# Discrete MathShow that if A \not{\subset} B^{c}, then A \cap B=\phi. Hint

Discrete Math
Show that if $$A \not{\subset} B^{c}$$, then $$\displaystyle{A}\cap{B}=\phi$$. Hint:Use the contrapositive.

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Step 1
Given: $$A \not{\subset} B^{c}$$
We need to prove that $$\displaystyle{A}\cap{B}=\phi$$.
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 $$\displaystyle{A}\cap{B}=\phi$$.