Step 1

Given: \(A \not{\subset} B^{c}\)

We need to prove that \(\displaystyle{A}\cap{B}=\phi\).

Proof by contradiction:

Suppose \(\displaystyle{A}\cap{B}\ne{}\phi\), then \(\displaystyle\exists{x}\in{A}\cap{B}\)

Therefore \(\displaystyle{x}\in{A}\) and \(\displaystyle{x}\in{B}\)

Step 2

Since \(\displaystyle{A}\subset{B}^{{{c}}}\) and \(\displaystyle{x}\in{A}\), therefore

\(\displaystyle{x}\in{A}\subset{B}^{{{c}}}\Rightarrow{x}\in{B}^{{{c}}}\). But \(\displaystyle{x}\in{B}\) as well. Which is not true because no element of a set B can be found in both B and \(\displaystyle{B}^{{{c}}}\), as \(\displaystyle{B}\cap{B}^{{{c}}}=\phi\). Therefore our assumption that \(\displaystyle{A}\cap{B}\ne{}\phi\) must be false.

Hence \(\displaystyle{A}\cap{B}=\phi\).