given 2 sets A and B, use membership table to show that (A-B)∪(B-A)=(A∪B) -(A∩B)

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\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} A & B & A-B & B-A & (A-B\cup (B-A) \\ \hline 1 & 1 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & 0 & 1\\ \hline 0 & 1 & 0 & 1 & 1\\ \hline 0 & 0 & 0 & 0 & 0 \end{array}

\def\arraystretch{1.5} \begin{array}{c:c:c} A\cup B & A\cap B & (A\cup B)-(A\cap B) \\ \hline 1 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 0 & 1 \\ \hline 0 & 0 & 0 \end{array}

Hence

(A-B)\cup(B-A)=(A\cup B)-(A\cap B)

Expert's answer

\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} A & B & A-B & B-A & (A-B\cup (B-A) \\ \hline 1 & 1 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & 0 & 1\\ \hline 0 & 1 & 0 & 1 & 1\\ \hline 0 & 0 & 0 & 0 & 0 \end{array}

\def\arraystretch{1.5} \begin{array}{c:c:c} A\cup B & A\cap B & (A\cup B)-(A\cap B) \\ \hline 1 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 0 & 1 \\ \hline 0 & 0 & 0 \end{array}

Hence

(A-B)\cup(B-A)=(A\cup B)-(A\cap B)