Demonstration with symmetric difference of sets

Question

Be the symmetric difference of A and B sets, $$ A\triangle B = (A-B) \cup (B-A) $$ Show that for any sets $A$, $B$ and $C$ $$ (A \triangle B) \cup (B \triangle C)=(A \cup B \cup C)-(A \cap B \cap C) $$

I tried to solve it using the definition \begin{align*} (A \triangle B) \cup (B \triangle C) &= (A-B) \cup (B-A) \cup (B-C) \cup (C-B) \\ & =(A \cap B^c) \cup (B \cap A^c) \cup (B \cap C^c) \cup (C \cap B^c) \\ \end{align*} But I do not know how to follow

Answer 1

$(A\triangle B)\cup(B\triangle C)=(A\cap B^c)\cup (B\cap A^c)\cup (B\cap C^c)\cup (C\cap B^c)\\ =[((A\cap B^c)\cup B)\cap ((A\cap B^c)\cup A^c))]\cup [((B\cap C^c)\cup C)\cap ((B\cap C^c)\cup B^c))]\\=[((A\cup B)\cap U)\cap(U\cap(B^c\cup A^c))]\cup [((B\cup C)\cap U)\cap(U\cap(C^c\cup B^c))]\\=[(A\cup B)\cap (A^c\cup B^c)]\cup [(B\cup C)\cap (B^c\cup C^c)]\\=\{[(A\cup B)\cap (A^c\cup B^c)]\cup (B\cup C)\}~~\cap~~\{[(A\cup B)\cap (A^c\cup B^c)]\cup (B^c\cup C^c)\}\\=[(A\cup B\cup C)\cap U]\cap [U\cap (A^c\cup B^c\cup C^c)]\\=[(A\cup B\cup C)\cap (A^c\cup B^c\cup C^c)\\=(A\cup B\cup C)\cap (A\cap B\cap C)^c\\=(A\cup B\cup C)-(A\cap B\cap C)$

where $U$ is universal set.

Answer 2

Hint:

It's easier if you use $$A\mathbin\Delta B=(A\cup B)\smallsetminus(A\cap B),\qquad B\mathbin\Delta C=(B\cup C)\smallsetminus(B\cap C).$$

Answer 3

Use Distribution (with complementation and identity), then of course deMorgan's Rule and he definition of set difference:

\begin{align*} (A \triangle B) \cup (B \triangle C) &= (A\smallsetminus B) \cup (B\smallsetminus A) \cup (B\smallsetminus C) \cup (C\smallsetminus B) \\ & =(A \cap B^c) \cup (B \cap A^c) \cup (B \cap C^c) \cup (C \cap B^c) \\ & =((A\cup B)\cap(A\cup A^c)\cap (B^c\cup B)\cap (B^c\cap A^c)) ~\cup~ (B \cap C^c) \cup (C \cap B^c) \\ & =((A\cup B)\cap (B^c\cap A^c)) ~\cup~ (B \cap C^c) \cup (C \cap B^c) \\ & =((A\cup B)\cap (B^c\cap A^c)) ~\cup~ ((B \cup C) \cap (B^c\cup C^c)) \\ & ~~\vdots \\ & = (A\cup B\cup C)\cap(A^c\cup B^c\cup C^c) \\ & = (A\cup B\cup C)\cap(A\cap B\cap C)^c \\ & = (A\cup B\cup C)\smallsetminus(A\cap B\cap C) \end{align*}