Question

$F=FE\cup F{E}^c$and$E\cup F=E\cup E^{c}F$

Short answer

Use associative and distributive laws.

$F=FE\cup F{E}^c\text{and}E\cup F=E\cup E^{c}F$

Step-by-step solution

Given Information.

$F=FE\cup F{E}^c\text{and}E\cup F=E\cup E^{c}F$

Explanation.

$\begin{array}{rrr}F=FE\cup F{E}^c& \text{and}E\cup F=E\cup E^{c}F& \\ F& =F\cap S& A\subseteq B\Rightarrow A\cap B=A\text{(from def. of intersection)}\\ & =F\cap \left(E\cup E^c\right)& S=E\cup E^c\text{(from def. of complement)}\\ & =FE\cup F{E}^c& \text{distributive law}\end{array}$

This proves:

$F=FE\cup F{E}^c$

$E\cup F=E\cup \left(FE\cup F{E}^c\right)$by previously proven identity

$=(E\cup FE)\cup F{E}^c$ associative law

role="math" localid="1649044670063" $=E\cup F{E}^{c}A\subseteq B\Rightarrow A\cup B=B$(from def. of the union)

This proves:

$E\cup F=E\cup F{E}^c$