [(B \cup C) \setminus (B \cap C)])\\
\end{array}
$$
$B + (A + C )$
$B + (A \cup C) \backslash (A \cap C)$
$$
\begin{array}{c}
B+(A+C)\\
B+[(A \cup C) \setminus (A \cap C)]\\
(B \cup [(A \cup C) \setminus (A \cap C)]) \setminus (B \cap [(A \cup C) \setminus (A \cap C)])\\
\end{array}
$$
$C + (A+B) $
$C + (A \cup B) \backslash (A \cap B)$
$$
\begin{array}{c}
C+(A+B)\\
C+[(A \cup B) \setminus (A \cap B)]\\
(C \cup [(A \cup B) \setminus (A \cap B)]) \setminus (C \cap [(A \cup B) \setminus (A \cap B)])\\
\end{array}
$$
Proof for commutative law which claims that for all statements $P$ and $Q$
$P \land Q \leftrightarrow Q \land P$ and $P \lor Q \leftrightarrow Q \lor P$
$$
\begin{array}{c}
A+(B+C) = (B+C)+A \\
A + (B \cup C) \backslash (B \cap C) = (B \cup C) \backslash (B \cap C) + A\\
\end{array}
$$
$$
\begin{array}{c}
B+(A+C) = (A+C)+B \\
B + (A \cup C) \backslash (A \cap C) = (A \cup C) \backslash (A \cap C) + B\\
\end{array}
$$
$$
\begin{array}{c}
C+(A+B) = (A+B)+C \\
C + (A \cup B) \backslash (A \cap B) = (A \cup B) \backslash (A \cap B) + C\\
\end{array}
$$
I don't know what to do next…I'm not even sure if this is correct either.
Best Answer
Perhaps the simplest proof is via Venn diagrams.
The usual method of showing $A+(B+C) = B + (A+C)$ is to prove