Question

Comment on the following statements
$A\Delta B=(A\cup B)-(A\cap B)$

Answer 1

(iv) We have $=(A\cup B)-(A\cap B)=(A\cup B)\cap {(A\cap B)}^{\prime }$ by part (i)
$=(A\cup B)\cap (A^{\prime }\cup B^{\prime })$ [De-Morgan Law]
$=\{(A\cap A^{\prime })\cup (A\cap B^{\prime })\}\cup \{(B\cap A^{\prime })\cup (B\cap B^{\prime })\}$ (By Distributive law)
$=\varphi \cup (A\cap B^{\prime })\cup (B\cap A^{\prime })\cup \varphi$
$=(A\cap B^{\prime })\cup (B\cap A^{\prime })$
$=(A-B)\cup (B-A)=A\Delta B$, (by definition of $\Delta$ operation)